Elementary Topology
Short Description
Elementary Topology...
Description
Elementary Topology A First Course
Textbook in Problems O. Y. Viro, O. A. Ivanov, N. Y. Netsvetaev, V. M. Kharlamov
This book in ludes basi material on general topology, introdu es algebrai topology via the fundamental group and overing spa es, and provides a ba kground on topologi al and smooth manifolds. It is written mainly for students with a limited experien e in mathemati s, but determined to study the subje t a tively. The material is presented in a on ise form, proofs are omitted. Theorems, however, are formulated in detail, and the reader is expe ted to treat them as problems. Abstra t.
Foreword Genre, Contents and Style of the Book The ore of the book is the material usually in luded in the Topology part of the two year Geometry le ture ourse at the Mathemati al Department of St. Petersburg University. It was omposed by Vladimir Abramovi h Rokhlin in the sixties and has almost not hanged sin e then. We believe this is the minimum topology that must be mastered by any student who has de ided to be ome a mathemati ian. Students with resear h interests in topology and related elds will surely need to go beyond this book, but it may serve as a starting point. The book in ludes basi material on general topology, introdu es algebrai topology via its most lassi al and elementary part, the theory of the fundamental group and overing spa es, and provides a ba kground on topologi al and smooth manifolds. It is written mainly for students with a limited experien e in mathemati s, but who are determined to study the subje t a tively. The ore material is presented in a on ise form; proofs are omitted. Theorems, however, are formulated in detail. We present them as problems and expe t the reader to treat them as problems. Most of the theorems are easy to nd elsewhere with omplete proofs. We believe that a serious attempt to prove a theorem must be the rst rea tion to its formulation. It should pre ede looking for a book where the theorem is proved. On the other hand, we want to emphasize the role of formulations. In the early stages of studying mathemati s it is espe ially important to take ea h formulation seriously. We intentionally for e a reader to think about ea h simple statement. We hope that this will make the book in onvenient for mere skimming. The ore material is enhan ed by many problems of various sorts and additional pie es of theory. Although they are losely related to the main material, they an be (and usually are) kept outside of the standard le ture ourse. These enhan ements an be re ognized by wider margins, as the next paragraph. iii
FOREWORD
iv
The problems, whi h do not omprise separate topi s and are intended ex lusively to be exer ises, are typeset with small fa e. Some of them are very easy and in luded just to provide additional examples. Few problems are diÆ ult. They are to indi ate relations with other parts of mathemati s, show possible dire tions of development of the subje t, or just satisfy an ambitious reader. Problems, whose solutions seem to be the most diÆ ult (from the authors' viewpoint), are marked with a star, as in many other books.
Further, we want to deliver additional pie es of theory (with respe t to the ore material) to more motivated and advan ed students. Maybe, a mathemati ian, who does not work in the elds geometri in avor, an a ord the luxury not to know some of these things. Maybe, students studying topology an postpone this material to their graduate study. We would like to in lude this in graduate le ture ourses. However, quite often it does not happen, be ause most of the topi s of this sort are rather isolated from the ontents of traditional graduate ourses. They are important, but more related to the material of the very rst topology ourse. In the book these topi s are intertwined with the ore material and exer ises, but are distinguishable: they are typeset, like these lines, with large fa e, large margins, theorems and problems in them are numerated in a spe ial manner des ribed below. Exer ises and illustrative problems to the additional topi s are typeset with even wider margins and marked in a di erent way.
Thus, the whole book ontains four layers: the ore material, exer ises and illustrative problems to the ore material, additional topi s, exer ises and illustrative problems to additional topi s. The text of the ore material is typeset with large fa e and smallest margins. The text of problems elaborating on the ore material is typeset with small fa e and larger margins.
The text of additional topi s is typeset is typeset with large fa e and slightly smaller margins as the problems elaborating on the ore material. The text of problems illustrating additional topi s is typeset with small fa e and the largest margins.
Therefore the book looks like a Russian folklore doll, matreshka omposed of several dolls sitting inside ea h other. We apologize for being non onventional in this and hope that it may help some readers and does not irritate the others too mu h.
FOREWORD
v
The whole text of the book is divided into se tions. Ea h se tion is divided into subse tions. Subse tions are not numerated. Ea h of them is devoted to a single topi and onsists of de nitions, ommentaries, theorems, exer ises, problems, and riddles. By a riddle we mean a problem of a spe ial sort: its solution is not
ontained in the formulation. One has to guess a solution, rather than dedu e it. 0.A. Theorems, exer ises, problems and riddles belonging to the ore material are marked with pairs onsisting of the number of se tion and a letter separated with a dot. The letter identi es the item inside the se tion. 0.1. Exer ises, problems, and riddles, whi h are not in luded in the ore, but are losely related to it (and typeset with small fa e) are marked with pairs
onsisting of the number of the se tion and the number of the item inside the se tion. The numbers in the pair are separated also by a dot.
Theorems, exer ises, problems and riddles related to additional topi s are enumerated independently inside ea h se tion and denoted similarly. 0:A. The only di eren e is that the omponents of pairs marking the
items are separated by a olon (rather than dot).
We assume that the reader is familiar with naive set theory, but anti ipate that this familiarity may be super ial. Therefore at points where set theory is espe ially ru ial we make set-theoreti digressions maintained in the same style as the rest of the book.
Advi e to the Reader Sin e the book ontains a summary of elementary topology, you may use the book while preparing for an examination (espe ially, if the exam redu es to solving a olle tion of problems). However, if you attend le tures on the subje t, it would be mu h wiser to read the book prior to the le tures and prove theorems before the le turer gives the proofs. We think that a reader who is able to prove statements of the ore of the book, does not need to solve all the other problems. It would be reasonable instead to look through formulations and on entrate on the most diÆ ult problems. The more diÆ ult the theorems of the main text seem to you, the more arefully you should onsider illustrative problems, and the less time you should waste with problems marked with stars. Keep in mind that sometimes a problem whi h seems to be diÆ ult is followed by easier problems, whi h may suggest hints or serve as te hni al lemmas. A hain of problems of this sort is often on luded with a
FOREWORD
vi
problem whi h suggests a return to the theorem, on e you are armed with the lemmas. Most of our illustrative problems are easy to invent, and, moreover, if you study the subje t seriously, it is always worthwhile to invent problems of this sort. To develop this style of studying mathemati s while solving our problems one should attempt to invent one's own problems and solve them (it does not matter if they are similar to ours or not). Of ourse, some problems presented in this book are not easy to invent.
Contents Foreword
iii iii v
Part 1. General Topology Chapter 1. Generalities 1. Topology in a Set
1
Genre, Contents and Style of the Book Advi e to the Reader
De nition of Topologi al Spa e Simplest Examples The Most Important Example: Real Line Using New Words: Points, Open and Closed Sets Set-Theoreti Digression. De Morgan Formulas Being Open or Closed Cantor Set Chara terization of Topology in Terms of Closed Sets Topology and Arithmeti Progressions Neighborhoods
2. Bases
De nition of Base Bases for Plane When a Colle tion of Sets is a Base Subbases In nity of the Set of Prime Numbers Hierar hy of Topologies
3. Metri Spa es
De nition and First Examples Further Examples Balls and Spheres Subspa es of a Metri Spa e Surprising Balls Segments (What Is Between) Bounded Sets and Balls Norms and Normed Spa es Metri Topology Metrizable Topologi al Spa es Equivalent Metri s Ultrametri vii
3 3 3 3 4 4 5 5 6 6 7 7 7 7 8 8 8 9 9 9 9 10 11 11 11 12 12 12 13 13 13 14
CONTENTS
Operations with Metri s Distan e Between Point and Set Distan e Between Sets
4. Subspa es
Relativity of Openness Agreement on Notations of Topologi al Spa es
5. Position of a Point with Respe t to a Set
Interior, Exterior and Boundary Points Interior and Exterior Closure Frontier Closure and Interior with Respe t to a Finer Topology Properties of Interior and Closure Chara terization of Topology by Closure or Interior Operations Dense Sets Nowhere Dense Sets Limit Points and Isolated Points Lo ally Closed Sets
6. Set-Theoreti Digression. Maps Maps and the Main Classes of Maps Image and Preimage Identity and In lusion Composition Inverse and Invertible Submappings
7. Continuous Maps
De nition and Main Properties of Continuous Maps Reformulations of De nition More Examples Behavior of Dense Sets Lo al Continuity Properties of Continuous Fun tions Spe ial About Metri Case Fun tions on Cantor Set and Square-Filling Curves Sets De ned by Systems of Equations and Inequalities Set-Theoreti Digression. Covers Fundamental Covers
8. Homeomorphisms
De nition and Main Properties of Homeomorphisms Homeomorphi Spa es Role of Homeomorphisms More Examples of Homeomorphisms Examples of Homeomorphi Spa es Examples of Nonhomeomorphi Spa es Homeomorphism Problem and Topologi al Properties
viii
14 15 15 16 16 17 17 17 18 18 19 19 19 20 21 21 22 22 22 22 23 24 24 25 25 25 25 26 26 27 27 28 28 29 30 31 31 32 32 32 32 33 34 37 37
CONTENTS
Information (Without Proof) Embeddings Information
Chapter 2. Topologi al Properties 9. Conne tedness
De nitions of Conne tedness and First Examples Conne ted Sets Properties of Conne ted Sets Conne ted Components Totally Dis onne ted Spa es Frontier and Conne tedness Behavior Under Continuous Maps Conne tedness on Line Intermediate Value Theorem and Its Genralizations Dividing Pan akes Indu tion on Conne tedness Appli ations to Homeomorphism Problem
10. Path-Conne tedness
Paths Path-Conne ted Spa es Path-Conne ted Sets Path-Conne ted Components Path-Conne tedness Versus Conne tedness Polygon-Conne tedness
11. Separation Axioms
Hausdor Axiom Limits of Sequen e Coin iden e Set and Fixed Point Set Hereditary Properties The First Separation Axiom The Third Separation Axiom The Fourth Separation Axiom Niemytski's Spa e Urysohn Lemma and Tietze Theorem
12. Countability Axioms
Set-Theoreti Digression. Countability Se ond Countability and Separability Embedding and Metrization Theorems Bases at a Point First Countability Sequential Approa h to Topology Sequential Continuity
13. Compa tness
De nition of Compa tness Terminology Remarks
ix
38 38 39 40 40 40 40 41 41 42 42 42 43 44 44 44 45 46 46 46 47 47 48 49 49 50 50 50 51 51 52 52 53 53 54 54 55 56 56 56 57 57 58 58 58
CONTENTS
Compa tness in Terms of Closed Sets Compa t Sets Compa t Sets Versus Closed Sets Compa tness and Separation Axioms Compa tness in Eu lidean Spa e Compa tness and Maps Norms in R n Closed Maps
14. Lo al Compa tness and Para ompa tness Lo al Compa tness One-Point Compa ti ation Proper Maps Lo ally Finite Colle tions of Subsets Para ompa t Spa es Para ompa tness and Separation Axioms Partitions of Unity Appli ation: Making Embeddings from Pie es
15. Sequential Compa tness
Sequential Compa tness Versus Compa tness In Metri Spa e Completeness and Compa tness Non-Compa t Balls in In nite Dimension p-Adi Numbers Indu tion on Compa tness Spa es of Convex Figures
Problems for Tests Chapter 3. Topologi al Constru tions 16. Multipli ation
Set-Theoreti Digression. Produ t of Sets Produ t of Topologies Topologi al Properties of Proje tions and Fibers Cartesian Produ ts of Maps Properties of Diagonal and Graph Topologi al Properties of Produ ts Representation of Spe ial Spa es as Produ ts
17. Quotient Spa es
Set-Theoreti Digression. Partitions and Equivalen e Relations Quotient Topology Topologi al Properties of Quotient Spa es Set-Theoreti Digression. Quotients and Maps Continuity of Quotient Maps Closed Partitions Open Partitions
18. Zoo of Quotient Spa es
Tool for Identifying a Quotient Spa e with a Known Spa e
x
59 59 59 60 60 61 62 62 62 62 63 64 64 65 65 65 66 66 66 66 67 67 68 68 69 69 72 72 72 73 73 74 74 75 75 76 76 77 78 78 79 79 79 80 80
CONTENTS
xi
Tools for Des ribing Partitions Entran e to the Zoo Transitivity of Fa torization Mobius Strip Contra ting Subsets Further Examples Klein Bottle Proje tive Plane You May Have Been Provoked to Perform an Illegal Operation Set-Theoreti Digression. Sums of Sets Sums of Spa es Atta hing Spa e Basi Surfa es
80 81 83 83 83 84 84 85 85 85 85 86 87 19. Proje tive Spa es 88 Real Proje tive Spa e of Dimension n 88 Complex Proje tive Spa e of Dimension n 89 Quaternion Proje tive Spa es and Cayley Plane 89 20. Topologi al Groups 89 Algebrai Digression. Groups 89 Topologi al Groups 90 Self-Homeomorphisms Making a Topologi al Group Homogeneous 91 Neighborhoods 92 Separaion Axioms 92 Countability Axioms 93 Subgroups 93 Normal Subgroups 94 Homomorphisms 95 Lo al Isomorphisms 95 Dire t Produ ts 96 21. A tions of Topologi al Groups 97 A tions of Group in Set 97 Continuous A tions 97 Orbit Spa es 97 Homogeneous Spa es 98 22. Spa es of Continuous Maps 98 Sets of Continuous Mappings 98 Topologi al Stru tures on Set of Continuous Mappings 98 Topologi al Properties of Spa es of Continuous Mappings 99 Metri Case 99 Intera tions With Other Constru tions 100 Mappings X Y ! Z and X ! C (Y; Z ) 101
Part 2. Algebrai Topology Chapter 4. Fundamental Group and Covering Spa es 23. Homotopy
102 103 104
CONTENTS
Continuous Deformations of Maps Homotopy as Map and Family of Maps Homotopy as Relation Straight-Line Homotopy Two Natural Properties of Homotopies Stationary Homotopy Homotopies and Paths Homotopy of Paths
24. Homotopy Properties of Path Multipli ation Multipli ation of Homotopy Classes of Paths Asso iativity Unit Inverse
25. Fundamental Group
De nition of Fundamental Group Why Index 1? High Homotopy Groups Cir ular loops The Very First Cal ulations Fundamental Group of Produ t Simply-Conne tedness Fundamental Group of a Topologi al Group
26. The Role of Base Point
Overview of the Role of Base Point De nition of Translation Maps Properties of Ts Role of Path High Homotopy Groups In Topologi al Group
27. Covering Spa es
De nition Lo al Homeomorphisms Versus Coverings Number of Sheets More Examples Universal Coverings Theorems on Path Lifting High-Dimensional Homotopy Groups of Covering Spa e
xii
104 104 105 105 106 106 107 107 108 108 108 109 109 110 110 110 111 111 112 113 113 114 114 114 115 115 115 116 116 117 117 117 118 118 119 119 121
28. Cal ulations of Fundamental Groups Using Universal Coverings 121
Fundamental Group of Cir le Fundamental Group of Proje tive Spa e Fundamental Groups of Bouquet of Cir les Algebrai Digression. Free Groups Universal Covering for Bouquet of Cir les
29. Fundamental Group and Continuous Maps
121 122 122 123 124 125
CONTENTS
Indu ed Homomorphisms Fundamental Theorem of High Algebra Generalization of Intermediate Value Theorem Winding Number Borsuk-Ulam Theorem
30.
Covering Spa es via Fundamental Groups
Homomorphisms Indu ed by Covering Proje tions Number of Sheets Hierar hy of Coverings Automorphisms of Covering Regular Coverings Existen e of Coverings Lifting Maps
Chapter 5. More Appli ations and Cal ulations 31. Retra tions and Fixed Points Retra tions and Retra ts Fundamental Group and Retra tions Fixed-Point Property.
32. Homotopy Equivalen es
Homotopy Equivalen e as Map Homotopy Equivalen e as Relation Deformation Retra tion Examples Deformation Retra tion Versus Homotopy Equivalen e Contra tible Spa es Fundamental Group and Homotopy Equivalen es
33. Cellular Spa es
De nition of Cellular Spa es First Examples More Two-Dimensional Examples Topologi al Properties of Cellular Spa es Embedding to Eu lidean Spa e One-Dimensional Cellular Spa es Euler Chara teristi
34. Fundamental Group of a Cellular Spa e
One-Dimensional Cellular Spa es Generators Relators Writing Down Generators and Relators Fundamental Groups of Basi Surfa es Seifert - van Kampen Theorem
35. One-Dimensional Homology and Cohomology Des ription of H1 (X ) in Terms of Free Cir ular Loops One-Dimensional Cohomology Cohomology and Classi ation of Regular Coverings
xiii
125 127 127 128 128 129 129 130 130 131 131 131 131 132 132 132 133 133 134 134 135 135 135 136 136 137 138 138 140 141 142 143 143 144 145 145 145 145 146 147 148 148 149 150 151
CONTENTS
Integer Cohomology and Maps to S 1 One-Dimensional Homology Modulo 2
Part 3. Manifolds Chapter 6. Bare Manifolds 36. Lo ally Eu lidean Spa es
De nition of Lo ally Eu lidean Spa e Dimension Interior and Boundary
37. Manifolds
De nition of Manifold Components of Manifold Making New Manifolds out of Old Ones Double Collars and Bites
38. Isotopy
Isotopy of Homeomorphisms Isotopy of Embeddings and Sets Isotopies and Atta hing Conne ted Sums
39. One-Dimensional Manifolds
Zero-Dimensional Manifolds Redu tion to Conne ted Manifolds Examples Statements of Main Theorems Lemma on 1-Manifold Covered with Two Lines Without Boundary With Boundary Consequen es of Classi ation Mapping Class Groups
40. Two-Dimensional Manifolds
Examples Ends and Odds Closed Surfa es Triangulations of Surfa es Two Properties of Triangulations of Surfa es S heme of Triangulation Examples Families of Polygons Operations on Family of Polygons Topologi al and Homotopy Classi ation of Closed Surfa es Re ognizing Closed Surfa es Orientations More About Re ognizing Closed Surfa es Compa t Surfa es with Boundary
xiv
151 152 154 156 156 156 157 157 159 159 160 160 161 161 162 162 162 164 164 164 164 165 165 165 166 166 167 167 167 167 167 168 169 170 170 171 172 172 173 174 175 175 176 176
CONTENTS
Simply Conne ted Surfa es 41. One-Dimensional mod2-Homology of Surfa es Polygonal Paths on Surfa e Subdivisions of Triangulation Bringing Loops to General Position Cutting Surfa e Along Curve Curves on Surfa es and Two-Fold Coverings One-Dimensional Z2-Cohomology of Surfa e One-Dimensional Z2-Homology of Surfa e Poin are Duality One-Sided and Two-Sided Simple Closed Curves on Surfa es Orientation Covering and First Stiefel-Whitney Class Relative Homology
42. Surfa es Beyond Classi ation
Genus of Surfa e Systems of disjoint urves on a surfa e Polygonal Jordan and S hon ies Theorems Polygonal Annulus Theorem Dehn Twists Coverings of Surfa es Bran hed Coverings Mapping Class Group of Torus Braid Groups
43. Three-Dimensional Manifolds Poin are Conje ture Lens Spa es Seifert Manifolds Fibrations over Cir le Heegaard Splitting and Diagrams
Chapter 7. Smooth Manifolds 44. Analyti Digression: Di erentiable Fun tions in Eu lidean Spa e Di erentiability and Di erentials Derivative Along Ve tor Main Properties of Di erential Higher Order Derivatives C r -Maps Di eomorphisms Inverse Fun tion Theorem Impli it Fun tion Theorem C r -Fun tions Useful C 1 -Fun tion Appli ations of Bell-Shape Fun tion C r -Maps
45. Di erential Spa es
xv
177 177 177 177 179 180 181 181 182 182 182 182 182 182 183 183 183 183 183 183 183 183 183 183 184 184 184 184 184 185 186 186 187 187 187 188 189 189 189 190 190 190 190 191
CONTENTS
Motivation: Topologi al Stru ture via Continuous Fun tions Di erential Spa es Di erential Stru ture of a Metri Spa e Di erential Subspa es C r -Stru tures on Subspa e of Metri Spa e Di erentiable Maps Di eomorphisms Di erentiale Embeddings Semi ubi Parabola
46. Constru ting Di erential Spa es
Multipli ation of Di erentiable Spa es Quotient Di erential Spa es Classi al Lie Groups and Homogeneous Spa es Spa e of n-Point Subsets of Surfa e Tori Varieties
47. Smooth Manifolds C r -Manifolds
Manifolds with Corners Traditional Approa h to Smooth Manifolds Equivalen e of the Two Approa hes Revision of Boundary Revision of Multipli ation Revision of Di erentiable Maps Rank of Mapping Di erential Topology Submanifolds
48. Immersions and Embeddings
Immersions Di erentiable Embeddings Immersions Versus Embeddings Embeddability to Eu lidean Spa es
49. Tangent Ve tors
Coordinate De nition Digression on Einstein Notations Di erentiation of Fun tions Di erential of Map Tangent Bundle Tangent Ve tors in Eu lidean Spa e Ve tors as Velo ities
50. Ve tor Bundles
General Terminology of Fibrations Trivial and Lo ally Trivial Indu ed Fibrations Ve tor Bundles Constru tions with Ve tor Bundles
xvi
191 192 193 194 195 195 196 196 197 197 197 198 199 199 199 199 199 200 200 202 203 203 203 204 204 204 205 205 206 207 207 208 209 210 210 210 210 211 211 211 211 211 211 211 211
CONTENTS
Tautologi al Bundles Homotopy Classi ation of Ve tor Bundles Low-Dimensional
51. Orientation
Linear Algebra Digression: Orientations of Ve tor Spa e Related Orientations Orientation of Ve tor Bundle Orientation and Orientability of Smooth Manifold Orientation of Boundary Orientation Covering Proje tive Spa es
52. Transversality and Cobordisms
Sard Theorem Transversality Embedding to R 2n+1 Normal Bundle and Tubular Neighborhood Pontryagin Constru tion Degree of Map Linking Numbers Hopf Invariant Thom Constru tion Cobordisms
xvii
211 211 211 211 212 212 212 212 212 212 212 212 212 212 212 212 212 212 212 212 212 212
Part 1
General Topology
Although it may seem unexpe ted, the goal of this part of the book is to tea h the language of mathemati s. More spe i ally, one of its most important omponents: the language of set-theoreti topology, whi h treats the basi notions related to ontinuity. The term general topology means: this is the topology that is needed and used by most mathemati ians. As a resear h eld, it was ompleted a long time ago. Its permanent usage in the apa ity of a ommon mathemati al language has polished its system of de nitions and theorems. Nowadays studying general topology really resembles studying a language rather than mathemati s: one needs to learn a lot of new words, while proofs of all theorems are extremely simple. On the other hand, the theorems are numerous. It is not surprising: they play the role of rules regulating usage of words. We have to warn students, for whom this is one of the rst mathemati al subje ts. Do not hurry to fall in love with it too seriously, do not let an imprinting happen. This eld may seam to be harming, but it is not very a tive. It hardly provides as mu h room for ex iting new resear h as most other elds.
CHAPTER 1
Generalities 1. Topology in a Set De nition of Topologi al Spa e Let X be a set. Let be a olle tion of its subsets su h that: (a) the union of a family of sets, whi h are elements of , belongs to ; (b) the interse tion of a nite family of sets, whi h are elements of , belongs to ; ( ) the empty set ? and the whole X belong to . Then is alled a topologi al stru ture or just a topology 1 in X ; the pair (X; ) is alled a topologi al spa e ; an element of X is alled a point of this topologi al spa e; an element of is alled an open set of the topologi al spa e (X; ). The onditions in the de nition above are alled the axioms of topologi al stru ture.
Simplest Examples A dis rete topologi al spa e is a set with the topologi al stru ture whi h
onsists of all the subsets. 1.A. Che k that this is a topologi al spa e, i.e., all axioms of topologi al stru ture hold true. An indis rete topologi al spa e is the opposite example, in whi h the topologi al stru ture is the most meager. It onsists only of X and ?. 1.B. This is a topologi al stru ture, is it not? Here are less trivial examples. 1 Thus
is important: it is alled by the same word as the whole bran h of mathemati s. Of ourse, this does not mean that oin ides with the subje t of topology, but everything in this subje t is related to . 3
1. TOPOLOGY IN A SET
4
Let X be the ray [0; +1), and onsists of ?, X , and all the rays (a; +1) with a 0. Prove that is a topologi al stru ture. 1.2. Let X be a plane. Let onsist of ?, X , and all open disks with enter at the origin. Is this a topologi al stru ture? 1.3. Let X onsist of four elements: X = fa; b; ; dg. Whi h of the following olle tions of its subsets are topologi al stru tures in X , i.e., satisfy the axioms of topologi al stru ture: (a) ?, X , fag, fbg, fa; g, fa; b; g, fa; bg; (b) ?, X , fag, fbg, fa; bg, fb; dg; ( ) ?, X , fa; ; dg, fb; ; dg?
1.1.
The spa e of 1.1 is alled an arrow. We denote the spa e of 1.3 (a) by 4pT . It is a sort of toy spa e made of 4 points. Both of these spa es, as well as the spa e of 1.2, are not important, but provide good simple examples.
The Most Important Example: Real Line Let X be the set R of all real numbers, be the set of unions of all intervals (a; b) with a; b 2 R . 1.C. Che k if satis es the axioms of topologi al stru ture. This is the topologi al stru ture whi h is always meant when R is onsidered as a topologi al spa e (unless other topologi al stru ture is expli itly spe i ed). This spa e is alled usually the real line and the stru ture is referred to as the anoni al or standard topology in R . Let X be R, and onsists of empty set and all the in nite subsets of R. Is a topologi al stru ture? 1.5. Let X be R, and onsists of empty set and omplements of all nite subsets of R. Is a topologi al stru ture?
1.4.
The spa e of 1.5 is denoted by RT1 and alled the line with T1 -topology . 1.6. Let (X; ) be a topologi al spa e and Y be the set obtained from X by adding a single element a. Is ffag [ U : U 2 g [ f?g a topologi al stru ture in Y ?
Using New Words: Points, Open and Closed Sets Re all that, for a topologi al spa e (X; ), elements of X are alled points, and elements of are alled open sets.2 2 The
letter stands for the letter O whi h is the initial of the words with the same meaning: Open in English, Otkrytyj in Russian, O en in German, Ouvert in Fren h.
1. TOPOLOGY IN A SET
5
1.D. Reformulate the axioms of topologi al stru ture using the words
open set wherever possible.
A set F 2 X is said to be losed in the spa e (X; ) if its omplement X r F is open (i.e., X r F 2 ).
Set-Theoreti Digression. De Morgan Formulas
1.E. Let fAg2 be an arbitrary family of subsets of a set X . Prove that (1)
Xr
(2)
Xr
[ 2
\ 2
A = A =
\ 2
[ 2
(X r A )
(X r A ):
Formula (2) is dedu ed from (1) in one step, is it not? These formulas are nonsymmetri ases of a single formulation, whi h ontains in a symmetri way sets and their omplements, unions and interse tions. 1.7. Riddle. Find su h a formulation.
Being Open or Closed
1.F Properties of Closed Sets. Prove that:
(a) the interse tion of any olle tion of losed sets is losed; (b) union of any nite number of losed sets is losed; ( ) empty set and the whole spa e (i.e., the underlying set of the topologi al stru ture) are losed.
Noti e that the property of being losed is not a negation of the property of being open. 1.G. Find examples of sets, whi h (a) are both open, and losed simultaneously; (b) are neither open, nor losed. 1.8.
(a) (b) ( ) (d) (e)
Give an expli it des ription of losed sets in a dis rete spa e; an indis rete spa e; the arrow; 4pT ; RT1 .
1.H. Prove that a losed segment [a; b℄ is losed in R .
1. TOPOLOGY IN A SET
6
Con epts of losed and open sets are similar in a number of ways. The main di eren e is that the interse tion of an in nite olle tion of open sets does not have to be ne essarily open, while the interse tion of any
olle tion of losed sets is losed. Along the same lines, the union of an in nite olle tion of losed sets is not ne essarily losed, while the union of any olle tion of open sets is open. Prove that the half-open interval [0; 1) is neither open nor losed in R, but an be presented as either the union of losed sets or interse tion of open sets. 1.10. Prove that every open set of the real line is a union of disjoint open intervals. 1 1 1.11. Prove that the set A = f0g [ is losed in R. n n=1
1.9.
Cantor Set Let K be the set of real numbers whi h an be presented as sums of series P ak of the form 1 k=1 3k with ak = 0 or 2. In other words, K is the set of real numbers whi h in the positional system with base 3 are presented as 0:a1 a2 : : : ak : : : without digit 1. 1:A. Find a geometri des ription of K . 1:A:1. Prove that (a) K is ontained in [0; 1℄, (b) K does not interse t 31 ; 23 , 3s+2 for any integers k and s. ( ) K does not interse t 3s3+1 k ; 3k 1:A:2. Present K as [0; 1℄ with an in nite family of open intervals removed. 1:A:3. Try to draw K . The set K is alled the Cantor set. It has a lot of remarkable properties and is involved in numerous problems below. 1:B. Prove that K is a losed set in the real line.
Chara terization of Topology in Terms of Closed Sets Prove that if a olle tion F of subsets of X satis es the following
onditions: (a) the interse tion of any family of sets from F belongs to F ; (b) the union of any nite number sets from F belongs to F ; ( ) ? and X belong to F , then F is the set of all losed sets of a topologi al spa e (whi h one?). 1.13. List all olle tions of subsets of a three-element set su h that there exist topologies, in whi h these olle tions are omplete sets of losed sets. 1.12.
2. BASES
7
Topology and Arithmeti Progressions 1.14*. Consider the following property of a subset F of the set N of natural numbers: there exists N 2 N su h that F does not ontain an arithmeti progression of length greater than N . Prove, that subsets with this property together with the whole N form a olle tion of losed subsets in some topology in N .
Solving this problem, you probably are not able to avoid the following ombinatorial theorem. 1.15 Van der Waerden's Theorem*. For every n 2 N there exists N 2 N su h that for any A f1; 2; : : : ; N g, either A or f1; 2; : : : ; N g r A ontains
an arithmeti progression of length n.
Neighborhoods By a neighborhood of a point one means any open set ontaining this point. Analysts and Fren h mathemati ians (following N. Bourbaki) prefer a wider notion of neighborhood: they use this word for any set
ontaining a neighborhood in the sense above. Give an expli it des ription of all neighborhoods of a point in a dis rete spa e; an indis rete spa e; the arrow; 4pT .
1.16.
(a) (b) ( ) (d)
2. Bases De nition of Base Usually the topologi al stru ture is presented by des ribing its part, whi h is suÆ ient to re over the whole stru ture. A olle tion of open sets is alled a base for a topology if ea h nonempty open set is a union of sets of . For instan e, all intervals form a base for the real line. 2.1.
Are there di erent topologi al stru tures with the same base?
Find some bases of topology of (a) a dis rete spa e; (b) an indis rete spa e; ( ) the arrow; (d) 4pT . Try to hoose the bases as small as possible.
2.2.
2.3.
Des ribe all topologi al stru tures having exa tly one base.
2. BASES
8
Bases for Plane 2.4.
Prove that any base of the anoni al topology of R an be diminished.
Consider the following three olle tions of subsets of R2 :
2 whi h onsists of all possible open disks (i.e., disks without its boundary ir les); 1 whi h onsists of all possible open squares (i.e., squares without their sides and verti es) with sides parallel to the oordinate axis; 1 whi h onsists of all possible open squares with sides parallel to the bise tors of the oordinate angles. (Squares of 1 and 1 are de ned by inequalities maxfjx aj; jy bjg < and jx aj + jy bj < respe tively.) 2.5. Prove that every element of 2 is a union of elements of 1. 2.6.
of
Prove that interse tion of any two elements of 1 is a union of elements
1.
Prove that ea h of the olle tions 2, 1, 1 is a base for some topologi al stru ture in R2 , and that the stru tures de ned by these olle tions
oin ide.
2.7.
When a Colle tion of Sets is a Base
2.A.
A olle tion of open sets is a base for the topology, i for any open set U and any point x 2 U there is a set V 2 su h that x 2 V U .
2.B.
A olle tion of subsets of a set X is a base for some topology in X , i X is a union of sets of and interse tion of any two sets of is a union of sets in .
2.C. Show that the se ond ondition in 2.B (on interse tion) is equiva-
lent to the following: the interse tion of any two sets of ontains, together with any of its points, some set of ontaining this point ( f. 2.A). Subbases Let (X; ) be a topologi al spa e. A olle tion of its open subsets is alled a subbase for , provided the olle tion = fV j V = \ki=1 Wi ; Wi 2 ; k 2 N g of all nite interse tions of sets belonging to is a base for . Prove that for any set X a olle tion of its subsets is a subbase of a topology in X , i 6= ? and X = [W 2 W .
2.8.
3. METRIC SPACES
9
In nity of the Set of Prime Numbers Prove that all in nite arithmeti progressions onsisting of natural numbers form a base for some topology in N . 2.10. Using this topology prove that the set of all prime numbers is in nite.
2.9.
(Hint: otherwise the set f1g would be open (?!) )
Hierar hy of Topologies If 1 and 2 are topologi al stru tures in a set X su h that 1 2 then
2 is said to be ner than 1 , and 1 oarser than 2 . For instan e, among all topologi al stru tures in the same set the indis rete topology is the oarsest topology, and the dis rete topology is the nest one, is it not? Show that T1 -topology (see Se tion 1) is oarser than the anoni al topology in the real line. 2.12. Riddle. Let 1 and 2 be bases for topologi al stru tures 1 and 2 in a set X . Find ne essary and suÆ ient ondition for 1 2 in terms of the bases 1 and 2 without expli it referring to 1 and 2 ( f. 2.7). 2.11.
Bases de ning the same topologi al stru ture are said to be equivalent. 2.D. Riddle. Formulate a ne essary and suÆ ient ondition for two bases to be equivalent without expli it mentioning of topologi al stru tures de ned by the bases. (Cf. 2.7: bases 2, 1, and 1 must satisfy the ondition you are looking for.)
3. Metri Spa es De nition and First Examples A fun tion : X X ! R + = f x 2 R j x 0 g is alled a metri (or distan e ) in X , if (a) (x; y ) = 0, i x = y ; (b) (x; y ) = (y; x) for every x; y 2 X ; ( ) (x; y ) (x; z ) + (z; y ) for every x; y; z 2 X . The pair (X; ), where is a metri in X , is alled a metri spa e. The
ondition ( ) is triangle inequality. 3.A. Prove that for any set X ( 0; if x = y ; : X X ! R + : (x; y ) 7! 1; if x 6= y
3. METRIC SPACES
10
is a metri .
3.B. Prove that R R ! R + : (x; y) 7! jx yj is a metri . p 3.C. Prove that R n R n ! R + : (x; y) 7! Pni=1(xi yi)2 is a metri . Metri s 3.B and 3.C are always meant when R and R n are onsidered as metri spa es unless another metri is spe i ed expli itly. Metri 3.B is a spe ial ase of metri 3.C. These metri s are alled Eu lidean.
Further Examples 3.1. 3.2.
Prove that Rn Rn
! R + : (x; y) 7! maxi=1;::: ;n jxi yi j is a metri . P Prove that Rn Rn ! R + : (x; y) 7! ni=1 jxi yi j is a metri .
Metri s in Rn introdu ed in 3.C{3.2 are in luded in in nite series of the metri s
(p) 3.3.
: (x; y) 7!
n X
i=1
jxi yi
1 p p j ;
p 1:
Prove that (p) is a metri for any p 1.
3.3.1 Holder Inequality. Prove that n X i=1
x i yi
if xi ; yi 0, p; q > 0
n X p
xi
i=1 and p1
!1=p
n X q i=1
yi
!1=q
+ 1q = 1.
Metri of 3.C is (2), metri of 3.2 is (1), and metri of 3.1 an be denoted by (1) and adjoined to the series sin e lim
p!+1
n X
i=1
1 p p ai
= max ai ;
for any positive a1 , a2 , : : : , an . 3.4. Riddle.
How is this related to 2, 1, and 1 from Se tion 2?
For a real number pP 11 denote by l(p) the set of sequen es x = fxi gi=1;2;::: su h that the series i=1 jxjp onverges. p (p) the series P1 jx 3.5. Prove that for any two elements x; y 2 l i=1 i yi j
onverges and that 1 1 X p p jxi yi j ; p 1 (x; y) 7! is a metri in l(p) .
i=1
3. METRIC SPACES
11
Balls and Spheres Let (X; ) be a metri spa e, let a be its point, and let r be a positive real number. The sets (3) Dr (a) = f x 2 X j (a; x) < r g; (4) Dr [a℄ = f x 2 X j (a; x) r g; (5) Sr (a) = f x 2 X j (a; x) = r g are alled, respe tively, open ball, losed ball, and sphere of the spa e (X; ) with enter at a and radius r.
Subspa es of a Metri Spa e If (X; ) is a metri spa e and A X , then the restri tion of metri to A A is a metri in A, and (A; AA) is a metri spa e. It is alled a subspa e of (X; ). The ball D1 [0℄ and sphere S1 (0) in R n (with Eu lidean metri , see 3.C) are denoted by symbols Dn and S n 1 and alled n-dimensional ball and (n 1)-dimensional sphere. They are onsidered as metri spa es (with the metri restri ted from R n ). 3.D. Che k that D1 is the segment [ 1; 1℄; D2 is a disk; S 0 is the pair of points f 1; 1g; S 1 is a ir le; S 2 is a sphere; D3 is a ball. The last two statements larify the origin of terms sphere and ball (in the ontext of metri spa es). Some properties of balls and spheres in arbitrary metri spa e resemble familiar properties of planar disks and ir les and spatial balls and spheres. 3.E. Prove that for points x and a of any metri spa e and any r > (a; x) Dr (a;x) (x) Dr (a): Surprising Balls However in other metri spa es balls and spheres may have rather surprising properties. 2 3.6. What are balls and spheres in R with metri s of 3.1 and 3.2 ( f. 3.4)? 3.7. Find D1 [a℄, D 1 [a℄, and S 1 (a) in the spa e of 3.A. 2 2 Find a metri spa e and two balls in it su h that the ball with the smaller radius ontains the ball with the bigger one and does not oin ide with it.
3.8.
3. METRIC SPACES
12
3.9. What is the minimal number of points in the spa e whi h is required to be onstru ted in 3.8. 3.10. Prove that in 3.8 the big radius does not ex eed double the smaller radius.
Segments (What Is Between) Prove that the segment with end points a; b 2 Rn an be des ribed as f x 2 Rn j (a; x) + (x; b) = (a; b) g; where is the Eu lidean metri . 3.12. How do the sets de ned as in 3.11 look like with of 3.1 and 3.2? (Consider the ase n = 2 if it appears to be easier.) 3.11.
Bounded Sets and Balls A subset A of a metri spa e (X; ) is said to be bounded, if there is a number d > 0 su h that (x; y ) < d for any x; y 2 A. The greatest lower bound of su h d is alled the diameter of A and denoted by diam(A). 3.F. Prove that a set A is bounded, i it is ontained in a ball. 3.13. What is the relation between the minimal radius of su h a ball and diam(A)?
Norms and Normed Spa es Let X be a ve tor spa e (over R). Fun tion X ! R + : x 7! j xj is alled a norm if (a) j xj = 0, i x = 0; (b) j xj = jjjjxj for any 2 R and x 2 X ; ( ) j x + yj j xj + j yj for any x; y 2 X . 3.14. Prove that if x 7! j xj is a norm then : X X ! R + : (x; y) 7! j x yj is a metri . The ve tor spa e equipped with a spe i ed norm is alled a normed spa e. The metri de ned by the norm as in 3.14 turns the normed spa e into the metri one in a anoni al way. 3.15. Look through the problems of this se tion and gure out whi h of the metri spa es involved are, in fa t, normed ve tor spa es. 3.16. Prove that every ball in the normed spa e is a onvex3 set symmetri with respe t to the enter of the ball. that a set A is said to be onvex if for any x; y 2 A the segment onne ting x; y is ontained in A. Of ourse, this de nition is based on the notion of segment, so it makes sense only for subsets of spa es, where the notion of segment onne ting two point is de ned. This is the ase in ve tor and aÆne spa es over R 3 Re all
3. METRIC SPACES
13
Prove that every onvex losed bounded set in Rn , whi h is symmetri with respe t to its enter and is not ontained in any aÆne spa e ex ept Rn itself, is the unit ball with respe t to some norm, and that this norm is uniquely de ned by this ball. 3.17*.
Metri Topology
3.G.
The olle tion of all open balls in the metri spa e is a base for some topology ( f. 2.A, 2.B and 3.E).
This topology is alled metri topology. It is said to be indu ed by the metri . This topologi al stru ture is always meant whenever the metri spa e is onsidered as a topologi al one (for instan e, when one says about open and losed sets, neighborhoods, et . in this spa e).
3.H. Prove that the standard topologi al stru ture in R introdu ed in Se tion 1 is indu ed by metri (x; y ) 7! jx y j. 3.18.
What topologi al stru ture is indu ed by the metri of 3.A?
3.I. A set is open in a metri spa e, i it ontains together with any its point a ball with enter at this point. 3.19.
Prove that a losed ball is losed (with respe t to the metri topology).
3.20.
Find a losed ball, whi h is open (with respe t to the metri topology).
3.21.
Find an open ball, whi h is losed (with respe t to the metri topol-
3.22.
Prove that a sphere is losed.
3.23.
Find a sphere, whi h is open.
ogy).
Metrizable Topologi al Spa es A topologi al spa e is said to be metrizable if its topologi al stru ture is indu ed by some metri .
3.J. An indis rete spa e is not metrizable unless it onsists of a single point (it has too few open sets).
3.K. A nite spa e is metrizable i it is dis rete. 3.24.
Whi h topologi al spa es des ribed in Se tion 1 are metrizable?
Equivalent Metri s Two metri s in the same set are said to be equivalent if they indu e the same topology.
3. METRIC SPACES
14
Are the metri s of 3.C, 3.1, and 3.2 equivalent? 3.26. Prove that metri s 1 , 2 in X are equivalent if there are numbers
; C > 0 su h that
1 (x; y) 2 (x; y) C1 (x; y) for any x; y 2 X . 3.27. Generally speaking the inverse is not true. 3.28. Riddle. Hen e the ondition of the equivalen e of metri s formulated in 3.26 an be weakened. How? 3.29*. Prove that the following two metri s 1 , C in the set of all ontinuous fun tions [0; 1℄ ! R are not equivalent:4
3.25.
Z 1 f (x)
C (f; g) = max f (x) g(x) : x2[0;1℄ 0 Is it true that topologi al stru ture de ned by one of them is ner than another? 1 (f; g) =
g(x) dx;
Ultrametri A metri is alled an ultrametri if it satis es to ultrametri triangle inequality : (x; y) maxf(x; z ); (z; y)g for any x; y, z . A metri spa e (X; ) with ultrametri is alled an ultrametri spa e. 3.30. Che k that only one metri in 3.A{3.2 is ultrametri . Whi h one? 3.31. Prove that in an ultrametri spa e all triangles are isos eles (i.e., for any three points a, b, two of the three distan es (a; b), (b; ), (a; ) are equal). 3.32. Prove that in a ultrametri spa e spheres are not only losed ( f. 3.22) but also open. The most important example of ultrametri is p-adi metri in the set Q of all rational numbers. Let p be a prime number. For x; y 2 Q , present the di eren e x y as rs p , where r, s, and are integers, and r, s are relatively prime with p. Put (x; y) = p . 3.33. Prove that this is an ultrametri .
Operations with Metri s 3.34. Prove that if : X X ! R + is a fun tion whi h satis es onditions (a) and ( ) of the de nition of metri then the fun tion (x; y) 7! (x; y) + (y; x) is a metri in X .
4 Indexes
in the notations allude to the spa es these metri s are de ning.
3. METRIC SPACES
15
Prove that if 1 , 2 are metri s in X then 1 + 2 and maxf1 ; 2 g are also metri s. Are the fun tions minf1 ; 2 g, 1 , and 1 2 metri s? 2 3.36. Prove that if : X X ! R + is a metri then (a) fun tion (x; y) (x; y) 7! 1 + (x; y) is a metri ; (b) fun tion (x; y) 7! f (x; y) is a metri , if f satis es the following onditions: (1) f (0) = 0, (2) f is a monotone in reasing fun tion, and (3) f (x + y) f (x) + f (y) for any x; y 2 R. 3.37. Prove that metri s and are equivalent. 1+ 3.35.
Distan e Between Point and Set Let (X; ) be a metri spa e, A X , b 2 X . The inf f (b; a) j a 2 A g is
alled a distan e from the point b to the set A and denoted by (b; A). 3.L. Let A be a losed set. Prove that (b; A) = 0, i b 2 A. Prove that j(x; A) of the same metri spa e.
3.38.
(y; A)j (x; y) for any set A and points x, y
Distan e Between Sets Let A and B be bounded subsets in the metri spa e (X; ). Put n
o
d (A; B ) = max sup (a; B ); sup (b; A) : a2A b 2B This number is alled the Hausdor distan e between A and B . 3.39. Prove that the Hausdor distan e in the set of all bounded subsets of a metri spa e satis es the onditions (b) and ( ) of the de nition of metri . 3.40. Prove that for every metri spa e the Hausdor distan e is a metri in the set of its losed bounded subsets. Let A and B be bounded polygons in the plane5 . Put d (A; B ) = S (A) + S (B ) 2S (A \ B ); where S (C ) is the area of polygon C . 3.41. Prove that d is a metri in the set of all plane bounded polygons. 5 Although we assume that the notion of bounded polygon is well-known from elementary geometry, re all the de nition. A bounded plane polygon is a set of the points of a simple losed polygonal line and the points surrounded by this line. By a simple
losed polygonal line we mean a y li sequen e of segments su h that ea h of them starts at the point where the previous one nishes and these are the only pairwise interse tions of the segments.
4. SUBSPACES
16
We will all d the area metri . 3.42. Prove that in the set of all bounded plane polygons the area metri is not equivalent to the Hausdor metri . 3.43. Prove that in the set of onvex bounded plane polygons the area metri is equivalent to the Hausdor metri .
4. Subspa es Let (X; ) be a topologi al spa e, and A X . Denote by A the olle tion of sets A \ V , where V 2 . 4.A. A is a topologi al stru ture in A. The pair (A; A ) is alled a subspa e of the spa e (X; ). The olle tion
A is alled the subspa e topology or the relative topology or the topology indu ed on A by , and its elements are alled open sets in A. 4.B. The anoni al topology in R 1 and the topology indu ed on R 1 as a subspa e of R 2 oin ide. How to onstru t a base for the topology indu ed on A using the base for the topology in X ? 4.2. Des ribe the topologi al stru tures indu ed (a) on the set N of natural numbers by the topology of the real line; (b) on N by the topology of the arrow; ( ) on the two-point set f1; 2g by the topology of RT1 ; (d) on the same set by the topology of the arrow. 4.3. Is the half-open interval [0; 1) open in the segment [0; 2℄ onsidered as a subspa e of the real line?
4.1. Riddle.
4.C. A set is losed in a subspa e, i it is the interse tion of the subspa e and a losed subset of the ambient spa e. Relativity of Openness
Sets, whi h are open in the subspa e, are not ne essarily open in the ambient spa e. 4.D. The unique open set in R 1 , whi h is also open in R 2 , is the empty set ?. However: 4.E. Open sets of an open subspa e are open in the ambient spa e, i.e., if A 2 then A . The same relation holds true for losed sets. Sets, whi h are losed in the subspa e, are not ne essarily losed in the ambient spa e. However:
5. POSITION OF A POINT WITH RESPECT TO A SET
17
4.F. Closed sets of the losed subspa e are losed in the ambient spa e. Prove that a set U is open in X , i every its point has a neighborhood V in X su h that U \ V is open in V .
4.4.
It allows one to say that the property of being open is a lo al property. 4.5.
Show that the property of being losed is not a lo al property.
4.G Transitivity of Indu ed Topology. Let (X; ) be a topologi al spa e, and X A B . Then ( A )B = B , i.e., the topology indu ed on
B by the topology indu ed on A oin ides with the topology indu ed on B dire tly.
4.6. Let (X; ) be a metri spa e, and A X . Then the topology in A generated by metri AA oin ides with the topology indu ed on A by the topology in X generated by metri . (To prove this statement you need to prove two in lusions. Whi h of them is less obvious?)
Agreement on Notations of Topologi al Spa es Di erent topologi al stru tures in the same set are not onsidered simultaneously very often. That is why a topologi al spa e is usually denoted by the same symbol as the set of its points, i.e., instead of (X; ) one writes just X . The same is applied for metri spa es: instead of (X; ) one writes just X .
5. Position of a Point with Respe t to a Set This se tion is devoted to a further expansion of the vo abulary needed when one speaks of phenomena in a topologi al spa e.
Interior, Exterior and Boundary Points Let X be a topologi al spa e, A X , and b 2 X . The point b is alled an interior point of the set A if it has a neighborhood ontained in A; an exterior point of the set A if it has a neighborhood disjoint with A; a boundary point of the set A if any its neighborhood interse ts both A and the omplement of A.
5. POSITION OF A POINT WITH RESPECT TO A SET
18
Interior and Exterior The interior of a set A in a topologi al spa e X is the maximal (with respe t to in lusion) open in X set ontained in A, i.e., an open set, whi h ontains any other open subset of A. It is denoted Int A or, going into details, IntX A.
5.A.
Every subset of a topologi al spa e has interior. It is the union of all open sets ontained in this set.
5.B. The interior of a set is the union of its interior points. 5.C. A set is open, i it oin ides with its interior. 5.D. Prove that in R : (a) Int[0; 1) = (0; 1), (b) Int Q = ? and ( ) Int(R r Q ) = ?. 5.1.
Find the interior of fa; b; dg in spa e 4pT .
The exterior of a set is the maximal open set disjoint from A. It is obvious that the exterior of A is Int(X r A).
Closure The losure of a set A is the minimal losed set ontaining A. It is denoted Cl A or, going into details, ClX A.
5.E.
Every subset of topologi al spa e has losure. It is the interse tion of all losed sets ontaining this set. 5.2. Prove that if A is a subspa e of X , and B A, then ClA B = (ClX B ) \ A. Is it true that IntA B = (IntX B ) \ A?
A point b is alled an adherent point for a set A if all of its neighborhood interse t A.
5.F. The losure of a set is the set of its adherent points. 5.G. A set A is losed, i A = Cl A. 5.H. The losure of a set is the omplement of its exterior. In formulas: Cl A = X r Int(X r A), where X is the spa e and A X . 5.I. Prove that in R : (a) Cl[0; 1) = [0; 1℄, (b) Cl Q = R , ( ) Cl(R r Q ) = R . 5.3.
Find the losure of fag in 4pT .
5. POSITION OF A POINT WITH RESPECT TO A SET
19
Frontier The frontier of a set A is the set Cl A r Int A. It is denoted by Fr A or, more pre isely, FrX A. 5.4.
In 4pT nd the frontier of fag.
5.J. The frontier of a set is the set of its boundary points. 5.K. Prove that a set A is losed, i Fr A A.
5.5. Prove that Fr A = Fr(X r A). Find a formula for Fr A, whi h is symmetri with respe t to A and X r A. 5.6. The frontier of a set A equals the interse tion of the losure of A and the losure of the omplement of A: Fr A = Cl A \ Cl(X r A):
Closure and Interior with Respe t to a Finer Topology 5.7. Let 1 , 2 be topologi al stru ture in X , and 1 2 . Let Cli denote the losure with respe t to i . Prove that Cl1 A Cl2 A for any A X . 5.8. Formulate and prove an analogous statement about interior.
Properties of Interior and Closure Prove that if A B then Int A Int B . 5.10. Prove that Int Int A = Int A. 5.11. Is it true that for any sets A and B the following equalities hold true: Int(A \ B ) = Int A \ Int B; Int(A [ B ) = Int A [ Int B ? 5.12. Give an example in whi h one of that equalities does not hold true. 5.13. In the example that you have found solving the previous problem an in lusion of one hand side into another one holds true. Does this in lusion hold true for any A and B ? 5.14. Study the operator Cl in a way suggested by the investigation of Int undertaken in 5.9{5.13. 5.15. Find Clf1g, Int[0; 1℄, and Fr(2; +1) in the arrow. 1 5.16. Find Int (0; 1℄ [ f2g , Cl f n j n 2 N g , and Fr Q in R. 5.17. Find Cl N , Int(0; 1), and Fr[0; 1℄ in RT1 . How to nd the losure and interior of a set in this spa e? 5.18. Prove that a sphere ontains the frontier of the open ball with the same enter and radius. 5.19. Find an example in whi h a sphere is disjoint from the losure of the open ball with the same enter and radius.
5.9.
(6) (7)
5. POSITION OF A POINT WITH RESPECT TO A SET
20
Let A be a subset, and b be a point of the metri spa e (X; ). Re all (see Se tion 3) that the distan e (b; A) from the point b to the set A is the inf f (b; a) j a 2 A g. 5.L. Prove that b 2 Cl A, i (b; A) = 0. How many pairwise distin t sets an one obtain out of a single set using operators Cl and Int?
5.20 The Kuratowski Problem.
The following problems will help you to solve problem 5.20. 5.20.1. Find a set A R su h that the sets A, Cl A, and Int A would be pairwise distin t. 5.20.2. Is there a set A R su h that (a) A, Cl A, Int A, Cl Int A are pairwise distin t; (b) A, Cl A, Int A, Int Cl A are pairwise distin t; ( ) A, Cl A, Int A, Cl Int A, Int Cl A are pairwise distin t? If you nd su h sets, keep on going in the same way, and when fail, try to formulate a theorem explaining the failure. 5.20.3. Prove that Cl Int Cl Int A = Cl Int A. Find three sets in the real line, whi h have the same frontier. Is it possible to in rease the number of su h sets?
5.21*.
Re all that a set A Rn is said to be onvex if together with any two points it ontains the whole interval onne ting them (i.e., for any x; y 2 A any point z belonging to the segment [x; y℄ belongs to A). Let A be a onvex set in Rn . 5.22. Prove that Cl A and Int A are onvex. 5.23. Prove that A ontains a ball, unless A is not ontained in an (n 1)dimensional aÆne subspa e of Rn . 5.24. When is Fr A onvex?
Chara terization of Topology by Closure or Interior Operations 5.25*. Let in the set of all subset of a set X exist an operator Cl whi h has the following properties: (a) Cl ? = ?; (b) Cl A A; ( ) Cl (A [ B ) = Cl A [ Cl B ; (d) Cl Cl A = Cl A. Prove that = f U X j Cl (X r U ) = X r U g is a topologi al stru ture, and Cl A is the losure of a set A in the spa e (X; ). 5.26. Find an analogous system of axioms for Int.
5. POSITION OF A POINT WITH RESPECT TO A SET
21
Dense Sets Let A and B be sets in a topologi al spa e X . A is said to be dense in B if Cl A B , and everywhere dense if Cl A = X .
5.M. A set is everywhere dense, i it interse ts any nonempty open set. 5.N. The set Q is everywhere dense in R .
Give a hara terization of everywhere dense sets in an indis rete spa e, in the arrow and in RT1 . 5.28. Prove that a topologi al spa e is a dis rete spa e, i it has a unique everywhere dense set (whi h is the entire spa e, of ourse). 5.29. Is it true that the union of everywhere dense sets is everywhere dense, and that the interse tion of everywhere dense sets is everywhere dense? 5.30. Prove that the interse tion of two open everywhere dense sets is everywhere dense. 5.31. Whi h ondition in the previous problem is redundant? 5.32*. Prove that in R a ountable interse tion of open everywhere dense sets is everywhere dense. Is it possible to repla e R here by an arbitrary topologi al spa e? 5.33*. Prove that Q annot be presented as a ountable interse tion of open sets dense in R. 5.34. Formulate a ne essary and suÆ ient ondition on the topology of a spa e whi h has an everywhere dense point. Find spa es satisfying the ondition in Se tion 1. 5.27.
Nowhere Dense Sets A set is alled nowhere dense if its exterior is everywhere dense. 5.35.
Can a set be everywhere dense and nowhere dense simultaneously?
5.O. A set A is nowhere dense in X , i any neighborhood of any point x 2 X ontains a point y su h that the omplement of A ontains y together with one of its neighborhoods.
What an you say about the interior of a nowhere dense set? 5.37. Is R nowhere dense in R2 ? 5.38. Prove that if A is nowhere dense then Int Cl A = ?. 5.39. Prove that the frontier of a losed set is nowhere dense. Is this true for the boundary of an open set; boundary of an arbitrary set? 5.40. Prove that a nite union of nowhere dense sets is nowhere dense. 5.41. Prove that in Rn (n 1) every proper algebrai set (i.e., a set de ned by algebrai equations) is nowhere dense. 5.42. Prove that for every set A there exists a maximal open set B in whi h A is dense. The extreme ases B = X and B = ? mean that A is either everywhere dense or nowhere dense respe tively. 5.36. Riddle.
6. SET-THEORETIC DIGRESSION. MAPS
22
Limit Points and Isolated Points A point b is alled a limit point of a set A if any neighborhood of b interse ts A r fbg. 5.P. Every limit point of a set is its adherent point. 5.43.
one.
Give an example proving that an adherent point may be not a limit
A point b is alled an isolated point of a set A if b 2 A and there exists a neighborhood of b disjoint with A r fbg. 5.Q. A set A is losed, i it ontains all its limit points. 1 5.44. Find limit and isolated points of the sets (0; 1℄ [ f2g, f n j n 2 N g in Q and in R. 5.45. Find limit and isolated points of the set N in RT . 1
Lo ally Closed Sets A subset A of a topologi al spa e X is alled lo ally losed if ea h of its points has a neighborhood U su h that A \ U is losed in U ( f. 4.4{4.5). 5.46. Prove that the following onditions are equivalent: (a) A is lo ally losed in X ; (b) A is an open subset of its losure ClX A; ( ) A is the interse tion of open and losed subsets of X .
6. Set-Theoreti Digression. Maps Maps and the Main Classes of Maps A mapping f of a set X to a set Y is a triple onsisting of X , Y , and a rule,6 whi h assigns to every element of X exa tly one element of Y . There are other words with the same meaning: map, fun tion. f If f is a mapping of X to Y then one writes f : X ! Y , or X ! Y . The element b of Y assigned by f to an element a of X is denoted by f (a) f and alled the image of a under f . One writes b = f (a), or a 7! b, or f : a 7! b.
A mapping f : X ! Y is alled a surje tive map , or just a surje tion if every element of Y is an image of at least one element of X . A mapping 6 Of
ourse, the rule (as everything in the set theory) may be thought of as a set. Namely, one onsiders a set of ordered pairs (x; y) with x 2 X , y 2 Y su h that the rule assigns y to x. This set is alled the graph of f . It is a subset of the set X Y of all ordered pairs (x; y).
6. SET-THEORETIC DIGRESSION. MAPS
23
f : X ! Y is alled an inje tive map , inje tion , or one-to-one map if every element of Y is an image of not more than one element of X . A mapping is alled a bije tive map , bije tion , or invertible if it is surje tive and inje tive.
Image and Preimage The image of a set A X under a map f : X ! Y is the set of images of all points of A. It is denoted by f (A). Thus
f (A) = ff (x) : x 2 Ag: The image of the entire set X (i.e., f (X )) is alled the image of f . The preimage of a set B Y under a map f : X ! Y is the set of elements of X whose images belong to B . It is denoted by f 1 (B ). Thus f 1 (B ) = fa 2 X : f (a) 2 B g:
Be areful with these terms: their etymology an be misleading. For example, the image of the preimage of a set B an di er from B . And even if it does not di er, It may happen that the preimage is not the only set with this property. Hen e, the preimage annot be de ned as a set whose image is a given set.
6.A. f f 1(B ) = B , i B is ontained in the image of f . 6.B. f f 1(B ) B for any map f : X ! Y and B Y . 6.C. Let f : X ! Y and B Y su h that f f 1(B ) = B . Then the following statements are equivalent: (a) f 1 (B ) is the unique subset of X whose image equals B ; (b) for any a1 ; a2 2 f 1 (B ) the equality f (a1 ) = f (a2 ) implies a1 = a2 .
6.D. A map f : X ! Y is an inje tion, i for any B Y su h that f f 1 (B ) = B the preimage f 1 (B ) is the unique subset of X whose image equals B .
6.E. f 1 f (A) A for any map f : X ! Y and A X . 6.F. f 1 f (A) = A, i f (A) \ f (X r A) = ?. 6.1.
Y:
(8) (9) (10)
Do the following equalities hold true for any A; B Y and any f : X !
f 1 (A [ B ) = f 1 (A) [ f 1(B ); f 1 (A \ B ) = f 1 (A) \ f 1(B ); f 1(Y r A) = X r f 1(A)?
6. SET-THEORETIC DIGRESSION. MAPS
6.2.
Y:
24
Do the following equalities hold true for any A; B X and any f : X !
f (A [ B ) = f (A) [ f (B ); f (A \ B ) = f (A) \ f (B ); f (X r A) = Y r f (A)?
(11) (12) (13) 6.3.
Give examples in whi h two of the equalities above are false.
6.4.
Repla e the false equalities of 6.2 by orre t in lusions.
What simple ondition on f : X ! Y should be imposed in order to make orre t all the equalities of 6.2 for any A; B X ?
6.5.
6.6.
Prove that for any map f : X ! Y , and subsets A X , B Y :
B \ f (A) = f f 1 (B ) \ A :
Identity and In lusion The identity map of a set X is the map X ! X de ned by formula x 7! x. It is denoted by idX , or just id, when there is no ambiguity. If A is a subset of X then the map A ! X de ned by formula x 7! x is
alled an in lusion map , or just in lusion , of A into X and denoted by in : A ! X , or just in, when A and X are lear.
6.G. The preimage of a set B under an in lusion in : A ! X is B \ A. Composition The omposition of mappings f : X ! Y and g : Y g Æ f : X ! Z de ned by formula x 7! g f (x) .
! Z is the mapping
6.H. h Æ (g Æ f ) = (h Æ g) Æ f for any maps f : X ! Y , g : Y h : Z ! U.
! Z , and
6.I. f Æ (idX ) = f = (idX ) Æ f for any f : X ! Y . 6.J. The omposition of inje tions is inje tive. 6.K. If the omposition g Æ f is inje tive then f is inje tive. 6.L. The omposition of surje tions is surje tive. 6.M. If the omposition g Æ f is surje tive then g is surje tive. 6.N. The omposition of bije tions is a bije tion. 6.7.
Let a omposition g Æ f be bije tive. Is then f or g ne essarily bije tive?
7. CONTINUOUS MAPS
25
Inverse and Invertible A map g : Y ! X is said to be inverse to a map f : X ! Y if g Æ f = idX and f Æ g = idY . A map, for whi h an inverse map exists, is said to be invertible. 6.O. A mapping is invertible, i it is a bije tion. 6.P. If an inverse map exists then it is unique.
Submappings If A X and B Y then for every f : X ! Y su h that f (A) B there is mapping ab(f ) : A ! B de ned by formula x 7! f (x) and alled an abbreviation of the mapping f to A; B , or submapping, or submap . If B = Y then ab f : A ! Y is denoted by f A and alled the restri tion of f to A. If B 6= Y then ab f : A ! B is denoted by f A;B or even simply f j. 6.Q. The restri tion of a map f : X ! Y to A X is the omposition of in lusion in A :! X and f . In other words, f A = f Æ in. 6.R. Any abbreviation (in luding any restri tion) of inje tions is inje tive. 6.S. If a restri tion of a mapping is surje tive then the original mapping is surje tive.
7. Continuous Maps De nition and Main Properties of Continuous Maps Let X , Y be topologi al spa es. A map f : X ! Y is said to be
ontinuous if the preimage of any open subset of Y is an open subset of X. 7.A. A map is ontinuous, i the preimage of any losed set is losed. 7.B. The identity map of any topologi al spa e is ontinuous. 7.1. Let 1 , 2 be topologi al stru tures in X . Prove that the identity mapping of X id : (X; 1 ) ! (X; 2 ) is ontinuous, i 2 1 . 7.2. Let f : X ! Y be a ontinuous map. Is it ontinuous with respe t to (a) a ner topology in X and the same topology in Y , (b) a oarser topology in X and the same topology in Y , ( ) a ner topology in Y and the same topology in X , (d) a oarser topology in Y and the same topology in X ?
7. CONTINUOUS MAPS
26
7.3. Let X be a dis rete spa e and Y an arbitrary spa e. Whi h maps X ! Y and Y ! X are ontinuous? 7.4. Let X be an indis rete spa e and Y an arbitrary spa e. Whi h maps X ! Y and Y ! X are ontinuous?
7.C. Let A be a subspa e of X . The in lusion in : A ! X is ontinuous. 7.D. The topology A indu ed on A X by the topology of X is the
oarsest topology in A su h that the in lusion mapping in : A ! X is
ontinuous with respe t to it.
7.5. Riddle. The statement 7.D admits a natural generalization with the in lusion map repla ed by an arbitrary map f : A ! X of an arbitrary set A. Find this generalization.
7.E. A omposition of ontinuous maps is ontinuous. 7.F. A submap of a ontinuous map is ontinuous. 7.G. A map f : X ! Y is ontinuous, i ab f : X ! f (X ) is ontinuous.
7.H. Any onstant map (i.e., a map with image onsisting of a single point) is ontinuous.
Reformulations of De nition Prove that a mapping f : X ! Y is ontinuous, i Cl f 1 (A) f 1(Cl A) for any A Y . 7.7. Formulate and prove similar riteria of ontinuity in terms of Int f 1 (A) and f 1 (Int A). Do the same for Cl f (A) and f (Cl A). 7.8. Let be a base for topology in Y . Prove that a map f : X ! Y is
ontinuous, i f 1(U ) is open for any U 2 .
7.6.
More Examples Is the mapping f : [0; 2℄ ! [0; 2℄ de ned by formula ( x; if x 2 [0; 1); f (x) = 3 x; if x 2 [1; 2℄
ontinuous (with respe t to the topology indu ed from the real line)? 7.10. Is the map f of segment [0; 2℄ (with the topology indu ed by the topology of the real line) into the arrow (see Se tion 1) de ned by formula ( x; if x 2 [0; 1℄; f (x) = x + 1; if x 2 (1; 2℄
ontinuous? 7.11. Give an expli it hara terization of ontinuous mappings of RT1 (see Se tion 1) to R. 7.9.
7. CONTINUOUS MAPS
7.12.
Whi h maps RT1
! RT
1
27
are ontinuous?
7.13. Give an expli it hara terization of ontinuous mappings of the arrow to itself.
Let f be a mapping of the set Z + of nonnegative numbers onto R de ned by formula
7.14.
(
1
; f (x) = x 0;
if x 6= 0; if x = 0:
Let g : Z + ! f (Z +) be its submap. Indu e topology on Z + and f (Z +) from R. Are f and the map g 1 , inverse to g, ontinuous?
Behavior of Dense Sets Prove that the image of an everywhere dense set under a surje tive
ontinuous map is everywhere dense.
7.15.
7.16. Is it true that the image of nowhere dense set under a ontinuous map is nowhere dense. 7.17*. Does there exist a nowhere dense set A of [0; 1℄ (with the topology indu ed out of the real line) and a ontinuous map f : [0; 1℄ ! [0; 1℄ su h that f (A) = [0; 1℄?
Lo al Continuity A map f of a topologi al spa e X to a topologi al spa e Y is said to be
ontinuous at a point a 2 X if for every neighborhood U of f (a) there exists a neighborhood V of a su h that f (V ) U .
7.I.
A map f : X of X .
!Y
is ontinuous, i it is ontinuous at ea h point
Let X , Y be metri spa es, and a 2 X . A map f : X ! Y is
ontinuous at a, i for every ball with enter at f (a) there exists a ball with enter at a whose image is ontained in the rst ball.
7.J.
Let X , Y be metri spa es, and a 2 X . A mapping f : X ! Y is
ontinuous at the point a, i for every " > 0 there exists Æ > 0 su h that for every point x 2 X inequality (x; a) < Æ implies f (x); f (a) < ".
7.K.
Theorem 7.K means that ontinuity introdu ed above oin ides with the one that is usually studied in Cal ulus.
7. CONTINUOUS MAPS
28
Properties of Continuous Fun tions Let f; g : X ! R be ontinuous. Prove that the mappings X ! R de ned by formulas x 7!f (x) + g(x); x 7!f (x)g(x); x 7!f (x) g(x); x 7! f (x) ; x 7! maxff (x); g(x)g; x 7! minff (x); g(x)g are ontinuous. 7.19. Prove that if 0 2 = g(X ) then a mapping X ! R de ned by formula f (x) x 7! g(x) is ontinuous. 7.20. Find a sequen e of ontinuous fun tions fi : R ! R, (i 2 N ) su h that the formula x 7! supf fi (x) j i 2 N g de nes a fun tion R ! R whi h is not ontinuous. 7.21. Let X be any topologi al spa e. Prove that a fun tion f : X ! Rn : x 7! (f1 (x); : : : ; fn (x)) is ontinuous, i all the fun tions fi : X ! R with i = 1; : : : ; n are ontinuous. 7.18.
(14) (15) (16) (17) (18) (19)
Real p q-matri es omprise a spa e Mat(p q; R), whi h di ers from Rpq only in the way of numeration of its natural oordinates (they are numerated by pairs of indi es). 7.22. Let f : X ! Mat(p q; R) and g : X ! Mat(q r; R) be ontinuous maps. Prove that then X ! Mat(p r; R) : x 7! g(x)f (x) is a ontinuous map. Re all that GL(n; R) is the subspa e of Mat(n n; R) onsisting of all the invertible matri es. 7.23. Let f : X ! GL(n; R) be a ontinuous map. Prove that X ! GL(n; R) : x 7! (f (x)) 1 is ontinuous.
Spe ial About Metri Case
7.L.
For every subset A of a metri spa e X the fun tion de ned by formula x 7! (x; A) (see Se tion 3) is ontinuous. Prove that a topology of a metri spa e is the oarsest topology, with respe t to whi h for every A X the fun tion X ! R de ned by formula x 7! (x; A) is ontinuous.
7.24.
7. CONTINUOUS MAPS
29
A mapping f of a metri spa e X into a metri spa e Y is alled an isometri embedding if f (a); f (b) = (a; b) for every a; b 2 X . A bije tion whi h is an isometri embedding is alled an isometry.
7.M. Every isometri embedding is inje tive. 7.N. Every isometri embedding is ontinuous.
A mapping f : X ! X of a metri spa e X is alled ontra tive if there exists 2 (0; 1) su h that f (a); f (b) (a; b) for every a, b 2 X . 7.25.
Prove that every ontra tive mapping is ontinuous.
Let X , Y be metri spa es. A mapping f : X ! Y is said to be Holder if there exist C > 0 and > 0 su h that f (a); f (b) C(a; b) for every a, b 2 X . 7.26.
Prove that every Holder mapping is ontinuous.
Fun tions on Cantor Set and Square-Filling Curves Re all that Cantor set K is thePset of real numbers whi h an be presented ak as sums of series of the form 1 k=1 3k with ak = 0 or 2. 7:A. Let 1 be a map K ! I de ned by 1 a 1 a X X k k ! 7 k k+1 : 3 2 k=1 k=1 Prove that 1 : K ! I is a ontinuous surje tion. Draw the graph of '. 7:B. Prove that the fun tion K ! K de ned by 1 a 1 a X X k 2k ! 7 k k 3 3 k=1 k=1
is ontinuous.
Denote by K 2 the set f(x; y) 2 R2 : x 2 K; y 2 K g. 7:C. Prove that the map 2 : K ! K 2 de ned by ! 1 1 a 1 a X X 2k 1 X a2k k 7! ; 3k 3k k=1 3k k=1 k=1
is a ontinuous surje tion. 7:D. Prove that the map 3 : K ! I 2 de ned as the omposition of
2 : K ! K 2 and K 2 ! I 2 : (x; y) 7! ( 1 (x); 1 (y)) is a ontinuous surje tion. 7:E. Prove that the map 3 : K ! I 2 is a restri tion of a ontinuous map. (Cf. 1:A:2.)
7. CONTINUOUS MAPS
30
The latter map is a ontinuous surje tion I ! I 2 . Thus, this is a urve lling the square. A urve with this property was rst onstru ted by G. Peano in 1890. Though the onstru tion sket hed above is based on the same ideas as the original Peano's onstru tion, they are slightly di erent. Sin e then a lot of other similar examples have been found. You may nd a ni e survey of them in a book by Hans Sagan, Spa e-Filling Curves, Springer-Verlag 1994. Here is a sket h of Hilbert's onstru tion. 7:F. Prove that there exists a sequen e of polygonal maps fk : I
! I2
su h that (a) fk onne ts all enters of the squares forming the obvious subdivision of I 2 into 4k equalpsquares with side 1=2k ; (b) dist(fk (x); fk 1 (x)) 2=2k+1 for any x 2 I (here dist means the metri indu ed on I 2 from the standard Eu lidean metri of R2 ). 7:G. Prove that any sequen e of paths fk : I ! I 2 satisfying the onditions of 7:F onverges to a map f : I ! I 2 (i.e. for any x 2 I there exists a limit f (x) = limk!1 fk (x)) and this map is ontinuous and its
image is dense in I 2 .
7:H. 7 Prove that any ontinuous map I
surje tive.
! I 2 with dense image is
7:I. Generalize 7:C { 7:E 7:F { 7:H to obtain a ontinuous surje tion of
I onto I n .
Sets De ned by Systems of Equations and Inequalities
7.O. Let fi (i = 1; : : : ; n) be ontinuous mappings X ! R . Then the subset of X onsisting of solutions of the system of equations f1 (x) = 0; : : : ; fn (x) = 0 is losed.
7.P. Let fi (i = 1; : : : ; n) be ontinuous mappings X ! R. Then the subset of X onsisting of solutions of the system of inequalities f1 (x) 0; : : : ; fn (x) 0
is losed, while the set onsisting of solutions of the system of inequalities
f1 (x) > 0; : : : ; fn (x) > 0 is open. 7.27.
one.
7 Although
Where in 7.O and 7.P a nite system an be repla ed by an in nite
this problem an be solved using theorems well-known from Cal ulus, we have to mention that it would be more appropriate after Se tion 13. Cf. Problems 13.O, 13.T, 13.K.
7. CONTINUOUS MAPS
31
Set-Theoreti Digression. Covers A olle tion of subsets of a set X is alled a overS or a overing of X if X is a union of sets of belonging to , i.e., X = A2 A. In this ase elements of are said to over X . There is also a more general meaning of these words. A olle tion of subsets of a set Y is alled a over or a overing of a set X S Y if X is
ontained in the union of the sets belonging to , i.e., X A2 A. In this ase, sets belonging to are also said to over X .
Fundamental Covers Consider a over of a topologi al spa e X . Ea h element of inherits from X a topologi al stru ture. When are these stru tures suÆ ient for re overing the topology of X ? In parti ular, under what onditions on does ontinuity of a map f : X ! Y follow from ontinuity of its restri tions to elements of . To answer these questions, solve the problems 7.28{7.29 and 7.Q{7.V. Is this true for the following overings: (a) X = [0; 2℄, = f[0; 1℄; (1; 2℄g; (b) X = [0; 2℄, = f[0; 1℄; [1; 2℄g; ( ) X = R, = fQ ; R r Q g; (d) X = R, is a set of all one-point subsets of R? 7.29. A over of a topologi al spa e onsisting of one-point subsets has the property des ribed above, i the spa e is dis rete. 7.28.
A over of a spa e X is said to be fundamental if a set U X is open, i for every A 2 the set U \ A is open in A. 7.Q. A overing of a spa e X is fundamental, i a set U X is open provided U \ A is open in A for every A 2 . 7.R. A overing of a spa e X is fundamental, i a set F X is losed provided F \ A is losed A for every A 2 . A over of a topologi al spa e is said to be open if it onsists of open sets, and losed if it onsists of losed sets. A over of a topologi al spa e is said to be lo ally nite if every point of the spa e has a neighborhood interse ting only a nite number of elements of the over.
7.S. Every open over is fundamental. 7.T. Every nite losed over is fundamental. 7.U. Every lo ally nite losed over is fundamental.
8. HOMEOMORPHISMS
32
7.V.
Let be a fundamental over of a topologi al spa e X . If the restri tion of a mapping f : X ! Y to ea h element of is ontinuous then f is ontinuous. A over 0 is said to be a re nement of a over if every element of 0 is
ontained in some element of . 0 is a re nement of a over , and 0 is funda7.30. Prove that if a over mental then is also fundamental. 7.31. Let be a fundamental over of a topologi al spa e X , and be a
over of X su h that A = f U \ A j U 2 g is a fundamental over for subspa e A X for every A 2 . Prove that is a fundamental over. 7.32. Prove that the property of being fundamental is lo al, i.e., if every point of a spa e X has a neighborhood V su h that V = f U \ V j U 2 g is fundamental, then is fundamental.
8. Homeomorphisms De nition and Main Properties of Homeomorphisms An invertible mapping is alled a homeomorphism if both this mapping and its inverse are ontinuous. 8.A. Find an example of a ontinuous bije tion, whi h is not a homeomorphism. 8.B. Find a ontinuous bije tion [0; 1) ! S 1, whi h is not a homeomorphism. 8.C. The identity map of a topologi al spa e is a homeomorphism.
8.D. A omposition of homeomorphisms is a homeomorphism. 8.E. The inverse of a homeomorphism is a homeomorphism.
Homeomorphi Spa es A topologi al spa e X is said to be homeomorphi to spa e Y if there exists a homeomorphism X ! Y . 8.F. Being homeomorphi is an equivalen e relation. (Cf. 8.C{8.E.)
Role of Homeomorphisms
8.G. Let f : X ! Y be a homeomorphism. Then U X is open (in X ), i f (U ) is open (in Y ). 8.H. f : X ! Y is a homeomorphism, i f is a bije tion and de nes a bije tion between the topologi al stru tures of X and Y .
8. HOMEOMORPHISMS
33
8.I. Let f : X ! Y be a homeomorphism. Then for every A X
A is losed in X , i f (A) is losed in Y ; f (Cl A) = Cl f (A); f (Int A) = Int f (A); f (Fr A) = Fr f (A); A is a neighborhood of a point x 2 X , i f (A) is a neighborhood of the point f (x); (f) et .
(a) (b) ( ) (d) (e)
Therefore from the topologi al point of view homeomorphi spa es are
ompletely identi al: a homeomorphism X ! Y establishes one-to-one
orresponden e between all phenomena in X and Y whi h an be expressed in terms of topologi al stru tures. This phenomenon was used as a basis for a de nition of the subje t of topology in the rst stages of its development, when the notion of topologi al spa e had not been developed yet. Then mathemati ians studied only subspa es of Eu lidean spa es, their ontinuous mappings and homeomorphisms. Felix Klein in his famous Erlangen Program,8 where he lassi ed various geometries that had emerged up to that time, like Eu lidean, Loba hevsky, aÆne, and proje tive geometries, de ned topology as a part of geometry whi h deals with the properties preserved by homeomorphisms.
More Examples of Homeomorphisms
8.J. Let f : X ! Y be a homeomorphism. Prove that for every A X the redu tion ab(f ) : A ! f (A) is also a homeomorphism. 8.K. Prove that every isometry (see Se tion 7) is a homeomorphism. 8.L. Prove that every nondegenerate aÆne transformation of R n is a homeomorphism.
Prove that inversion Rx x 7! 2 : Rn r f0g ! Rn r f0g jxj is a homeomorphism. 8.2. Let H = f z 2 C j Imz > 0 g be the upper half-plane. Prove that az + b mapping f : H ! H de ned by f (z ) = , where a; b; ; d 2 R, is a
z +d a b homeomorphism if d > 0. 8.1.
8 In fa t it was not assumed to be a program in the sense of being planned, although it
be ame a kind of program. It was a sort of dissertation presented by Klein for getting the position as a professor at Erlangen University.
8. HOMEOMORPHISMS
34
8.3. Prove that a bije tion R ! R is a homeomorphism, i it is a monotone fun tion. 8.4. Prove that every bije tion of an indis rete spa e onto itself is a homeomorphism. Prove that the same holds true for a dis rete spa e and RT1 . 8.5. Find all homeomorphisms of the spa e 4pT (see Se tion 1) to itself. 8.6. Prove that every ontinuous bije tion of the arrow onto itself is a homeomorphism. 8.7. Find two homeomorphi spa es X and Y and a ontinuous bije tion X ! Y , whi h is not a homeomorphism. 2 8.8. Is 2 : K ! K onsidered in Problem 7:C a homeomorphism? Re all that K is the Cantor set, K 2 = f(x; y) 2 R2 : x 2 K; y 2 K g and 2 is de ned by ! 1 1 a 1 a X X 2 k 1 X a2 k k k 7! k ; k k=1 3 k=1 3 k=1 3
Examples of Homeomorphi Spa es Below the homeomorphism relation is denoted by =. It is not a ommonly a
epted notation. In other textbooks any sign lose to, but distin t from =, e. g. , ', , is used. 8.M. [0; 1℄ = [a; b℄ for any a < b. 8.N. [0; 1) = [a; b) = (0; 1℄ = (a; b℄ for any a < b. 8.O. (0; 1) = (a; b) for any a < b. 8.P. ( 1; 1) = R. 8.Q. [0; 1) = (0; +1). = [0; +1) and (0; 1) 8.R. S 1 r f(0; 1)g = R1 . 8.S. S n r fpointg = Rn . 8.9.
(a) (b) ( ) (d) (e) (f) (g) (h) (i)
Prove that the following plane gures are homeomorphi : the whole plane R2 ; open square f (x; y) 2 R2 j x; y 2 (0; 1) g; open strip f (x; y) 2 R2 j x 2 (0; 1) g; half-plane f (x; y) 2 R2 j y > 0 g; open half-strip f (x; y) 2 R2 j x > 0; y 2 (0; 1) g; open disk f (x; y) 2 R2 j x2 + y2 < 1 g; open re tangle f (x; y) 2 R2 j a < x < b; < y < d g; open quadrant f (x; y) 2 R2 j x; y > 0 g; f (x; y) 2 R2 j y2 + jxj > x g, i.e., plane ut along the ray f y = 0; x 0 g.
8.T. Prove that
(a) losed disk D2 is homeomorphi to square I 2 = f (x; y ) 2 R 2 j x; y 2 [0; 1℄ g;
8. HOMEOMORPHISMS
35
(b) open dis Int D2 = f (x; y ) 2 R 2 j x2 + y 2 < 1 g is homeomorphi to open square Int I 2 = f (x; y ) 2 R 2 j x; y 2 (0; 1) g; ( ) ir le S 1 is homeomorphi to the boundary of square I 2 = I 2 r Int I 2 . 8.U. Prove that (a) every bounded losed onvex set in the plane with nonempty interior is homeomorphi to D2 ; (b) every bounded open onvex nonempty set in the plane is homeomorphi to the plane; ( ) boundary of every bounded onvex set in the plane with nonempty interior is homeomorphi to S 1 . In whi h of the situations onsidered in 8.U an the assumption that the set is bounded be omitted? 8.11. Classify up to homeomorphism all losed onvex sets in the plane. (Make a list without repeats; prove that every su h set is homeomorphi to one in the list; postpone a proof of nonexisten e of homeomorphisms till Se tion 9.) 8.10.
Generalize the previous three problems to the ase of sets in Rn with arbitrary n. 8.12*.
The latter four problems show that angles are not essential in topology, i.e., for a line or boundary of a domain the property of having angles is not preserved by homeomorphism. And now two more problems on this. Prove that every losed simple (i.e., without self-interse tions) polygon in R2 (and in Rn with n > 2) is homeomorphi to the ir le S 1 . 8.14. Prove that every non- losed simple nite unit polyline in R2 (and in Rn with n > 2) is homeomorphi to the segment [0; 1℄.
8.13.
8.15.
Prove that R2 r f jxj; jyj > 1 g = I 2 r f(1; 1); (1; 1)g.
Prove that the following plane gures are homeomorphi to ea h other: f (x; y) j 0 x; y < 1 g; f (x; y) j 0 < x < 1; 0 y < 1 g; f (x; y) j 0 x 1; 0 y < 1 g; f (x; y) j x; y 0 g; f (x; y) j x 0 g; f (x; y) j x y 0 g; f (x; y) j x2 + y2 1; x 6= 1 g.
8.16.
(a) (b) ( ) (d) (e) (f) (g)
Prove that the following plane gures are homeomorphi to ea h other: pun tured plane R2 r f(0; 0)g; pun tured dis f (x; y) j 0 < x2 + y2 < 1 g; annulus f (x; y) j a < x2 + y2 < b g where 0 < a < b; plane without dis f (x; y) j x2 + y2 > 1 g; plane without square f (x; y) j 0 x; y 1 g; plane without segment R2 r [0; 1℄.
8.17.
(a) (b) ( ) (d) (e) (f)
8. HOMEOMORPHISMS
36
Let X R2 be an union of several segments with a ommon end point. Prove that the omplement R2 r X is homeomorphi to the pun tured plane. 8.19. Let X R2 simple non- losed nite polyline. Prove that its omplement R2 r X is homeomorphi to the pun tured plane. 2 8.20. Let D1 ; : : : ; Dn R be pairwise disjoint losed dis s. The omplement of the union of its interior is said to be plane with n holes. Prove that any two planes with n holes are homeomorphi , i.e., dislo ation of dis s D1 , : : : , Dn does not a e t on the topologi al type of R2 r [ni=1 Int Di . 8.21. Prove that for ontinuous fun tions f; g : R ! R su h that f < g , the spa e between their graphs f (x; y) 2 R2 j f (x) y g(x) g is homeomorphi to a losed strip f (x; y) j y 2 [0; 1℄ g. 8.22. Prove that a mug (with handle) is homeomorphi to a doughnut. 8.23. Arrange the following items to homeomorphism lasses: a up, a sau er, a glass, a spoon, a fork, a knife, a plate, a oin, a nail, a s rew, a bolt, a nut, a wedding ring, a drill, a ower pot (with hole in the bottom), a key. 8.24. In a spheri al shell (the spa e between two on entri spheres) one drilled out a ylindri al hole onne ting the boundary spheres. Prove that the rest is homeomorphi to D3 . 8.25. In a spheri al shell one made a hole onne ting the boundary spheres and having the shape of a knotted tube (see Figure 1.). Prove that the rest of the shell is homeomorphi to D3 .
8.18.
Figure 1 8.26. Prove that surfa es shown in Figure 2 are homeomorphi (they are
alled handles ).
Prove that surfa es shown in the Figure 3 are homeomorphi . (They are homeomorphi to Klein bottle with two holes. More details about this is given in Se tion 18.)
8.27.
8.28*.
Prove that R3 r S 1 = R3 r R1 [ f(1; 1; 1)g .
8. HOMEOMORPHISMS
37
Figure 2
Figure 3 8.29.
Prove that subset of the sphere S n de ned in standard oordinates in < x2k+1 + + x2n is homeomorphi to
Rn+1 by inequality x21 + x22 + + x2k Rn r Rn k .
Examples of Nonhomeomorphi Spa es
8.V. Spa es onsisting of di erent number of points are not homeomorphi .
8.W. A dis rete spa e and an indis rete spa e (whi h have more than one point) are not homeomorphi . 8.30. Prove that the spa es Z, Q (with topology indu ed from R), R, RT
1
and the arrow are pairwise non-homeomorphi . 8.31. Find two non-homeomorphi spa es X and Y for whi h there exist
ontinuous bije tions X ! Y and Y ! X .
Homeomorphism Problem and Topologi al Properties One of the lassi problems of topology is the homeomorphism problem: to nd out whether two given topologi al spa es are homeomorphi . In ea h spe ial ase the hara ter of solution depends mainly on the answer. To prove that spa es are homeomorphi , it is enough to present a homeomorphism between them. Essentially this is what one usually does in this ase. To prove that spa es are not homeomorphi , it does not suÆ e to onsider any spe ial mapping, and usually it is impossible to review all the mappings. Therefore for proving non-existen e of a homeomorphism one uses indire t arguments. In parti ular, one nds a property or a
hara teristi shared by homeomorphi spa es and su h that one of the spa es has it, while the other does not. Properties and hara teristi s whi h are shared by homeomorphi spa es are alled topologi al properties and invariants. Obvious examples of them are the ardinality (i.e.,
8. HOMEOMORPHISMS
38
the number of elements) of the set of points and the set of open sets ( f. Problems 8.29 and 8.V). Less obvious examples are the main obje t of the next hapter.
Information (Without Proof) Eu lidean spa es of di erent dimensions are not homeomorphi . The balls Dp, Dq with p 6= q are not homeomorphi . The spheres S p , S q with p 6= q are not homeomorphi . Eu lidean spa es are homeomorphi neither to balls, nor to spheres (of any dimension). Letters A and B are not homeomorphi (if the lines are absolutely thin!). Pun tured plane R 2 rfpointg is not homeomorphi to the plane with hole R 2 rf x2 + y2 < 1 g. These statements are of di erent degrees of diÆ ulty. Some of them will be onsidered in the next se tion. However some of them an not be proven by te hniques of this ourse. (See, e.g., D. B. Fu hs, V. A. Rokhlin. Beginner's ourse in topology: Geometri hapters. Berlin; New York: Springer-Verlag, 1984.)
Embeddings Continuous mapping f : X ! Y is alled a (topologi al ) embedding if the submapping ab(f ) : X ! f (X ) is a homeomorphism.
8.X. The in lusion of a subspa e into a spa e is an embedding. 8.Y. Composition of embeddings is an embedding. 8.Z. Give an example of ontinuous inje tive map, whi h is not a topologi al embedding. (Find su h an example above and reate a new one.)
Find topologi al spa es X and Y su h that X an be embedded into Y , Y an be embedded into X , but X 6 Y. = 8.33. Prove that Q annot be embedded into Z. 8.34. Can a dis rete spa e be embedded into an indis rete spa e? How about vi e versa? 8.35. Prove that spa es R, RT1 , and the arrow annot be embedded into ea h other. 8.36 Corollary of Inverse Fun tion Theorem. Dedu e from the Inverse Fun tion Theorem (see, e.g., any ourse of advan ed al ulus) the following statement: 8.32.
fi ) does For any di erentiable fun tion f : Rn ! Rn whose Ja obian det( x j not vanish at the origin 0 2 Rn there exists a neighborhood U of the origin su h that f jU : U ! Rn is an embedding and f (U ) is open.
8. HOMEOMORPHISMS
39
Embeddings f1 ; f2 : X ! Y are said to be equivalent if there exist homeomorphisms hX : X ! X and hY : Y ! Y su h that f2 Æ hX = hY Æ f1 (the latter equality mayb stated as follows: the diagram f1
X
f2
? ? hX y
is ommutative).
! Y?
X
? yhY
!Y
An embedding of the ir le S 1 into R3 is alled a knot. 1 3 1 1 8.37. Prove that knots f1 ; f2 : S ! R with f1 (S ) = f2 (S ) are equivalent. 8.38.
Prove that knots
are equivalent.
Information There are nonequivalent knots. For instan e,
and
.
CHAPTER 2
Topologi al Properties 9. Conne tedness De nitions of Conne tedness and First Examples A topologi al spa e X is said to be onne ted if it has only two subsets whi h are both open and losed: ? and the entire X . A partition of a set is a over of this set with pairwise disjoint sets. To partition a set means to onstru t su h a over.
9.A.
A topologi al spa e is onne ted, i it annot be partitioned into two nonempty open sets, i it annot be partitioned into two nonempty
losed sets.
Is an indis rete spa e onne ted? The same for the arrow and RT1 . 9.2. Des ribe expli itly all onne ted dis rete spa es. 9.3. Is the set Q of rational numbers (with the topology indu ed from R)
onne ted? The same about the set of irrational numbers. 9.4. Let 1 , 2 be topologi al stru tures in a set X , and 2 be ner than
1 (i.e., 1 2 ). If (X; 1 ) is onne ted, is (X; 2 ) onne ted? If (X; 2 ) is onne ted, is (X; 1 ) onne ted? 9.1.
Conne ted Sets When one says that a set is onne ted, it means that this set lies in some topologi al spa e (whi h should be lear from the ontext), and, with the indu ed topology, is a onne ted topologi al spa e. 9.5. Give a de nition of dis onne ted subset without relying on the indu ed topology. 9.6. Is the set f0; 1g onne ted in R, in the arrow, in RT1 ? 9.7. Des ribe expli itly all onne ted subsets of the arrow, of RT1 . 9.8. Show that the set [0; 1℄ [ (2; 3℄ is dis onne ted in R. 9.9. Prove that every non- onvex subset of the real line is dis onne ted. 9.10. Let A be a subset of a topologi al spa e X . Prove that A is dis onne ted, i there exist non-empty sets B and C su h that A = B [ C , B \ ClX C = ?, and C \ ClX B = ?. 40
9. CONNECTEDNESS
41
9.11. Find a topologi al spa e X and dis onne ted subset A X su h that for any disjoint open sets U and V , whi h form a over of X , either U A, or V A. n 9.12. Prove that for every dis onne ted set A in R there exist disjoint open sets U and V su h that A U [ V , U \ A 6= ?, and V \ A 6= ?.
Compare 9.10{9.12 with 9.5.
Properties of Conne ted Sets
9.B.
The losure of a onne ted set is onne ted. 9.13. Prove that if a set A is onne ted and A
onne ted.
B
Cl A, then B is
Let fA g2 be a family of onne ted subsetsSof a spa e X . Assume that any two sets of this family interse t. Then 2 A is onne ted. (In other words: the union of pairwise interse ting onne ted sets is
onne ted.)
9.C.
9.D. Let fAk gk2Z be a family S of onne ted sets su h that Ak \ Ak+1 6= ? for any k 2 Z. Prove that k2Z Ak is onne ted.
9.14. Let A, B be onne ted sets, and A \ Cl B 6= ?. Prove that A [ B is
onne ted. 9.15. Let A be a onne ted subset of a onne ted spa e X , and B X r A be an open and losed set in the topology of the subspa e X r A of the spa e X . Prove that A [ B is onne ted. 9.16. Does onne tedness of A [ B and A \ B imply onne tedness of A and B? 9.17. Prove that if A and B are either both losed or both open sets, and their union and interse tion are onne ted then A and B are onne ted, too. 9.18. Let A1 A2 be an in nite des ending sequen e of onne ted T spa es. Is 1 k=1 Ak a onne ted set?
Conne ted Components A onne ted omponent of a spa e X is its maximal onne ted subset, that is a onne ted subset, whi h is not ontained in any other (stri tly) larger onne ted subset of X .
9.E.
Every point belongs to some onne ted omponent. Moreover, this
omponent is unique. It is the union of all onne ted sets ontaining this point.
9.F. Conne ted omponents are losed. 9.G. Two onne ted omponents either are disjoint or oin ide.
9. CONNECTEDNESS
42
A onne ted omponent of a spa e X is alled just a omponent of X . Theorems 9.E and 9.G mean that onne ted omponents omprise a partition of the whole spa e. The next theorem des ribes the orresponding equivalen e relation.
9.H.
Prove that two points are in the same omponent, i they belong to the same onne ted set. 9.19. Let x and y belong to the same omponent. Prove that any set, whi h is losed and open, either ontains both x and y or does not ontain either of them ( f. 9.29). 9.20. Let a spa e X has a group stru ture, and the multipli ation by an element of the group is a ontinuous map. Prove that the omponent of unity is a normal subgroup.
Totally Dis onne ted Spa es A topologi al spa e is alled totally dis onne ted if ea h of its omponents
onsists of a single point. 9.I Obvious Example. Any dis rete spa e is totally dis onne ted. 9.J. The spa e Q (with the topology indu ed from R ) is totally dis onne ted. Note that Q is not dis rete. Give an example of an un ountable losed totally dis onne ted subset of the line. 9.22. Prove that Cantor set (see 1:A) is totally dis onne ted.
9.21.
Frontier and Conne tedness 9.23. Prove that if A is a proper nonempty subset of a onne ted topologi al spa e then Fr A 6= ?. 9.24. Let F be a onne ted subset of X . Prove that if A X , F \ A, and F \ (X r A) 6= ? then F \ Fr A 6= ?. 9.25. Let A be a subset of onne ted topologi al spa e. Prove that if Fr A is a onne ted set then Cl A is also onne ted.
Behavior Under Continuous Maps A ontinuous image of a spa e is its image under a ontinuous mapping.
9.K.
A ontinuous image of a onne ted spa e is onne ted. (In other words if f : X ! Y is a ontinuous map, and X is onne ted then f (X ) is also onne ted.)
9. CONNECTEDNESS
43
9.L Corollary. Conne tedness is a topologi al property. The number
of onne ted omponents is a topologi al invariant.
9.M.
A spa e X is not onne ted, i there is a ontinuous surje tion
X ! S 0.
Conne tedness on Line
9.N.
The segment I = [0; 1℄ is onne ted.
There are several ways to prove 9.N. One is suggested by 9.M, but refers to a famous Intermediate Value Theorem from al ulus, see 9.S. Basi ally the same proof as a ombination of 9.M with a traditional proof of Intermidiate Value Theorem is sket hed in the following two problems. Cf. also 9.26 below. 9.N.1. Let U , V be subsets of I with V = U r V . Let a 2 U , b 2 V and a > b. Prove that there exists a des ending sequen e an with a1 = a, an 2 U and an as ending sequen e bn with b1 = b, bn 2 V su h that both an and bn have the same limit . 9.N.2. If under assumptions of 9.N.1 U and V are open, then in whi h of them an be ? 9.26. Prove that every open subset of the real line is a union of disjoint open intervals (do not use 9.N). Dedu e 9.N from this.
9.O. Prove that the set of onne ted omponents of an open subset of R is ountable. 9.P. Prove that R1 is onne ted. 9.Q. Des ribe expli itly all onne ted subsets of the line. 9.R. Prove that every onvex set in Rn is onne ted. 9.27.
Consider the union of spiral
r = exp
1 ; with ' 0 1 + '2
(r; ' are the polar oordinates) and ir le S 1 . Is this set onne ted? Would the answer hange, if the entire ir le was repla ed by some its subset? (Cf. 9.13) Consider the subset of the plane R2 onsisting of points with both
oordinates rational or both oordinates irrational. Is it onne ted?
9.28.
Find a spa e and two points belonging to its di erent omponents su h that ea h simultaneously open and losed set ontains either both of the points, or neither of them ( f. 9.19).
9.29.
9. CONNECTEDNESS
44
Intermediate Value Theorem and Its Genralizations The following theorem is usually in luded in Cal ulus. You an easily dedu e it from the matterial of this se tion. In fa t, in a sense it is equivalent to onne tedness of interval. 9.S Intermediate Value Theorem. A ontinuous fun tion f : [a; b℄ ! R takes every value between f (a) and f (b). Many problems whi h an be solved using Intermediate Value Theorem an be found in Cal ulus textbooks. Here are few of them. 9.30. Prove that any polynomial of odd degree in one variable with real
oeÆ ients has at least one real root.
9.T Generalization. Let X be a onne ted spa e and f : X ! R a
ontinuous fun tion. Then f (X ) is a onvex subset of R . Dividing Pan akes
Any irregularly shaped pan ake an be ut in half by one stroke of the knife made in any pres ribed dire tion. In other words, if A is a bounded open set in the plane and l is a line in the plane, then there exists a line L parallel to l whi h divides A in half by area. 9.32. If, under the onditions of 9.31, A is onne ted then L is unique. 9.33. Suppose two irregularly shaped pan akes lie on the same platter; show that it is possible to ut both exa tly in half by one stroke of the knife. In other words: if A and B are two bounded regions in the plane, then there exists a line in the plane whi h divides ea h region in half by area. 9.34 Dividing Pan ake. Prove that a plane pan ake of any shape an be divided to four pie es of equal area by two straight uts orthogonal to ea h other. In other words, if A is a bounded onne ted open set in the plane, then there are two perpendi ular lines whi h divide A into four parts having equal areas. 9.35. Riddle. What if the knife is not makes uts of a shape di erent from straight line? For whi h shapes of the blade you an formulate and solve problems similar to 9.31 { 9.34? 9.36. Riddle. Formulate and solve ounter-parts of Problems 9.31 { 9.34 for regions in the three-dimensional spa e. Can you in rease the number of regions in the ounter-part of 9.31 and 9.33? 9.37. Riddle. What about pan akes in Rn ? 9.31.
Indu tion on Conne tedness A fun tion is said to be lo ally onstant if ea h point of its sour e spa e has a neighborhood su h that the restri tion of the fun tion to this neighborhood is onstant.
9. CONNECTEDNESS
45
9.U. A lo ally onstant fun tion on a onne ted set is onstant.
How are 9.24 and 9.U related? 9.39. Let G be a group equipped with a topology su h that for any g 2 G the map G ! G de ned by x 7! xgx 1 is ontinuous, and let G with this topology be onne ted. Prove that if the topology indu ed in a normal subgroup H of G is dis rete, then H is ontained in the enter of G (i.e., hg = gh for any h 2 H and g 2 G). 9.40 Indu tion on Conne tedness. Let E be a property of subsets of a topologi al spa e su h that the union of sets with nonempty pairwise interse tions inherits this property from the sets involved. Prove that if the spa e is onne ted and ea h its point has a neighborhood with property E , then the spa e has property E . 9.41. Prove 9.U and solve 9.39 using 9.40. 9.38. Riddle.
For more appli ations of indu tion on onne tedness see 10.R, 10.14, 10.16 and 10.18.
Appli ations to Homeomorphism Problem Conne tedness is a topologi al property, and the number of onne ted
omponents is a topologi al invariant (see Se tion 8). 9.V. [0; 2℄ and [0; 1℄ [ [2; 3℄ are not homeomorphi . Simple onstru tions, whi h assign homeomorphi spa es to homeomorphi ones (e.g. deleting one or several points), allow one to use onne tedness for proving that some onne ted spa es are not homeomorphi . 9.W. I , R 1 , S 1 and [0; 1) are pairwise nonhomeomorphi . 9.42. Prove that a ir le is not homeomorphi to any subspa e of R1 . Give a topologi al lassi ation of the letters: A, B, C, D, : : : , onsidered as subsets of the plane (the ar s omprising the letters are assumed to have zero thi kness). 9.44. Prove that square and segment are not homeomorphi .
9.43.
Re all that there exist ontinuous surje tions of the segment onto square and these maps are alled Peano urves, see Se tion 7. 9.X. R 1 and R n are not homeomorphi if n > 1. Information. R p and R q are not homeomorphi unless p = q . It follows, for instan e, from the Lebesgue-Brower Theorem on invarian e of dimension (see, e.g., W. Hurewi z and H. Wallman, Dimension Theory Prin eton, NJ, 1941). 9.45. The statement \Rp is not homeomorphi to Rq unless p = q " implies that S p is not homeomorphi to S q unless p = q.
10. PATH-CONNECTEDNESS
46
10. Path-Conne tedness Paths A path in a topologi al spa e X is a ontinuous mapping of the interval I = [0; 1℄ to X . The point s(0) is alled the initial point of a path s : I ! X , while s(1) is alled its nal point. One says that path s
onne ts s(0) with s(1). This terminology is inspired by an image of moving point: at the moment t 2 [0; 1℄ it is in s(t). To tell the truth, this is more than what is usually alled a path, sin e besides an information on traje tory of the point it ontains a omplete a
ount on the movement: the s hedule saying when the point goes through ea h point. A onstant map s : I ! X is alled a stationary path and denoted by ea where a = s(I ). For a path s the inverse path is the path de ned by t 7! s(1 t). It is denoted by s 1 . Although, stri tly speaking, this notation is already used (for the inverse mapping), the ambiguity of notations does not lead to onfusion: in the ontext involving paths, inverse mappings, as a rule, do not appear. Let u : I ! X , v : I ! X be paths su h that u(1) = v (0). Set (
uv (t) =
(20)
u(2t); v (2t 1);
if t 2 [0; 12 ℄ if t 2 [ 12 ; 1℄:
10.A. Prove that the map uv : I ! X de ned by (10) is ontinuous (i.e., it is a path). Cf. 7.T and 7.V.
Path uv is alled the produ t of paths u and v . Re all that it is de ned only if the nal point u(1) of u oin ides with the initial point v (0) of v .
Path-Conne ted Spa es A topologi al spa e is said to be path- onne ted or pathwise onne ted , if any two points an be onne ted in it by a path. 10.B. Prove that I is pathwise onne ted. 10.C. Prove that the Eu lidean spa e of any dimension is pathwise onne ted. 10.D. Prove that sphere of dimension n > 0 is path- onne ted. 10.E. Prove that the zero-dimensional sphere S 0 is not path- onne ted. Whi h of the following topologi al spa es are path- onne ted: (a) a dis rete spa e; (b) an indis rete spa e;
10.1.
10. PATH-CONNECTEDNESS
47
( ) the arrow; (d) RT1 ; (e) 4pT ?
Path-Conne ted Sets By a path- onne ted set or pathwise onne ted set one alls a subset of a topologi al spa e (whi h should be lear from the ontext) path onne ted as a spa e with the topology indu ed from the ambient spa e. Prove that a subset A of a topologi al spa e X is path- onne ted, i any two points in it an be onne ted by a path s : I ! X with s(I ) A.
10.2.
10.3.
Prove that a onvex subset of Eu lidean spa e is path- onne ted.
10.4. Prove that the set of plane onvex polygons with topology de ned by the Hausdor metri is path- onne ted.
Path- onne tedness is very similar to onne tedness. Further, in some important situations it is even equivalent to onne tedness. However, some properties of onne tedness do not arry over path- onne tedness (see 10.O, 10.P). For properties, whi h arry over, proofs are usually easier in the ase of path- onne tedness.
10.F.
The union of a family of pairwise interse ting path- onne ted sets is path- onne ted. Prove that if sets A and B are both losed or both open and their union and interse tion are path- onne ted, then A and B are also path- onne ted.
10.5.
Prove that interior and frontier of a path- onne ted set may not be path- onne ted and that onne tedness shares this property.
10.6.
Let A be a subset of Eu lidean spa e. Prove that if Fr A is onne ted then Cl A is also onne ted.
10.7.
Prove that the same holds true for a subset of an arbitrary path onne ted spa e.
10.8.
Path-Conne ted Components A path- onne ted omponent or pathwise onne ted omponent of a topologi al spa e X is a path- onne ted subset of X su h that no other path onne ted subset of X ontains it.
10.G. Every point belongs to a path- onne ted omponent. 10.H. Two path- onne ted omponents either oin ide or are disjoint. 10.I. Prove that two points belong to the same path- onne ted omponent, i they an be onne ted by a path.
10. PATH-CONNECTEDNESS
48
Unlike to the ase of onne tedness, path- onne ted omponents may be non- losed. (See 10.O, f. 10.N, 10.P.)
10.J.
A ontinuous image of a pathwise onne ted spa e is pathwise
onne ted. 10.9. Let s : I ! X be a path onne ting a point of a set A with a point of X r A. Prove that s(I ) \ Fr(A) 6= ?.
Path-Conne tedness Versus Conne tedness
10.K.
Any path- onne ted spa e is onne ted.
Put
A = (x; y ) 2 R
and X = A [ f(0; 0)g. 10.10.
2
1 : x > 0; y = sin x
Draw A.
10.L. Prove that A is path- onne ted and X is onne ted. 10.M. Prove that deleting any point from A makes A and X dis on-
ne ted (and hen e, not path- onne ted).
10.N. X is not path- onne ted. 10.O. Find an example of a path- onne ted set, whose losure is not path- onne ted.
10.P. Find an example of a path- onne ted omponent that is not
losed.
10.Q. If ea h point of a spa e has a path- onne ted neighborhood, then ea h path- onne ted omponent is open. 10.R. If ea h point of a spa e has a path- onne ted neighborhood, then the spa e is path- onne ted, i it is onne ted. 10.S. For an open subset of Eu lidean spa e onne tedness is equivalent to path- onne tedness. 10.11. For subsets of the real line path- onne tedness and onne tedness are equivalent.
Prove that for any " > 0 an "-neighborhood of a onne ted subset of Eu lidean spa e is path- onne ted.
10.12.
Prove that any neighborhood of a onne ted subset of Eu lidean spa e ontains a path- onne ted neighborhood of the same set.
10.13.
11. SEPARATION AXIOMS
49
Polygon-Conne tedness A subset A of Eu lidean spa e is said to be polygon- onne ted if any two points of A an be onne ted by a nite polygonal line ontained in A. 10.14. Prove that for open subsets of Eu lidean spa e onne tedness is equivalent to polygon- onne tedness.
Constru t a path- onne ted subset A of Eu lidean spa e su h that A
onsists of more than one point and no two distin t points an be onne ted with a polygon in A. 10.15.
Let X R2 be a ountable set. Prove that then R2 r X is polygon onne ted.
10.16.
Let X Rn be a union of a ountable olle tion of aÆne subspa es with dimensions not greater than n 2. Prove that then Rn r X is polygon onne ted. 10.17.
Let X C n be a union of a ountable olle tion of algebrai subsets (i.e., subsets de ned by systems of algebrai equations in the standard
oordinates of C n ) Prove that then C n r X is polygon- onne ted. 10.18.
Re all, that real n n-matri es omprise a spa e, whi h di ers from Rn only in the way of enumeration of its natural oordinates (they are numerated by pairs of indi es). The same relation holds between the set of omplex 2 2 n n-matrix and C n (homeomorphi to R2n ). 2
10.19. Find onne ted and path- onne ted omponents of the following subspa es of the spa e of real n n-matri es:
GL(n; R) = fA : det A 6= 0g; O(n; R) = fA : A (t A) = 1g; Symm(n; R) = fA : t A = Ag; Symm(n; R) \ GL(n; R); fA : A2 = 1g.
10.20. Find onne ted and path- onne ted omponents of the following subspa es of the spa e of omplex n n-matri es:
GL(n; C ) = fA : det A 6= 0g; U (n; C ) = fA : A (t A) = 1g; Herm(n; C ) = fA : t A = Ag; Herm(n; C ) \ GL(n; C ).
11. Separation Axioms The aim of this se tion is to onsider natural restri tions on topologi al stru ture making the stru ture loser to being metrizable.
11. SEPARATION AXIOMS
50
Hausdor Axiom A lot of separation axioms are known. We restri t ourselves to the most important four of them. They are numerated and denoted by T1 , T2 , T3 , and T4 respe tively. Let us start with the most important se ond axiom. Besides the notation T2 it has a name, the Hausdor axiom. A topologi al spa e satisfying it is alled a Hausdor spa e. This axiom is stated as follows: any two distin t points possess disjoint neighborhoods.
11.A.
Any metri spa e is Hausdor . Whi h of the following spa es are Hausdor : a dis rete spa e; an indis rete spa e; the arrow; RT1 ; 4pT ?
11.1.
(a) (b) ( ) (d) (e)
If the next problem holds you up even for a minute, we advise you to think over all de nitions and solve all simple problems. 11.B. Is the segment [0; 1℄ with the topology indu ed from R a Hausdor spa e? Do the points 0 and 1 possess disjoint neighborhoods? Whi h if any?
Limits of Sequen e Let fan g be a sequen e of points of a topologi al spa e X . A point b 2 X is alled its limit, if for any neighborhood U of b there exists a number N su h that an 2 U for any n > N . The sequen e is said to onverge or tend to b as n tends to in nity. 11.2. Explain the meaning of the statement \ b is not a limit of sequen e an " avoiding as mu h as you an negations (i.e., the words no, not, none, et ..)
11.C. In a Hausdor spa e any sequen e has at most one limit. 11.D. Prove that in the spa e R T1 ea h point is a limit of the sequen e fan = ng. Coin iden e Set and Fixed Point Set
Let f; g : X ! Y be maps. Then the set fx 2 X : f (x) = g(x)g is alled the
oin iden e set of f and g. 11.3. Prove that the oin iden e set for two ontinuous maps of an arbitrary topologi al spa e to a Hausdor spa e is losed. 11.4. Constru t an example proving that the Hausdor ondition in 11.3 is essential.
11. SEPARATION AXIOMS
51
A point x 2 X is alled a xed point of a map f : X ! X if f (x) = x. The set of all xed points of a map f is alled the xed point set of f . 11.5. Prove that the xed point set of a ontinuous map of a Hausdor spa e to itself is losed. 11.6. Constru t an example proving that the Hausdor ondition in 11.5 is essential. 11.7. Prove that if f; g : X ! Y are ontinuous maps, Y is Hausdor , A is everywhere dense in X , and f jA = gjA then f = g. 11.8. Riddle. How are problems 11.3, 11.5, and 11.7 related?
Hereditary Properties A topologi al property is alled hereditary if it is arried over from a spa e to its subspa es, i.e. if a spa e X possesses this property then any subspa e of X possesses it. Whi h of the following topologi al properties are hereditary: niteness of the set of points; niteness of the topologi al stru ture; in niteness of the set of points;
onne tedness; path- onne tedness?
11.9.
11.E.
The property of being Hausdor spa e is hereditary.
The First Separation Axiom A topologi al spa e is said to satisfy the rst separation axiom T1 if ea h of any two points of the spa e has a neighborhood whi h does not ontain the other point. 11.F. A topologi al spa e X satis es the rst separation axiom, i all one-point sets in X are losed, i all nite sets in X are losed. Prove that a spa e X satis es the rst separation axiom, i any point of X oin ides with the interse tion of all its neighborhoods. 11.11. Any Hausdor spa e satis es the rst separation axiom.
11.10.
11.G. In a Hausdor spa e any nite set is losed. 11.H. A metri spa e satis es the rst separation axiom.
Find an example showing that the rst separation axiom does not imply the Hausdor axiom.
11.12.
11.I. Show that R T1 meets the rst separation axiom, but is not a Haus-
dor spa e ( f. 11.12).
11.J.
The rst separation axiom is hereditary.
11. SEPARATION AXIOMS
52
11.13. Prove that if for any two distin t points a and b of a topologi al spa e X there exists a ontinuous map f of X to a spa e with the rst separation axiom su h that f (a) 6= f (b) then X possesses the rst separation axiom. 11.14. Prove that a ontinuous mapping of an indis rete spa e to a spa e satisfying axiom T1 is onstant. 11.15. Prove that in every set there exists a oarsest topologi al stru ture satisfying the rst separation axiom. Des ribe this stru ture.
The Third Separation Axiom A topologi al spa e X is said to satisfy the third separation axiom if any
losed set and a point of its omplement have disjoint neighborhoods, i.e., for any losed set F X and point b 2 X n F there exist open sets U; V X su h that U \ V = ?, F U , and b 2 V . A topologi al spa e is alled regular if it satis es the rst and third separation axioms. 11.K. A regular spa e is Hausdor spa e. 11.L. A spa e is regular, i it satis es the se ond and third separation axioms. 11.16. 11.17.
iom.
Find a Hausdor spa e whi h is not regular. Find a spa e satisfying the third, but not the se ond separation ax-
Prove that a spa e satis es the third separation axiom, i any neighborhood of any point ontains the losure of some neighborhood of the same point. 11.19. Prove that the third separation axiom is hereditary.
11.18.
11.M.
Any metri spa e is regular.
The Fourth Separation Axiom A topologi al spa e X is said to satisfy the fourth separation axiom if any two disjoint losed sets have disjoint neighborhoods, i.e., for any losed sets A; B X su h that A \ B = ? there exist open sets U; V X su h that U \ V = ?, A U , and B V . A topologi al spa e is alled normal if it satis es the rst and fourth separation axioms. 11.N. A normal spa e is regular (and hen e Hausdor ). 11.O. A spa e is normal, i it satis es the se ond and fourth separation axioms.
11. SEPARATION AXIOMS
11.20.
axiom.
53
Find a spa e whi h satis es the fourth, but not se ond separation
Prove that a spa e satis es the fourth separation axiom, i in any neighborhood of any losed set ontains the losure of some neighborhood of the same set. 11.22. Prove that any losed subspa e of a normal spa e is normal. 11.23. Find losed disjoint subsets A and B of some metri spa e su h that inf f(a; b) j a 2 A; b 2 B g = 0.
11.21.
11.P.
Any metri spa e is normal.
Let f : X ! Y be a ontinuous surje tion su h that the image of any
losed set is losed. Prove that if X is normal then Y is normal.
11.24.
Niemytski's Spa e Denote by H the open upper half-plane f(x; y) 2 R2 : y > 0g equipped with the topology indu ed by the Eu lidean metri . Denote by X the union of H and its boundary line L = f(x; y) 2 R2 : y = 0g, but equip it with the topology, whi h is obtained by adjoining to the Eu lidean topology the sets of the form x [ D, where x 2 R1 and D is an open dis in H whi h is tangent to L at the point x. This is the Niemytski spa e. It an be used to larify properties of the fourth separation axiom. 11.25. Prove that the Niemytski spa e is Hausdor . 11.26. Prove that the Niemytski spa e is regular. 11.27. What topologi al stru ture is indu ed on L from X ? 11.28. Prove that the Niemytski spa e is not normal. 11.29 Corollary. There exists a regular spa e, whi h is not normal. 11.30. Embed the Niemytski spa e into a normal spa e in su h a way that the omplement of the image would be a single point. 11.31 Corollary. Theorem 11.22 does not extend to non- losed subspa es, i.e., the property of being normal is not hereditary?
Urysohn Lemma and Tietze Theorem 11:A*. Let Y be a topologi al spa e satisfying the rst separation axiom. Let T be a subbase1 of the topology of Y . Let be an open over of a spa e X . Prove that if there exists a bije tion : ! T whi h preserves in lusions then there exists a ontinuous map f : X ! Y su h that f 1 (V ) = 1 (V ) for any V 2 T . 11:B. Prove that intervals [0; r) and (r; 1℄ where r = 2nq , n; q 2 N form a subbase for [0; 1℄, i.e., a olle tion of open sets in [0; 1℄, whose nite interse tions form a base of the standard topology in [0; 1℄. 1 Re all
that a subbase of the topology of Y is a olle tion T of open sets of Y su h that all nite interse tions of sets from T form a base of topology of Y , see Se tion 2.
12. COUNTABILITY AXIOMS
54
11:C Urysohn Lemma. Let A and B be disjoint losed subsets of a normal spa e X . Then there exists a ontinuous fun tion f : X ! I su h that f (A) = 0 and f (B ) = 1. 11:D. Let A be a losed subset of a normal spa e X . Let f : A ! [ 1; 1℄
be a ontinuous fun tion. Prove that there exists a ontinuous fun tion g : X ! 13 ; 13 su h that jf (x) g(x)j 32 for any x 2 A. 11:E. Prove that under the onditions of 11:D for any " > 0 there exists
a ontinuous fun tion : X ! [ 1; 1℄ su h that jf (x) (x)j " for any x 2 A. 11:F Tietze Extension Theorem. Prove that under the onditions of 11:D there exists a ontinuous fun tion F : X ! [ 1; 1℄ su h that F A = f.
11:G. Would the statement of Tietze Theorem remain true if in the
hypothesis the segment [ 1; 1℄ was repla ed by R,
Rn ,
S 1 , or S 2 ?
12. Countability Axioms In this se tion we ontinue to study topologi al properties whi h are imposed additionally on a topologi al stru ture to make the abstra t situation under onsideration loser to spe ial situations and hen e ri her in ontents. Restri tions studied in this se tion bound a topologi al stru ture from above: they require something to be ountable.
Set-Theoreti Digression. Countability Re all that two sets are said to be of equal ardinality if there exists a bije tion of one of them onto the other. A set of the same ardinality as a subset of the set N of natural numbers is said to be ountable. Sometimes this term is used only for in nite ountable sets, i.e. for set of the ardinality of the whole set N of natural numbers, while a set ountable in the sense above is alled at most ountable. This is less onvenient. In parti ular, if we adopted this terminology, then this se tion would have to be alled \At Most Countability Axioms". This would lead to other more serious in onvenien es as well. Our terminology has the following advantageous properties.
12.A. Any subset of a ountable set is ountable. 12.B. The image of a ountable set under any mapping is ountable. 12.C. The union of a ountable family of ountable sets is ountable.
12. COUNTABILITY AXIOMS
55
Se ond Countability and Separability In this se tion we study three restri tions on topologi al stru ture. Two of them have numbers (one and two), the third one has no number. As in the previous se tion, we start from the restri tion having number two. A topologi al spa e is said to satisfy the se ond axiom of ountability or to be se ond ountable if it has a ountable base. A spa e is alled separable if it ontains a ountable dense set. (This is the ountability axiom without a number mentioned above.)
12.D. The se ond axiom of ountability implies separability. 12.E. The se ond axiom of ountability is hereditary. 12.1. 12.2. 12.3.
Are the arrow and RT1 se ond ountable? Are the arrow and RT1 separable? Constru t an example proving that separability is not hereditary.
12.F. A metri separable spa e is se ond ountable. 12.G Corollary. For metri spa es, separability is equivalent to the
se ond axiom of ountability. 12.H. (Cf. 12.3.) Prove that for metri spa es separability is hereditary. 12.I. Prove that Eu lidean spa es and all their subspa es are separable and se ond ountable. 12.4.
12.J.
Constru t a metri spa e whi h is not se ond ountable.
A ontinuous image of a separable spa e is separable. 12.5. Constru t an example proving that a ontinuous image of a se ond
ountable spa e may be not se ond ountable.
12.K Lindelof Theorem. Any open over of a se ond ountable spa e
ontains a ountable part, whi h also overs the spa e.
Prove that any base of a se ond ountable spa e ontains a ountable part whi h is also a base. 12.7. Prove that in a separable spa e any olle tion of pairwise disjoint open sets is ountable. n 12.8. Prove that the set of omponents of an open set A R is ountable. 12.9. Prove that any set of disjoint gure eight urves in the plane is ountable. 12.10 Brower Theorem*. Let fK g be a family of losed sets of a se ond
ountable spa e and let for any des ending sequen e K1 K2 : : : of sets belonging to this family the interse tion \1 1 Kn also belongs to the family. Then the family ontains a minimal set, i.e., a set su h that no proper its subset belongs to the family. 12.6.
12. COUNTABILITY AXIOMS
56
Embedding and Metrization Theorems 12:A. Prove that the spa e l2 is separable and se ond ountable. 12:B. Prove that a regular se ond ountable spa e is normal. 12:C. Prove that a normal se ond ountable spa e an be embedded into l2 . (Use Urysohn Lemma 11:C.) 12:D. Prove that a se ond ountable spa e is metrizable, i it is regular.
Bases at a Point Let X be a topologi al spa e, and a its point. A neighborhood base at a or just base of X at a is a olle tion of neighborhoods of a su h that any neighborhood of a ontains a neighborhood from this olle tion.
12.L. If is a base of a spa e X then fU 2 : a 2 U g is a base of X at a.
In a metri spa e the following olle tions of balls are neighborhood bases at a point a: the set of all open balls of enter a; the set of all open balls of enter a and rational radii; the set of all open balls of enter a and radii rn , where frn g is any sequen e of positive numbers onverging to zero.
12.11.
12.12.
spa es?
What are the minimal bases at a point in the dis rete and indis rete
First Countability A topologi al spa e X is says to satisfy the rst axiom of ountability or to be a rst ountable spa e if it has a ountable neighborhood base at ea h point.
12.M. Any metri spa e is rst ountable. 12.N. The se ond axiom of ountability implies the rst one. 12.O. Find a rst ountable spa e whi h is not se ond ountable. (Cf. 12.4.)
Whi h of the following spa es are rst ountable: the arrow; RT1 ; a dis rete spa e; an indis rete spa e?
12.13.
(a) (b) ( ) (d)
12. COUNTABILITY AXIOMS
57
Sequential Approa h to Topology Spe ialists in Mathemati al Analysis love sequen es and their limits. Moreover they like to talk about all topologi al notions relying on the notions of sequen e and its limit. This tradition has almost no mathemati al justi ation, ex ept for a long history des ending from the XIX
entury studies on the foundations of analysis. In fa t, almost always2 it is more onvenient to avoid sequen es, provided you deal with topologi al notions, ex ept summing of series, where sequen es are involved in the underlying de nitions. Paying a tribute to this tradition we explain here how and in what situations topologi al notions an be des ribed in terms of sequen es. Let A be a subset of a topologi al spa e X . The set of limits of all sequen es an with an 2 A is alled a sequential losure of A and denoted by SCl A. 12.P. Prove that SCl A Cl A. 12.Q. If a spa e X is rst ountable then the for any A X the opposite in lusion Cl A SCl A holds also true, and hen e SCl A = Cl A. Therefore, in a se ond ountable spa e (in parti ular, any metri spa es) one an re over (hen e, de ne) the losure of a set provided it is known whi h sequen es are onvergent and what the limits are. In turn, knowledge of losures allows one to re over whi h sets are losed. As a onsequen e, knowledge of losed sets allows one to re over open sets and all other topologi al notions. 12.14. Let X be the set of real numbers equipped with the topology onsisting of ? and omplements of all ountable subsets. Des ribe onvergent sequen es, sequential losure and losure in X . Prove that in X there exists a set A with SCl A 6= Cl A.
Sequential Continuity Consider now ontinuity of maps along the same lines. A map f : X ! Y is said to be sequentially ontinuous if for any b 2 X and a sequen e an 2 X , whi h onverges to b, the sequen e f (an ) onverges to f (b).
12.R. Any ontinuous map is sequentially ontinuous. 12.S. The preimage of a sequentially losed set under
a sequentially
ontinuous map is sequentially losed. 12.T. If X is a rst ountable spa e then any sequentially ontinuous map f : X ! Y is ontinuous.
2 The
ex eptions whi h one may nd in the standard urri ulum of a mathemati al department an be ounted on two hands.
13. COMPACTNESS
58
Thus for mappings of a rst ountable spa e ontinuity and sequential
ontinuity are equivalent. 12.15.
12.14)
Constru t a sequentially ontinuous, but dis ontinuous map. (Cf.
13. Compa tness De nition of Compa tness This se tion is devoted to a topologi al property, whi h plays a very spe ial role in topology and its appli ations. It is sort of topologi al
ounter-part for the property of being nite in the ontext of set theory. (It seems though that this analogy has never been formalized.) Topologi al spa e is said to be ompa t if any of its open overs ontains a nite part whi h overs the spa e. If is a over of X and is a over of X then GS is alled a sub over (or sub overing ) of . Thus, a topologi al spa e is ompa t if every open
over has a nite sub over. 13.A. Any nite topologi al spa e and indis rete spa e are ompa t. 13.B. Whi h dis rete topologi al spa es are ompa t? 13.1. Let 1 2 be topologi al stru tures in X . Does ompa tness of (X; 2 ) imply ompa tness of (X; 1 )? And vi e versa?
13.C. Prove that the line R is not ompa t. 13.D. Prove that a topologi al spa e X is not ompa t i there exists an open overing whi h ontains no nite sub overing. 13.2. Is the arrow ompa t? Is RT ompa t? 1
Terminology Remarks Originally the word ompa tness was used for the following weaker property: any ountable open over ontains a nite sub over. 13.E. Prove that for a se ond ountable spa e the original de nition of
ompa tness is equivalent to the modern one. The modern notion of ompa tness was introdu ed by P. S. Alexandro (1896{1982) and P. S. Urysohn (1898{1924). They suggested for it the term bi ompa tness. This notion appeared to be so su
essful that it has displa ed the original one and even took its name, i.e. ompa tness. The term bi ompa tness is sometimes used (mainly by topologists of Alexandro s hool).
13. COMPACTNESS
59
Another deviation from the terminology used here omes from Bourbaki: we do not in lude the Hausdor property into the de nition of ompa tness, whi h Bourbaki in ludes. A
ording to our de nition, R T1 is
ompa t, a
ording to Bourbaki it is not.
Compa tness in Terms of Closed Sets A olle tion of subsets of a set is said to be entered if the interse tion of any nite sub olle tion is not empty. 13.F. A olle tion of subsets of a set X is entered, i there exists no nite 1 su h that the omplements of the sets belonging to 1
over X .
13.G.
A topologi al spa e is ompa t, i any entered olle tion of its
losed sets has nonempty interse tion.
Compa t Sets By a ompa t set one means a subset of a topologi al spa e (the latter must be lear from the ontext) provided it is ompa t as a spa e with the topology indu ed from the ambient spa e. 13.H. A subset A of a topologi al spa e X is ompa t, i any over whi h onsists of sets open in X ontains a nite sub over. 13.3. Is [1; 2) R ompa t?
Is the same set [1; 2) ompa t in the arrow? 13.5. Find a ne essary and suÆ ient ondition (formulated not in topologi al terms) for a subset of the arrow to be ompa t? 13.6. Prove that any subset of RT1 is ompa t. 13.7. Let A and B be ompa t subsets of a topologi al spa e X . Does it follow that A [ B is ompa t? Does it follow that A \ B is ompa t? 1 13.8. Prove that the set A = f0g [ f n g1 n=1 in R is ompa t.
13.4.
Compa t Sets Versus Closed Sets
13.I. Is ompa tness hereditary? 13.J. Any losed subset of a ompa t spa e is ompa t. 13.K. Any ompa t subset of a Hausdor spa e is losed. 13.L Lemma to 13.K, but not only : : : . Let A be a ompa t sub-
set of a Hausdor spa e X and b a point of X whi h does not belong to A. Then there exists open sets U; V X su h that b 2 V , A U and U \ V = ?. Constru t a non losed ompa t subset of some topologi al spa e. What is the minimal number of points needed?
13.9.
13. COMPACTNESS
60
Compa tness and Separation Axioms
13.M. A ompa t Hausdor spa e is regular. 13.N. Prove that a ompa t Hausdor spa e is normal. Prove that the interse tion of any family of ompa t subsets of a Hausdor spa e is ompa t. (Cf. 13.7.) 13.11. Let X be a Hausdor spa e, let fK g 2 be a family of its ompa t subsets, and let U be an open set ontaining \ 2 K . Prove that U \ 2A K for some nite A . 13.12. Let fKn g be a de reasing sequen e of ompa t nonempty onne ted subset of a Hausdor spa e. Prove that the interse tion \1 n=1 Kn is nonempty and onne ted. 13.13. Constru t a de reasing sequen e of onne ted subsets of the plane with non onne ted interse tion. 13.14. Let K be a onne ted omponent of a ompa t Hausdor spa e X and let U be an open set ontaining K . Prove that there exists an open and
losed set V su h that K V U . 13.10.
Compa tness in Eu lidean Spa e
13.O.
The interval I is ompa t.
Re all that n-dimensional ube is the set I n = fx 2 R n j xi 2 [0; 1℄ for i = 1; : : : ; ng: 13.P. The ube I n is ompa t.
13.Q.
Any ompa t subset of a metri spa e is bounded.
Therefore, any ompa t subset of a metri spa e is losed and bounded, see 13.K and 13.Q. 13.R. Constru t a losed and bounded, but non ompa t set of a metri spa e. 13.15.
13.S.
Are the metri spa es of Problem 3.A ompa t?
A subset of a Eu lidean spa e is ompa t, i it is losed and bounded. Whi h of the following sets are ompa t: [0; 1); ray R+ = fx 2 R j x 0g; S1; Sn; one-sheeted hyperboloid; ellipsoid; [0; 1℄ \ Q ?
13.16.
(a) (b) ( ) (d) (e) (f) (g)
13. COMPACTNESS
61
Matrix (aij ) with 1 i n, 1 j k with real ai j an be onsidered as a point of Rnk . For this, one needs to enumerate somehow (e.g, lexi ographi ally) its elements by numbers from 1 till nk. This identi es the set L(nk) of all matri es like that with Rnk and endows it with a topologi al stru ture. (Cf. Se tion 10.) 13.17. Whi h of the following subsets of L(n; n) are ompa t: (a) GL(n) = fA 2 L(n; n) j det A 6= 0g; (b) SL(n) = fA 2 L(n; n) j det A = 1g; ( ) O(n) = fA 2 L(n; n) : j A is an orthogonal matrixg; (d) fA 2 L(n; n) j A2 = 1g, here 1 is the unit matrix?
Compa tness and Maps
13.T.
A ontinuous image of a ompa t set is ompa t. (In other words, if X is a ompa t spa e and f : X ! Y is a ontinuous map then f (X ) is ompa t.) 13.U. On a ompa t set any ontinuous fun tion is bounded and attains its maximal and minimal values. (In other words, if X is a ompa t spa e and f : X ! R is a ontinuous fun tion, then there exist a; b 2 X su h that f (a) f (x) f (b) for any x 2 X .) Cf. 13.T and 13.S. 13.18.
interval.
Prove that if f : I
!R
is a ontinuous fun tion then f (I ) is an
13.19. Prove that if F and G are disjoint subsets of a metri spa e, F is
losed and G ompa t then (F; G) = inf f(x; y) j x 2 F; y 2 Gg > 0. 13.20. Prove that any open set ontaining a ompa t set A of a metri spa e X ontains an "-neighborhood of A. (i.e., the set fx 2 X j (x; A) < "g for some " > 0). 13.21. Let A be a losed onne ted subset of Rn and let V be its losed "-neighborhood (i.e., V = fx 2 Rn j (x; A) < "g). Prove that V is path onne ted. 13.22. Prove that if in a ompa t metri spa e the losure of any open ball is the losed ball with the same enter and radius then any ball of this spa e is onne ted. 13.23. Let X be a ompa t metri spa e and f : X ! X be a map su h that (f (x); f (y)) < (x; y) for any x; y 2 X with x 6= y. Prove that f has a unique xed point. (Re all that a xed point of f is a point x su h that f (x) = x.) 13.24. Prove that for any open over of a ompa t metri spa e there exists a number r > 0 su h that any open ball of radius r is ontained in some element of the over.
13.V Lebesgue Lemma. Let f : X ! Y be a ontinuous map of a
ompa t metri spa e X to a topologi al spa e Y , and let be an open
over of Y . Then there exists a number Æ > 0 su h that for any set A X with diameter diam(A) < Æ the image f (A) is ontained in some element of .
14. LOCAL COMPACTNESS AND PARACOMPACTNESS
62
Norms in Rn Prove that any norm Rn ! R (see Se tion 3) is a ontinuous fun tion (with respe t to the standard topology of Rn ). n 13.26. Prove that any two norms in R are equivalent (i.e. de ne the same topologi al stru ture). See 3.26, f. 3.29. n 13.27. Does the same hold true for metri s in R ?
13.25.
Closed Maps A ontinuous map is said to be losed if the image of any losed set under this map is losed.
13.W.
losed.
A ontinuous map of a ompa t spa e to a Hausdor spa e is
Here are two important orollaries of this theorem.
13.X.
A ontinuous inje tion of a ompa t spa e to a Hausdor spa e is a topologi al embedding.
13.Y.
A ontinuous bije tion of a ompa t spa e to a Hausdor spa e is a homeomorphism. 13.28. Show that none of the hypothesis in 13.Y an be omitted without making the statement false. 13.29. Does there exist a non ompa t subspa e of Eu lidian spa e su h that any its map to a Hausdor spa e is losed? (Cf. 13.U and 13.W.)
14. Lo al Compa tness and Para ompa tness Lo al Compa tness A topologi al spa e X is alled lo ally ompa t if ea h of its points has a neighborhood with ompa t losure. 14:A. Prove that lo al ompa tness is a lo al property, i.e., a spa e is lo ally ompa t, i ea h of its points has a lo ally ompa t neighborhood. 14:B. Is lo al ompa tness hereditary? 14:C. Prove that a losed subset of a lo ally ompa t spa e is lo ally
ompa t. 14:D. Prove that an open subset of a lo ally ompa t Hausdor spa e is lo ally ompa t. : Whi h of the following spa es are lo ally ompa t: (a) R; (b) Q ;
14 1.
14. LOCAL COMPACTNESS AND PARACOMPACTNESS
63
( ) Rn ; (d) a dis rete spa e? : Find two lo ally ompa t sets on the line su h that their union is not lo ally ompa t.
14 2.
One-Point Compa ti ation Let X be a Hausdor topologi al spa e. Let X be the set obtained by adding a point to X (of ourse, the point does not belong to X ). Let be the olle tion of subsets of X onsisting of sets open in X and sets of the form X r C , where C X is a ompa t set. 14:E. Prove that is a topologi al stru ture. 14:F. Prove that the spa e (X ; ) is ompa t. 14:G. Prove that the in lusion X ,! X is a topologi al embedding (with respe t to the original topology of X and ). 14:H. Prove that if X is lo ally ompa t then the spa e (X ; ) is
Hausdor . (Re all that X is assumed to be Hausdor .)
A topologi al embedding of a spa e X into a ompa t spa e Y is alled a ompa ti ation of X if the image of X is dense in Y . In this situation Y is also alled a ompa ti ation of X . 14:I. Prove that if X is a lo ally ompa t Hausdor spa e and Y is its
ompa ti ation with Y r X onsisting of a single point then there exists a homeomorphism Y ! X whi h is the identity on X .
The spa e Y of Problem 14:I is alled a one-point ompa ti ation or Alexandro ompa ti ation of X . 14:J. Prove that the one-point ompa ti ation of the plane is homeo-
morphi to S 2 .
: Prove that the one-point ompa ti ation of Rn is homeomorphi to S n . 14 3.
14:4. Give expli it des riptions of one-point ompa ti ations of the following spa es:
(a) (b) ( ) (d)
annulus f(x; y) 2 R2 j 1 < x2 + y2 < 2g; square without verti es f(x; y) 2 R2 j x; y 2 [ 1; 1℄; jxyj < 1g; strip f(x; y) 2 R2 j x 2 [0; 1℄g; a ompa t spa e.
14:K. Prove that a lo ally ompa t Hausdor spa e is regular.
14. LOCAL COMPACTNESS AND PARACOMPACTNESS
64
Proper Maps A ontinuous map f : X ! Y is said to be proper if the preimage of any
ompa t subset of Y is ompa t. Let X , Y be Hausdor spa es. Any ontinuous map f : X ! Y is naturally extended to a map X ! Y de ned by the following formula:
f (x) =
(
f (x); Y r Y;
if x 2 X otherwise, i.e., if x = X r X:
14:L. Prove that f is ontinuous, i f is proper. 14:M. Prove that any proper map of a Hausdor spa e to a Hausdor lo ally ompa t spa e is losed.
Problem 14:M is related to Theorem 13.W. 14:N. Extend this analogy: formulate and prove statements orresponding to theorems 13.X and 13.Y.
Lo ally Finite Colle tions of Subsets A olle tion of subsets of a spa e X is said to be lo ally nite if ea h point b 2 X has a neighborhood U su h that A \ U = ; for all but nite number of A 2 . 14:O. Any lo ally nite over of a ompa t spa e is nite. 14:5. If a olle tion is fCl A j A 2 g.
of subsets of a spa e X is lo ally nite then so
14:6. If a olle tion of subsets of a spa e X is lo ally nite and Cl A is ompa t for ea h A 2 then ea h A 2 interse ts only nite number of elements of .
:
14 7.
Any lo ally nite over of a sequentially ompa t spa e is nite.
14:P. Find an example of an open over of Rn whi h does not possess
a lo ally nite sub over.
Let and be overs of a set X . Then is said to be a re nement of if for ea h A 2 there exists B 2 su h that B A. 14:Q. Prove that any open over of Rn has a lo ally nite open re ne-
ment.
14:R. Let fUi gi2N be a lo ally nite open over of Rn . Prove that there exist an open over fVi gi2N su h that Cl Vi Ui for ea h i 2 N .
14. LOCAL COMPACTNESS AND PARACOMPACTNESS
65
Para ompa t Spa es A spa e X is said to be para ompa t if any its open over has a lo ally nite open re nement. 14:S. Any ompa t spa e is para ompa t. 14:T. Rn is para ompa t. 14:U. Let X = [1 i=1 Xi and Xi are ompa t sets. Then X is para ompa t. 14:V. Any losed subspa e of a para ompa t spa e is para ompa t. : A disjoint union of para ompa t spa es is para ompa t. 14:9. If X is a para ompa t spa e and Y ompa t then X para ompa t.
14 8.
Y
is
Para ompa tness and Separation Axioms : Any Hausdor para ompa t spa e is regular. 14:11. Any Hausdor para ompa t spa e is normal. 14:12. Let X be a normal spa e and its lo ally nite open over. Then there exists a lo ally nite open over su h that fCl V j V 2 g is a re nement of .
14 10.
Information. Any metrizable spa e is para ompa t. Partitions of Unity For a fun tion f : X ! R, the set Clfx 2 X j f (x) 6= 0g is alled the support of f and denoted by supp f . 14:W. Let ff g 2 be a family of ontinuous fun tions X ! R su h that the sets supp(f ) omprise a lo ally nite over of the spa e X . Prove that the relation X f (x) = f (x) 2 de nes a ontinuous fun tion f : X ! R. A family of nonnegative fun tions f X ! R+ is alled a partition of unityPif the sets supp(f ) omprise a lo ally nite over of the spa e X and 2 f (x) = 1: A partition of unity ff g is said to be subordinate to a over if ea h supp(f ) is ontained in an element of . 14:X. For every normal spa e X there exists a partition of unity whi h is subordinate to a given lo ally nite open over of X . 14:Y. A Hausdor spa e is para ompa t, i any its open over admits a partition of unity whi h is subordinate to this over.
15. SEQUENTIAL COMPACTNESS
66
Appli ation: Making Embeddings from Pie es 14:Z. Let hi Ui ! Rn , i = 1; : : : ; k, be embeddings, where Ui omprise
an open over of a spa e X . Then X an be embedded in Rk(n+1) . 14:Z:1. Show that the map x 7! (fi (x)h^ i (x)), where fi X ! R omrise a partition of unity, whi h is subordinate to the given over and h^ i (x) = (hi (x); 1) 2 Rn+1 , is an embedding.
15. Sequential Compa tness Sequential Compa tness Versus Compa tness A topologi al spa e is said to be sequentially ompa t if every sequen e of its points ontains a onvergent subsequen e.
15.A.
Any ompa t rst ountable spa e is sequentially ompa t.
A point b is alled an a
umulation point of a set A if every neighborhood of b ontains in nitely many points of A. 15.A.1. Prove that in a rst ountable spa e the notions of a
umulation point and limit point oin ide. 15.A.2. In a ompa t spa e any in nite set has an a
umulation point. 15.A.3. Dedu e Theorem 15.A from 15.A.2.
15.B.
A sequentially ompa t se ond ountable spa e is ompa t.
15.B.1. In a sequentially ompa t spa e a de reasing sequen e of nonempty
losed sets has a nonempty interse tion. 15.B.2. Prove that in a topologi al spa e every de reasing sequen e of nonempty losed sets has nonempty interse tion, i any entered ountable
olle tion of losed sets has nonempty interse tion.
15.C.
For se ond ountable spa es ompa tness and sequential ompa tness are equivalent.
In Metri Spa e A subset A of a metri spa e X is alled an "-net (where " is a positive number) if (x; A) < " for ea h point x 2 X . 15.D. Prove that in any ompa t metri spa e for any " > 0 there exists a nite "-net. 15.E. Prove that in any sequentially ompa t metri spa e for any " > 0 there exists a nite "-net.
15. SEQUENTIAL COMPACTNESS
67
15.F. Prove that a subset of a metri spa e is everywhere dense, i it is
an "-net for any " > 0.
15.G. Any sequentially ompa t metri spa e is separable. 15.H. Any sequentially ompa t metri spa e is se ond ountable. 15.I. For metri spa es ompa tness and sequential ompa tness equivalent.
are
15.1. Prove that a sequentially ompa t metri spa e is bounded. (Cf. 15.E and 15.I.) 15.2. Prove that in any metri spa e for any " > 0 there exists (a) a dis rete "-net and even (b) an "-net su h that the distan e between any two of its points is greater than ".
Completeness and Compa tness A sequen e fxn gn2N of points of a metri spa e is alled a Cau hy sequen e if for any " > 0 there exists a number N su h that (xn ; xm ); " for any n; m > N . A metri spa e is said to be omplete if ea h Cau hy sequen e in it is onvergent. 15:A. A Cau hy sequen e, whi h ontains a onvergent subsequen e,
onverges. 15:B. Prove that a metri spa e is omplete, i any de reasing sequen e of its losed balls with radii tending to 0 has nonempty interse tion. 15:C. Prove that a ompa t metri spa e is omplete? 15:D. Is any lo ally ompa t, but not ompa t metri spa e omplete? 15:E. Prove that a omplete metri spa e is ompa t, i for any " > 0 it ontains a nite "-net. 15:F. Prove that a omplete metri spa e is ompa t i for any " > 0 it
ontains a ompa t "-net.
Non-Compa t Balls in In nite Dimension By l1 denote the set of all bounded sequen es of real numbers. This is a ve tor spa e with respe t to the omponent-wise operations. There is a natural norm in it: j xj = supfjxn j : n 2 N g. 15.3. Are losed balls of l 1 ompa t? What about spheres? 15.4. Is the set fx 2 l 1 : jxn j 2 n ; n 2 N g ompa t? 15.5. Prove that the set fx 2 l 1 : jxn j = 2 n ; n 2 N g is homeomorphi to the Cantor set K introdu ed in Se tion 1. 15.6*. Does there exist an in nitely dimensional normed spa e, in whi h
losed balls are ompa t?
15. SEQUENTIAL COMPACTNESS
68
p-Adi Numbers Fix a prime integer p. By Zp denote the set of series of the form a0 + a1 p + with 0 an < p, an 2 N . For x; y 2 Zp put (x; y) = 0 if x = y and (x; y) = p m , if m is the smallest number su h that the m-th
oeÆ ients in the series x and y di er.
+ an pn + : : : 15.7.
Prove that is a metri in Zp.
This metri spa e is alled the spa e of integer p-adi numbers. There is an inje tion Z ! Zp assigning to a0 + a1 p + + an pn 2 Z with 0 ak < p the series
a0 + a1 p + + an pn + 0pn+1 + 0pn+2 + 2 Zp
and to (a0 + a1 p + + an pn ) 2 Z with 0 ak < p the series
b0 + b1 p + + bn pn + (p 1)pn+1 + (p 1)pn+2 + : : : ;
where
b0 + b1 p + + bn pn = pn+1
(a0 + a1 p + + an pn ):
Cf. 3.33. 15.8.
Prove that the image of the inje tion Z ! Zp is dense in Zp.
15.9.
Is Zp a omplete spa e?
15.10.
Is Zp ompa t?
Indu tion on Compa tness A fun tion f : X ! R is lo ally bounded if for any point a 2 X there exists a neighborhood U and a number M > 0 su h that jf (x)j M for x 2 U (i.e., ea h point has a neighborhood su h that the restri tion of f to this neighborhood is bounded). 15.11. Prove that if a spa e X is ompa t and a fun tion f : X lo ally bounded then f is bounded.
!R
is
This statement is one of the simplest appli ations of a general prin iple formulated below in 15.12. This prin iple may be alled indu tion on ompa tness ( f. indu tion on onne tedness dis ussed in Se tion 9). Let X be a topologi al spa e, C a property of subsets of X . We say that C is additive if the union of any nite family of sets having C also has C . The spa e X is said to possess C lo ally if ea h point of X has a neighborhood with property C . 15.12. Prove that a ompa t spa e whi h possesses lo ally an additive property has this property itself. 15.13. Dedu e from this prin iple the statements of problems 13.Q, 15:E, and 15:F.
PROBLEMS FOR TESTS
69
Spa es of Convex Figures Let D R2 be a losed dis of radius p. Consider the set of all onvex polygons P with the following properties: the perimeter of P is at most p; P is ontained in D; P has n verti es (the ases of one and two verti es are not ex luded). See 3.39, f. 3.41. 15.14. Equip this set with a natural topologi al stru ture. For instan e, de ne a natural metri . 15.15. Prove that this spa e is ompa t. 15.16. Prove that there exists a polygon belonging to this set and having the maximal area. 15.17. Prove that this is a regular n-gon. Consider now the set of all onvex polygons of perimeter p ontained in D. In other words, onsider the union of the sets of n-gons onsidered above. 15.18. Constru t a topologi al stru ture in this set su h that it indu es the stru tures introdu ed above in the spa es of n-polygons. 15.19. Prove that the spa e provided by the solution of Problem 15.18 is not ompa t. Consider now the set of all onvex subsets of the plane of perimeter p
ontained in D. 15.20. Constru t a topologi al stru ture in this set su h that it indu es the stru ture introdu ed above in the spa es of polygons. 15.21. Prove that the spa e provided by the solution of Problem 15.20 is
ompa t. 15.22. Prove that there exists a onvex plane set with perimeter p having a maximal area. p 15.23. Prove that this is a dis of radius 2 . 15.24. Consider the set of all bounded subsets of a ompa t metri spa e. Prove that this set endowed with the Hausdor metri (see 3.40) is a ompa t spa e.
Problems for Tests Test.1. Let X be a topologi al spa e. Fill Table 1 with pluses and minuses a
ording to your answers to the orresponding questions. Test.2. Let X be a topologi al spa e. Fill Table 2 with pluses and minuses a
ording to your answers to the orresponding questions. Test.3. Give as many proves as you an for non-existen e of a homeomorphism between
PROBLEMS FOR TESTS
If X is: Has Y the same property, if: Y X
70
nonnonse ond
onne ted Hausdor Hausdor separable ompa t ompa t ountable
Y is open subset of X Y is losed subset of X X is dense in Y Y is quotient spa e of X Y = X as sets,
X Y Y is open subset of Rn Y is antidis rete Table 1
(a) (b) ( ) (d)
S 1 and R1 , I and I 2 , R and RT1 R and R+ = fx 2 R : x 0g.
PROBLEMS FOR TESTS
If X is: Has Y the same property, if: X =Y Z
71
nonnonse ond
onne ted Hausdor Hausdor separable ompa t ompa t ountable
Y =X Z Y is open dense in X X is open dense in Y X is quotient spa e of Y Y = X as sets,
X Y Y is losed and bounded subset of Rn Y is dis rete Table 2
CHAPTER 3
Topologi al Constru tions 16. Multipli ation Set-Theoreti Digression. Produ t of Sets Let X and Y be sets. The set of ordered pairs (x; y ) with x 2 X and y 2 Y is alled a dire t produ t or Cartesian produ t or just produ t of X and Y and denoted by X Y . If A X and B Y then A B X Y . Sets X fbg with b 2 Y and fag Y with a 2 X are alled bers of the produ t X Y . 16.A. Prove that for any A1 ; A2 X and B1; B2 Y (A1 [ A2 ) (B1 [ B2 ) = (A1 B1 ) [ (A1 B2 ) [ (A2 B1 ) [ (A2 B2 ); (A1 B1 ) \ (A2 B2 ) = (A1 \ A2 ) (B1 \ B2 ): There are natural maps of X Y onto X and Y de ned by formulas (x; y ) 7! x and (x; y ) 7! y . They are denoted by prX and prY and are
alled (natural) proje tions. 16.B. Prove that prX1(A) = A Y for A X . Write down the orresponding formula for B Y To a map f : X ! Y there orresponds a subset f of X Y de ned by f = f(x; f (x)) : x 2 X g and alled the graph of f . 16.C. A set X Y is the graph of a map X ! Y , i for ea h a 2 X the interse tion \ (a Y ) ontains exa tly one point.
Prove that for any map f : X ! Y and any set A X , f (A) = prY ( f \ (A Y )) = prY ( f \ prX1 (A)) and f 1 (B ) = prX ( \ (X B )) for any B Y . 16.2. Let A and B be subsets of X and = f(x; y ) 2 X X : x = y g. Prove that (A B ) \ = ?, i A \ B = ? 16.3. Prove that the map prX is bije tive. f 16.1.
16.4.
Prove that f is inje tive, i prY
f
is inje tive.
Let T : X Y ! Y X be the map de ned by (x; y) 7! (y; x). Prove that f 1 = T ( f ) for any invertible map f : X ! Y .
16.5.
72
16. MULTIPLICATION
73
Produ t of Topologies Let X and Y be topologi al spa es. If U is an open set of X and B is an open set of Y , then we say that U V is an elementary set of X Y .
16.D.
The set of elementary sets of X stru ture in X Y .
Y
is a base of a topologi al
The produ t of topologi al spa es X and Y is the set X Y with the topologi al stru ture de ned by the base onsisting of elementary sets. Prove that for any subspa es A and B of spa es X and Y the topology of the produ t A B oin ides with the topology indu ed from X Y via the natural A B X Y . 16.6.
16.E. The produ t Y X is ( anoni ally) homeomorphi to X Y . The produ t X (Y Z ) is anoni ally homeomorphi to (X Y ) Z .
Prove that if A is losed in X and B is losed in Y then A B is losed in X Y .
16.7.
16.8. 16.9.
Prove that Cl(A B ) = Cl A Cl B for any A X and B Y . Is it true that Int(A B ) = Int A Int B ?
16.10. 16.11. 16.12. 16.13.
Is it true that Fr(A B ) = Fr A Fr B ?
Is it true that Fr(A B ) = (Fr A B ) [ (A Fr B )?
Prove that for losed A and B Fr(A B ) = (Fr A B ) [ (A Fr B )?
Find a formula for Fr(A B ) in terms of A, Fr A, B and Fr B .
Topologi al Properties of Proje tions and Fibers
16.F. The natural proje tions prX and prY are ontinuous. 16.G. Prove that the topology of produ t is the oarsest topology with respe t to whi h prX and prY are ontinuous.
16.H.
A ber of a produ t is anoni ally homeomorphi to the orresponding fa tor. The anoni al homeomorphism is the restri tion to the ber of the natural proje tion of the produ t onto the fa tor.
16.I. Prove that R 1 R 1 = R 2 , (R 1 )n = R n , (I )n = I n (re all that
I n is the n-dimensional ube).
16.14. Let X and Y be bases of topologi al spa es X and Y . Prove that sets U V with U 2 X and V 2 Y omprise a base for X Y . 16.15.
Prove that a map f : X ! Y is ontinuous i prX j
16.16.
Prove that if W is open in X Y then prX (W ) is open in X .
phism.
f
is a homeomor-
16. MULTIPLICATION
74
A map of a topologi al spa e X to a topologi al spa e Y is said to be open if the image of any open set under this map is open. Therefore 16.16 states that prX : X Y ! X is an open map. 16.17. Is prX a losed map? 16.18. Prove that for ea h topologi al spa e X and ea h ompa t topologi al spa e Y the map prX : X Y ! X is losed.
Cartesian Produ ts of Maps Let X , Y , and Z be sets. To a map f : Z ! X Y one assigns the
ompositions f1 = prX Æ f : Z ! X and f2 = prY Æ f : Z ! Y . They are alled fa tors of f . Indeed, f an be re overed from them as a sort of produ t. 16.19. Prove that for any maps f1 : Z ! X and f2 : Z ! Y there exists a unique map f : Z ! X Y with prX Æ f = f1 and prY Æ f = f2 16.20. Let X , Y , and Z be topologi al spa es. Prove that f is ontinuous i f1 and f2 are ontinuous. For any maps g1 : X1 ! Y1 and g2 : X2 ! Y2 there is a map X1 X2 ! Y1 Y2 de ned by formula (x1 ; x2 ) 7! (g1 (x1 ); g2 (x2 )). This map is alled a (Cartesian) produ t of g1 and g2 and denoted by g1 g2 . 16.21. Prove that the Cartesian produ t of ontinuous maps is ontinuous, and the Cartesian produ t of open maps is open. 16.22. Prove that a metri : X X ! R is ontinuous with respe t to the topology de ned by the metri .
Properties of Diagonal and Graph 16.23. Prove that a topologi al spa e is Hausdor i the set = f(x; x) : x 2 X g (whi h is alled the diagonal of X X ) is losed. 16.24. Prove that if Y is a Hausdor spa e and a map f : X ! Y is
ontinuous then the graph f is losed in X Y . 16.25. Let Y be a ompa t spa e and f be losed. Prove that then f is
ontinuous. 16.26. Prove that in 16.25 the hypothesis on ompa tness is ne essary. 16.27. Let f R ! R be a ontinuous fun tion. Prove that its graph is: (a) losed; (b) onne ted; ( ) path onne ted; (d) lo ally onne ted; (e) lo ally ompa t. 16.28. Does any of properties of the graph of a fun tion mentioned in 16.27 imply its ontinuity? 16.29. Let f be losed. Then the following assertions are equivalent: (a) f is ontinuous;
16. MULTIPLICATION
75
(b) f is lo ally bounded; ( ) the graph f of f is onne ted. 16.30.
Prove that if f is onne ted and lo ally onne ted then f is ontin-
16.31.
Prove that if f is onne ted and lo ally ompa t then f is ontinu-
uous.
ous.
16.32. Are some of assertions in problems 16.29 { 16.31 true for mappings f : R2 ! R?
Topologi al Properties of Produ ts
16.J.
The produ t of Hausdor spa es is Hausdor . 16.33.
Prove that the produ t of regular spa es is regular.
16.34.
The produ t of normal spa es is not ne essarily normal.
16.34.1. Prove that the set of real numbers with the topology de ned by the base whi h onsists of all the rays [a; 1) is normal. 16.34.2. Prove that in the Cartesian square of the spa e introdu ed in .1 the subspa e f(x; y) : x = yg is losed and dis rete. 16.34.3. Find two disjoint subsets of f(x; y) : x = yg whi h have no disjoint neighborhoods in the Cartesian square of the spa e of .1.
16.K. The produ t of separable spa es is separable. 16.L. First ountability of fa tors implies rst ountability of the produ t .
16.M. The produ t of se ond ountable spa es is se ond ountable. 16.N. The produ t of metrizable spa es is metrizable. 16.O. The produ t of onne ted spa es is onne ted. Prove that for onne ted spa es X and Y and any proper subsets A X , B Y the set X Y r A B is onne ted.
16.35.
16.P. The produ t of path- onne ted spa es is path- onne ted. 16.Q. The produ t of ompa t spa es is ompa t. 16.36.
Prove that the produ t of lo ally ompa t spa es is lo ally ompa t.
16.37. For whi h of the topologi al properties studied above, if X Y has the property then X also has?
17. QUOTIENT SPACES
76
Representation of Spe ial Spa es as Produ ts
16.R. Prove that R 2 r f0g is homeomorphi to S 1 R . 16.38.
Prove that Rn r Rk is homeomorphi to S n k
Prove that S n \ fx 2 Rn+1 : x21 + + x2k homeomorphi to S k 1 Dn k+1 . 16.39.
16.40.
1 Rk+1 .
x2k+1 + + x2n+1 g is
Prove that O(n) is homeomorphi to SO(n) O(1).
16.41.
Prove that GL(n) is homeomorphi to SL(n) GL(1).
16.42.
Prove that GL+ (n) is homeomorphi to SO(n) R
n(n+1) 2
, where
GL+ (n) = fA 2 L(n; n) : det A > 0g:
16.43.
Prove that SO(4) is homeomorphi to S 3 SO(3).
The spa e S 1 S 1 is alled a torus.
16.S. Constru t a topologi al embedding of the torus to R 3 The produ t S 1 S 1 of k fa tors is alled the k-dimensional torus.
16.T. Prove that the k-dimensional torus an be topologi ally embedded into R k+1 . 16.U. Find topologi al embeddings of S 1 D2 , S 1 S 1 I , and S 2 I into R 3 .
17. Quotient Spa es Set-Theoreti Digression. Partitions and Equivalen e Relations Re all that a partition of a set is its over onsisting of pairwise disjoint sets. Ea h partition of a set X gives rise to an equivalen e relation (i.e., a relation, whi h is re exive, symmetri and transitive): two elements of X are said to be equivalent if they belong to the same element of the partition. Vi e versa, ea h equivalen e relation in X gives rise to the partition of X to lasses of equivalent elements. Thus partitions of a set into nonempty subsets and equivalen e relations in the set are essentially the same. More pre isely, they are two ways of des ribing the same phenomenon. Let X be a set, and S be a partition. The set whose elements are members of the partition S (whi h are subsets of X ) is alled the quotient set or fa tor set of X by S and denoted by X=S .
17. QUOTIENT SPACES
77
How is this operation related to division of numbers? Why is there a similarity in terminology and notations?
17.1. Riddle.
At rst glan e, the de nition of quotient set ontradi ts one of the very profound prin iples of the set theory whi h states that a set is de ned by its elements. Indeed, a
ording to this prin iple, X=S = S , sin e S and X=S have the same elements. Hen e, there seems to be no need to introdu e X=S . The real sense of the notion of quotient set is not in its literal set-theoreti meaning, but in our way of thinking of elements of partitions. If we remember that they are subsets of the original set and want to keep tra k of their internal stru ture (at least, of their elements), we speak of a partition. If we think of them as atoms, getting rid of their possible internal stru ture then we speak on the quotient set. The set X=S is alled also the set of equivalen e lasses for the equivalen e relation orresponding to the partition S . The mapping X ! X=S that maps x 2 X to the element of S ontaining this point is alled a ( anoni al) proje tion and denoted by pr. A subset of X whi h is a union of elements of a partition is said to be saturated. The smallest saturated set ontaining a subset A of X is alled the saturation of A. 17.2. Prove that A X is an element of a partition S of X , i A = pr 1 (point) where pr : X ! X=S is the natural proje tion.
17.A. Prove that the saturation of a set A equals pr 1 pr(A) . 17.B. Prove that a set is saturated i it is equal to its saturation. Quotient Topology
A quotient set X=S of a topologi al spa e X with respe t to a partition S into nonempty subsets is provided with a natural topology: a set U X=S is said to be open in X=S if its preimage pr 1 (U ) under the anoni al proje tion pr : X ! X=S is open.
17.C.
The olle tion of these sets is a topologi al stru ture in the quotient set X=S .
This topologi al stru ture is alled the quotient topology. The set X=S with this topology is alled the quotient spa e of the spa e X by partition S . 17.3. Give an expli it des ription of the quotient spa e of the segment [0; 1℄ by the partition onsisting of [0; 13 ℄, ( 31 ; 23 ℄, ( 23 ; 1℄.
17. QUOTIENT SPACES
78
What an you say about a partition S of a topologi al spa e X if the quotient spa e X=S is known to be dis rete?
17.4.
17.D. A subset of a quotient spa e X=S is open i it is the image of an open saturated set under the anoni al proje tion pr.
17.E. A subset of a quotient spa e X=S is losed, i its preimage under pr is losed in X , i it is the image of a losed saturated set. 17.F. The anoni al proje tion pr : X ! X=S is ontinuous. 17.G. Prove that the quotient topology is the nest topology in X=S su h that the anoni al proje tion pr is ontinuous with respe t to it.
Topologi al Properties of Quotient Spa es
17.H. A quotient spa e of a onne ted spa e is onne ted. 17.I. A quotient spa e of a path- onne ted spa e is path- onne ted. 17.J. A quotient spa e of a separable spa e is separable. 17.K. A quotient spa e of a ompa t spa e is ompa t. 17.L. The quotient spa e of the real line by partition R + , R r R + is
not Hausdor . 17.M. The quotient spa e of a topologi al spa e X by a partition S is Hausdor , i any two elements of S possess disjoint saturated neighborhoods. 17.5. Formulate similar ne essary and suÆ ient onditions for a quotient spa e to satisfy other separation axioms and ountability axioms. 17.6. Give an example showing that se ond ountability may get lost when we go over to a quotient spa e.
Set-Theoreti Digression. Quotients and Maps Let S be a partition of a set X into nonempty subsets. Let f : X ! Y be a map whi h is onstant on ea h element of S . Then there is a map X=S ! Y whi h assigns to ea h element A of S the element f (A). This map is denoted by f=S and alled the quotient map or fa tor map of f (by partition S ). 17.N. Prove that a map f : X ! Y is onstant on ea h element of a partition S of X i there exists a map g : X=S ! Y su h that the following diagram is ommutative:
X
? pr? y
f
!Y %g
X=S Prove that su h a map g oin ides with f=S .
17. QUOTIENT SPACES
79
More generally, if S and T are partitions of sets X and Y then every map f : X ! Y , whi h maps ea h element of S into an element of T , gives rise to a map X=S ! Y=T whi h assigns to an element A of partition S the element of partition T ontaining f (A). This map is denoted by f=S; T and alled the quotient map or fa tor map of f (with respe t to S and T ). 17.O. Formulate and prove for f=S; T a statement whi h generalizes 17.N. A map f : X ! Y de nes a partition of the set X into nonempty preimages of the elements of Y . This partition is denoted by S (f ). 17.P. The map f=S (f ) : X=S (f ) ! Y is inje tive. This map is alled inje tive fa tor (or inje tive quotient ) of the map f .
Continuity of Quotient Maps
17.Q.
Let X , Y be topologi al spa es, S be a partition of X into nonempty sets, and f : X ! Y be a ontinuous map, whi h is onstant on ea h element of S . Then the fa tor f=S of f is ontinuous.
Let X , Y be topologi al spa es, S be a partition of X into nonempty sets. Prove that the formula f 7! f=S de nes a bije tion of the set of all
ontinuous maps X ! Y , whi h are onstant on ea h element of the partition S , onto the set of all ontinuous maps X=S ! Y .
17.7.
17.R. Let X , Y be topologi al spa es, S and T partitions of X and Y , and f : X ! Y a ontinuous map, whi h maps ea h element of S into an element of T . Then the map f=S; T : X=S ! Y=T is ontinuous. Closed Partitions A partition S of a topologi al spa e X is alled losed, if the saturation of ea h losed set is losed. 17:1. Prove that a partition is losed i the anoni al proje tion X ! X=S is a losed map. 17:2. Prove that a partition, whi h ontains only one element onsisting of more than one point, is losed if this element is a losed set.
17:A. The quotient spa e of a topologi al spa e satisfying the rst separation axiom with respe t to a losed partition satis es the rst separation axiom. 17:B. The quotient spa e of a normal topologi al spa e with respe t to a losed partition is normal.
18. ZOO OF QUOTIENT SPACES
80
Open Partitions A partition S of a topologi al spa e X is alled open, if the saturation of ea h open set is open. : Prove that a partition is open i the anoni al proje tion X ! X=S is an open map. 17:4. Prove that if a set A is saturated with respe t to an open partition, then Int A and Cl A are also saturated. 17 3.
17:C. The quotient spa e of a se ond ountable spa e with respe t to an open partition is se ond ountable. 17:D. The quotient spa e of a rst ountable spa e with respe t to an open partition is rst ountable. 17:E. Let S be an open partition of a topologi al spa e X and T be an open partition of a topologi al spa e Y . Denote by S T the partition of X Y onsisting of A B with A 2 S and B 2 T . Then the inje tive fa tor X Y=S T ! X=S Y=T of pr pr X Y ! X=S Y=T is a homeomorphism.
18. Zoo of Quotient Spa es Tool for Identifying a Quotient Spa e with a Known Spa e
18.A. If f : X ! Y
is a ontinuous map of a ompa t spa e X onto a Hausdor spa e Y then the inje tive fa tor f=S (f ) : X=S (f ) ! Y is a homeomorphism.
18.B.
The inje tive fa tor of a ontinuous map of a ompa t spa e to a Hausdor one is a topologi al embedding. 18.1. Des ribe expli itly partitions of a segment su h that the orresponding quotient spa es are all the onne ted letters of the alphabet. 18.2. Prove that there exists a partition of a segment I with the quotient spa e homeomorphi to square I I .
Tools for Des ribing Partitions Usually an a
urate literal des ription of a partition is umbersome, but
an be shortened and made more understandable. Of ourse, this requires a more exible vo abulary with lots of words with almost the same meanings. For instan e, the words fa torize and pass to a quotient an be repla ed by atta h, glue, identify, ontra t, and other words a
ompanying these ones in everyday life. Some elements of this language are easy to formalize. For instan e, fa torization of a spa e X with respe t to a partition onsisting of a set A
18. ZOO OF QUOTIENT SPACES
81
and one-point subsets of the omplement of A is alled a ontra tion (of the subset A to a point), and the result is denoted by X=A. Let A; B X omprise a fundamental over of a topologi al spa e X . Prove that the quotient map A=A \ B ! X=B of the in lusion A ,! X is a homeomorphism.
18.3.
If A and B are disjoint subspa es of a spa e X , and f : A ! B is a homeomorphism then passing to the quotient of the spa e X by the partition into one-point subsets of the set X r (A [ B ) and two-point sets fx; f (x)g, where x 2 A, is alled gluing or identifying (of sets A and B by homeomorphism f ). Rather onvenient and exible way for des ribing partitions is to des ribe the orresponding equivalen e relations. The main advantage of this approa h is that, due to transitivity, it suÆ es to spe ify only some pairs of equivalent elements: if one states that x y and y z then it is not needed to state x z , sin e this follows. Hen e, a partition is represented by a list of statements of the form x y , whi h are suÆ ient to re over the equivalen e relation. By su h a list en losed into square bra kets, we denote the orresponding partition. For example, the quotient of a spa e X obtained by identifying subsets A and B by a homeomorphism f : A ! B is denoted by X=[a f (a) for any a 2 A℄ or just X=[a f (a)℄. Some partitions are easy to des ribe by a pi ture, espe ially if the original spa e an be embedded into plane. In su h a ase, as in the pi tures below, one draws arrows on segments to be identi ed to show dire tions whi h are to be identi ed. Below we introdu e all these kinds of des riptions for partitions and give examples of their usage, providing simultaneously literal des riptions. The latter are not ni e, but they may help to keep the reader on dent about the meaning of the new words and, on the other hand, appre iating the improvement the new words bring in.
Entran e to the Zoo
18.C. Prove that I=[0 1℄ is homeomorphi to S 1. In other words, the quotient spa e of segment I by the partition onsisting of f0; 1g and fag with a 2 (0; 1) is homeomorphi to a ir le. 18.C.1. Find a surje tive ontinuous map I ! S 1 su h that the orresponding partition into preimages of points onsists of one-point subsets of the interior of the segment and the pair of boundary points of the segment.
18. ZOO OF QUOTIENT SPACES
82
18.D. Prove that Dn=S n 1 is homeomorphi to S n. In 18.D we deal with the quotient spa e of ball Dn by the partition into S n 1 and one-point subsets of its interior. Reformulation of 18.D: Contra ting the boundary of an n-dimensional ball to a point gives rise to an n-dimensional sphere. 18.D.1. Find a ontinuous map of ball Dn to the sphere S n that maps the boundary of the ball to a single point, and maps the interior of the ball bije tively onto the omplement of this point.
18.E. Prove that I 2 =[(0; t) (1; t) for t 2I℄ is homeomorphi to S 1 I .
Here the partition onsisits of pairs of points f(0; t); (1; t)g where t 2 I , and one-point subsets of (0; 1) I . Reformulation of 18.E: If we glue the side edges of a square identifying points on the same hight, we get a ylinder.
18.F.
Let X and Y be topologi al spa es, S a partition of X . Denote by T the partition of X Y into sets A y with A 2 S , y 2 Y . Then the natural bije tion X=S Y ! X Y=T is a homeomorphism.
18.G. Riddle. How are the problems 18.C, 18.E and 18.F related? 18.H. S 1 I=[(z; 0) (z; 1) for z 2 S 1 ℄ is homeomorphi to S 1 S 1 .
Here the partition onsists of one-point subsets of S 1 (0; 1), and pairs of points of the basis ir les lying on the same generatrix of the ylinder. Reformulation of 18.H: If we glue the basis ir les of a ylinder identifying points on the same generatrix, then we get a torus. 18.I. I 2 =[(0; t) (1; t); (t; 0) (t; 1)℄ is homeomorphi to S 1 S 1 . In 18.I the partition onsists of one-point subsets of the interior (0; 1) (0; 1) of the square, pairs of points on the verti al sides, whi h are the same distan e from the bottom side (i.e., pairs f(0; t); (1; t)g with t 2 (0; 1)), pairs of points on the horizontal sides whi h lie on the same verti al line (i.e., pairs f(t; 0); (t; 1)g with t 2 (0; 1)), the four verti es of the square
18. ZOO OF QUOTIENT SPACES
83
Reformulation of 18.I: Identifying the sides of a square a
ording to the pi ture , we get a torus .
Transitivity of Fa torization A solution of Problem 18.I an be based on Problems 18.E and 18.H and the following general theorem.
18.J Transitivity of Fa torization. Let S be a partition of a spa e
X , and let S 0 be a partition of the spa e X=S . Then the quotient spa e (X=S )=S 0 is anoni ally homeomorphi to X=T , where T is the partition of the spa e X into preimages of elements of the partition S 0 under proje tion X ! X=S .
Mobius Strip Mobius strip or Mobius band is I 2 =[(0; t) (1; 1 t)℄. In other words, this is the quotient spa e of square I 2 by the partition into pairs of points symmetri with respe t to the enter of the square and lying on the verti al edges and one-point set whi h do not lie on the verti al edges. Figuratively speaking, the Mobius strip is obtained by identifying the verti al sides of a square in su h a way that the dire tions shown on them by arrows are superimposed.
18.K. Prove that the Mobius strip is homeomorphi to the surfa e swept in R 3 by an interval, whi h rotates in a halfplane around the middle point
while the halfplane rotates around its boundary line. The ratio of the angular velo ities of these rotations is su h that rotation of the halfplane by 360Æ takes the same time as rotation of the interval by 180Æ. See Figure 1.
Figure 1
18. ZOO OF QUOTIENT SPACES
84
Contra ting Subsets Prove that [0; 1℄=[ 31 ; 23 ℄ is homeomorphi to [0; 1℄, and [0; 1℄=f 31 ; 1g is homeomorphi to letter P. 18.5. Prove that the following spa es are homeomorphi : (a) R2 ; (b) R2 =I ; ( ) R2 =D2 ; (d) R2 =I 2 ; (e) R2 =A where A is a union of several segments with a ommon end point; (f) R2 =B where B is a simple nite polygonal line, i.e., a union of a nite sequen e of segments I1 , : : : , In su h that the initial point of Ii+1
oin ides with the nal point of Ii ). 18.6. Prove that if f : X ! Y is a homeomorphism then the quotient spa es X=A and Y=f (A) are homeomorphi . 18.7. Prove that R2 =[0; +1) is homeomorphi to Int D 2 [ f(0; 1)g. 18.4.
Further Examples 18.8.
Prove that S 1 =[z e2i=3 z ℄ is homeomorphi to S 1 .
In 18.8 the partition onsists of triples of points whi h are verti es of equilateral ins ribed triangles. 18.9. Prove that the following quotient spa es of disk D 2 are homeomorphi to D2 : (a) D2 =[(x; y) ( x; y)℄, (b) D2 =[(x; y) (x; y)℄, ( ) D2 =[(x; y) ( y; x)℄. 18.10. Find a generalization of 18.9 with D n substituted for D 2 . 18.11. Des ribe expli itly the quotient spa e of line R1 by equivalen e relation x y , x y 2 Z. 18.12. Present the M obius strip as a quotient spa e of ylinder S 1 I .
Klein Bottle Klein bottle is I 2 =[(t; 0) (t; 1); (0; t) (1; 1 t)℄. In other words, this is the quotient spa e of square I 2 by the partition into one-point subsets of its interior, pairs of points (t; 0); (t; 1) on horizontal edges whi h lie on the same verti al line, pairs of points (0; t); (1; 1 t) symmetri with respe t to the enter of the square whi h lie on the verti al edges, and the quadruple of verti es.
18. ZOO OF QUOTIENT SPACES
85
Present the Klein bottle as a quotient spa e of (a) a ylinder; (b) the Mobius strip. 1 1 18.14. Prove that S S =[(z; w ) ( z; w )℄ is homeomorphi to the Klein bottle. (Here w denotes the omplex number onjugate to w.) 18.15. Embed the Klein bottle into R4 ( f. 18.K and 16.S). 18.16. Embed the Klein bottle into R4 so that the image of this embedding under the orthogonal proje tion R4 ! R3 would look as follows. 18.13.
Proje tive Plane Let us identify ea h boundary point of the disk D2 with the antipodal point, i.e., fa torize the disk by the partition onsisting of one-point subsets of the interior of the disk and pairs of points on the boundary
ir le symmetri with respe t to the enter of the disk. The result is
alled the proje tive plane. This spa e annot be embedded into R 3 , too. Thus we are not able to draw it. Instead, we present it in other way. 18.L. A proje tive plane is the result of gluing of a disk and the Mobius strip by homeomorphism between boundary ir le of the disk and boundary ir le of the Mobius strip.
You May Have Been Provoked to Perform an Illegal Operation Solving the previous problem you did something whi h does not t into the theory presented above. Indeed, the operation with two spa es alled gluing in 18.L has not appeared yet. It is a ombination of two operations: rst we must make a single spa e onsisting of disjoint opies of the original spa es, and then we fa torize this spa e identifying points of one
opy with points of another. Let us onsider the rst operation in details.
Set-Theoreti Digression. Sums of Sets A sum of a family of sets fX g 2A`is the set of pairs (x ; ) su h that x ` 2 X . The sum is denoted by 2A X . The map of X ( 2 A) to 2A X de ned by formula x 7! (x; ) is an inje tion and denoted by in . If only sets X and Y are involved and they are distin t, we an avoid indi es and de ne the sum by setting X q Y = f(x; X ) j x 2 X g [ f(y; Y ) j y 2 Y g:
18. ZOO OF QUOTIENT SPACES
86
Sums of Spa es
18.M. If fX` g 2A is a olle tion of topologi al spa es then the olle tion of subsets of 2A X whose preimages under all in lusions in ( 2 A)
are open, is a topologi al stru ture. `
The sum 2A X with this topology is alled the (disjoint) sum of topologi al spa es X , ( 2 A). 18.N. Topology des ribed in 18.M is the nest topology with respe t to whi h all in lusions in are ontinuous. `
The maps in : X ! 2A X` are topologi al embedding, and their images are both open and losed in 2A X . 18.18. Whi h topologi al properties are inherited from summands X by ` the sum 2A X ? Whi h are not?
18.17.
Atta hing Spa e Let X , Y be topologi al spa es, A a subset of Y , and f : A ! X a ontinuous map. The quotient spa e (X q Y )=[a f (a) for a 2 A℄ is denoted by X [f Y , and is said to be the result of atta hing or gluing the spa e Y to the spa e X by f . The latter is alled the atta hing map. Here the partition of X q Y onsists of one-point subsets of in2 (Y and in1 (X r f (A)), and sets in1 (x) [ in2 f 1 (x) with x 2 f (A). 18.19.
X qY
18.20.
r A)
Prove that the omposition of in lusion X ! X q Y and proje tion is a topologi al embedding. Prove that if X is a point then X [f Y is Y=A.
! X [f Y
18.O. Prove that atta hing a ball Dn to its opy by the identity map of
the boundary sphere S n
1
gives rise to a spa e homeomorphi to S n .
Prove that the Klein bottle an be obtained as a result of gluing two
opies of the Mobius strip by the identity map of the boundary ir le. 18.22. Prove that the result of gluing two opies of a ylinder by the identity map of the boundary ir les (of one opy to the boundary ir les of the other) is homeomorphi to S 1 S 1 . 18.23. Prove that the result of gluing two opies of solid torus S 1 D 2 by the identity map of the boundary torus S 1 S 1 is homeomorphi to S 1 S 2. 18.24. Obtain the Klein bottle by gluing two opies of the ylinder S 1 I to ea h other. 18.25. Prove that the result of gluing two opies of solid torus S 1 D 2 by the map S 1 S 1 ! S 1 S 1 : (x; y) 7! (y; x) of the boundary torus to its opy is homeomorphi to S 3 .
18.21.
18. ZOO OF QUOTIENT SPACES
87
18.P. Let X , Y be topologi al spa es, A a subset of Y , and f; g : A ! X
ontinuous maps. Prove that if there exists a homeomorphism h : X ! X su h that h Æ f = g then X [f Y and X [g Y are homeomorphi . 18.Q. Prove that Dn [h Dn is homeomorphi to S n for any homeomorphism h : S n 1 ! S n 1 . 18.26. Classify up to homeomorphism topologi al spa es, whi h an be obtained from a square by identifying a pair of opposite sides by a homeomorphism. 18.27. Classify up to homeomorphism the spa es whi h an be obtained from two opies of S 1 I by identifying of the opies of S 1 f0; 1g by a homeomorphism. 18.28. Prove that the topologi al type of the spa e resulting in gluing two
opies of the Mobius strip by a homeomorphism of the boundary ir le does not depend on the homeomorphism. 18.29. Classify up to homeomorphism topologi al spa es, whi h an be obtained from S 1 I by identifying S 1 0 with S 1 1 by a homeomorphism.
Basi Surfa es A torus S 1 S 1 with the interior of an embedded disk deleted is alled a handle. A two-dimensional sphere with the interior of n disjoint embedded disks deleted is alled a sphere with n holes. 18.R. A sphere with a hole is homeomorphi to disk D2. 18.S. A sphere with two holes is homeomorphi to ylinder S 1 I . A sphere with three holes has a spe ial name. It is alled pantaloons. The result of atta hing p opies of a handle to a sphere with p holes by embeddings of the boundary ir les of handles onto the boundary
ir les of the holes (the boundaries of the holes) is alled a sphere with p handles, or, more eremonial (and less understandable, for a while), orientable onne ted losed surfa e of genus p. 18.30. Prove that a sphere with p handles is well-de ned up to homeomorphism (i.e., the topologi al type of the result of gluing does not depend on the atta hing embeddings).
18.T. A sphere with one handle is homeomorphi to torus S 1 S 1. 18.U. A sphere with two handles is homeomorphi to the result of gluing two opies of a handle by the identity map of the boundary ir le.
A sphere with two handles is alled a pretzel. Sometimes this word denotes also a sphere with more handles.
19. PROJECTIVE SPACES
88
The spa e obtained from a sphere with q holes by atta hing q opies of the Mobius strip by embeddings of the boundary ir les of the Mobius strips onto the boundary ir les of the holes (the boundaries of the holes) is alled a sphere with q ross aps , or non-orientable onne ted losed surfa e of genus q . 18.31. Prove that a sphere with q ross aps is well-de ned up to homeomorphism (i.e., the topologi al type of the result of gluing does not depend on the atta hing embeddings).
18.V. A sphere with one ross ap is homeomorphi to the proje tive plane.
18.W. A sphere with two ross aps is homeomorphi to the Klein bottle. A sphere, spheres with handles, and spheres with ross aps are alled basi surfa es. 18.X. Prove that a sphere with p handles and q ross aps is homeomorphi to a sphere with 2p + q ross aps (here q > 0). 18.32. Classify up to homeomorphisms topologi al spa es, whi h an be obtained by atta hing to a sphere with 2p holes p opies of S 1 I by embeddings of the boundary ir les of the ylinders onto the boundary ir les of the sphere with holes.
19. Proje tive Spa es This se tion an be onsidered as a ontinuation of the previous one. The quotient spa es des ribed here are of too great importan e to onsider them just as examples of quotient spa es.
Real Proje tive Spa e of Dimension n This spa e is de ned as the quotient spa e of the sphere S n by the partition into pairs of antipodal points, and denoted by R P n . 19.A. The spa e R P n is homeomorphi to the quotient spa e of the ball Dn by the partition into one-point subsets of the interior of Dn , and pairs of antipodal point of the boundary sphere S n 1 . 19.B. R P 0 is a point. 19.C. The spa e R P 1 is homeomorphi to the ir le S 1. 19.D. The spa e R P 2 is homeomorphi to the proje tive plane de ned in the previous se tion. 19.E. The spa e RP n is anoni ally homeomorphi to the quotient spa e of R n+1 r f0g by the partition into one-dimensional ve tor subspa es of R n+1 pun tured at 0.
19. PROJECTIVE SPACES
89
A point of the spa e R n+1 r f0g is a sequen e of real numbers whi h are not all zeros. These numbers are alled homogeneous oordinates of the
orresponding point of R P n . The point with homogeneous oordinates x0 , x1 , : : : , xn is denoted by (x0 : x1 : : xn ). Homogeneous oordinates de ne a point of R P n , but are not de ned by this point: proportional ve tors of oordinates (x0 ; x1 ; : : : ; xn ) and (x0 ; x1 ; : : : ; xn ) de ne the same point of R P n . The spa e R P n is anoni ally homeomorphi to the metri spa e, whose points are lines of R n+1 passing through the origin 0 = (0; : : : ; 0) and the metri is de ned as the angle between lines (whi h takes values in [0; 2 ℄). Prove that this is really a metri .
19.F.
Complex Proje tive Spa e of Dimension n This spa e is de ned as the quotient spa e of unit sphere S 2n+1 of the spa e C n+1 by the partition into ir les whi h ut by ( omplex) lines of C n+1 passing through the point 0. It is denoted by C P n . 19:A. C P n is homeomorphi to the quotient spa e of the unit ball D2n
of the spa e C n by the partition whose elements are one-point subsets of the interior of D2n and ir les ut on the boundary sphere S 2n 1 by ( omplex) lines of the spa e C n passing through the origin 0 2 C n . 19:B. C P 0 is a point. 19:C. C P 1 is homeomorphi to S 2 . 19:D. The spa e C P n is anoni ally homeomorphi to the quotient spa e
of the spa e C n+1 pun tured at 0.
r
f0g by the partition into omplex lines of
C n+1
Hen e, C P n an be viewed as the spa e of omplex-proportional nonzero omplex sequen es (x0 ; x1 ; : : : ; xn ). Notation (x0 : x1 : : xn ) and term homogeneous oordinates introdu ed for the real ase are used in the same way for the omplex ase. 19:E. The spa e C P n is anoni ally homeomorphi to the metri spa e,
whose points are the ( omplex) lines of the spa e C n+1 passing through the origin 0 and the metri is de ned to be the angle between lines (whi h takes values in [0; 2 ℄).
Quaternion Proje tive Spa es and Cayley Plane
Must be written
20. TOPOLOGICAL GROUPS
90
20. Topologi al Groups Algebrai Digression. Groups Re all that a group is a set G equipped with a group operation. A group operation in set G is a map ! : G G ! G satisfying the following three
onditions (known as group axioms ): Asso iativity. !(a; !(b; )) = !(!(a; b); ) for any a; b; 2 G, Existen e of Neutral Element. There exists e 2 G su h that !(e; a) = !(a; e) = a for every a 2 G, Existen e of Inverse. For any a 2 G there exists b 2 G su h that !(a; b) = !(b; a) = e. :
In a group a neutral element is unique.
:
For any element of a group an inverse element is unique.
20 1. 20 2.
The notations above are never used. (The only ex eption may happen, as here, if the de nition of group is dis ussed.) Instead, one uses either multipli ative or aditive notations. Under multipli ative notations the group operations is alled multipli ation and denoted as multipli ation: (a; b) 7! ab. The neutral element is alled unity and denoted by 1. The element inverse to a is denoted by a 1 . These notations are borrowed from the ase, say, of group of nonzero rational numbers with the usual multipli ation. Under additive notations the group operations is alled addition and denoted as addition: (a; b) 7! a + b. The neutral element is alled zero and denoted by 0. The element inverse to a is denoted by a. These notations are borrowed from the ase, say, of group of integer numbers with the usual addition. An operation ! : G G ! G is ommutative provided that !(a; b) = !(b; a) for all a; b 2 G. A group with ommutative group operation is
alled ommutative or abelian. Traditionally the additive notations are used only in the ase of ommutative group, while the multipli ative notations are used both for ommutative and non- ommutative ases. Below we use mostly the multipli ative notations. :
Che k that in ea h of the following situations we have a group: the set Sn of bije tions of the set f1; 2; : : : ; ng of n rst natural numbers with omposition (symmetri group of degree n,) the set Homeo(X ) of all homeomorphisms of a topologi al spa e X with omposition, the set of invertible real n n-matri es GL(n; R) with matrix multipli ation, the set of all real p q-matri es with addition of matri es,
20 3.
(a) (b) ( ) (d)
20. TOPOLOGICAL GROUPS
91
(e) the set of all subsets of a set X with symmetri di eren e (A; B ) 7! (A [ B ) r (A \ B ) .
Topologi al Groups A topologi al group is a set G equipped with both topologi al and group stru tures su h that the maps G G ! G : (x; y) 7! xy and G ! G : x 7! x 1 are ontinuous. 20:4. Prove that if G is a group and a topologi al spa e then G G ! G : (x; y) 7! xy and G ! G : x 7! x 1 are ontinuous, i G G ! G : (x; y) 7! x 1 y is ontinuous. 20:5. Prove that for a topologi al group G the inversion G ! G : x 7! x 1 is a homeomorphism. 20:6. Let G be a topologi al group, X a topologi al spa e, and f; g : X ! G be maps ontinuous at a point x0 2 X . Prove that maps X ! G : x 7! f (x)g(x) and X ! G : x 7! (f (x)) 1 are ontinuous at x0 .
20:A. Any group equipped with the dis rete topologi al stru ture is a
topologi al group.
: Is a group equipped with the indis rete topologi al stru ture a topologi al group?
20 7.
20:B. The real line R with the addition is a topologi al group. 20:C. The pun tured real line R r 0 with the multipli ation is a topologi al group. 20:D. The pun tured omplex line C r 0 with the multipli ation is a topologi al group. : Che k that in ea h of the following situations we have a topologi al group: (a) the set GL(n; R) of invertible real n n-matri es with the matrix multipli ation and the topology indu ed by the in lusion to the 2 set of all real n n-matri es onsidered as Rn , (b) the set GL(n; C ) of invertible omplex n n-matri es with the matrix multipli ation and the topology indu ed by the2in lusion 2 to the set of all omplex n n-matri es onsidered as C n = R2n .
20 8.
Self-Homeomorphisms Making a Topologi al Group Homogeneous Re all that the maps of a group G to itself de ned by formula x 7! xa 1 and x 7! ax, respe tively, are alled (right and left ) translations and denoted by Ra and La . 20:E. Any translation of a topologi al group is a homeomorphism.
20. TOPOLOGICAL GROUPS
92
Re all that the onjugation of a group G by a 2 G is the map G ! G : x 7! a 1 xa. 20:F. Conjugation of a topologi al group by any its element is a homeomorphism.
Given subsets A, B of a group G, the set fab : a 2 A; b 2 B g is denoted by AB , and fa 1 : a 2 Ag is denoted by A 1 . 20:G. If U is an open set in a topologi al group G then for any x 2 G the sets xU , Ux and U 1 are open. : Does the same hold true for losed sets? 20:10. Prove that if U and V are subsets of a topologi al group G and U is open then UV and V U are open. 20:11. Does the same hold true if one repla es all the words open by
losed? 20 9.
20:11:1. Whi h of the following sugroups of the additive group are losed: (a) Z p, , (b) 2Zp ( ) Z + 2Z?
R
Neighborhoods 20:H. If is a neighborhood basis at the unity 1 in a topologi al group G then = faU : a 2 G; U 2 g is a basis for topology of G.
A subset A of a group G is said to be symmetri if A 1 = A. 20:I. Any neighborhood of unity of a topologi al group ontains a symmetri neighborhood of unity. 20:J. For any neihgborhood U of 1 of a topologi al group there exists a neighborhood V of 1 su h that V V U .
20:12. For any neihgborhood U of 1 of a topologi al group and any natural number n there exists a symmetri neighborhood V of 1 su h that V n U . 20:13. Let G be a group and be a olle tion of its subsets. Prove that there exists a unique topology on G su h that G with this topology is a topologi al group and is its neighborhood basis at the unity, i satis es the following ve onditions: (a) ea h U 2 ontains the unity of G, (b) for every x 2 U 2 there exists V 2 su h that xV U , ( ) for ea h U 2 there exists V 2 su h that V 1 U , (d) for ea h U 2 there exists V 2 su h that V V U , (e) for every x 2 G and U 2 there exists V 2 su h that V x 1 Ux. 20:K. Riddle. For what reasons 20:J is similar to the triangle inequal-
ity?
20. TOPOLOGICAL GROUPS
93
Separaion Axioms 20:L. A topologi al group is Hausdor , i it satis es the rst separation
axiom, i the unity is losed. 20:M. A topologi al group is Hausdor , i the unity is equal to the interse tion of its neighborhoods. 20:N. If the unity of a topologi al group G is losed, then G (as a topologi al spa e) is regular. Consequently, for topologi al groups the rst three separation axioms are equivalent.
Countability Axioms 20:O. If
is a neighborhood basis at the unity 1 in a topologi al group G and S G is dense in G, then = faU : a 2 S; U 2 g is a basis for topology of G. Cf. 20:H and 12.F. 20:P. A rst ountable separable topologi al group is se ond ountable.
Subgroups Re all that a subset H of a group G su h that HH = H and H 1 = H is alled a subgroup of G. It is a group with the operation de ned by the group operation of G. If G is a topologi al group, then H inherits also a topologi al stru ture from G. 20:Q. If H is a subgroup of a topologi al group G, then the topologi al and group stru tures indu ed from G make H a topologi al group. : Prove that a subgroup of a topologi al group is open, i it
ontains an interior point. 20:15. Prove that every open subgroup of a topologi al group is also
losed. 20:16. Find an example of a subgroup of a topologi al group, whi h (a) is losed, but not open, (b) is neither losed, nor open. 20:17. Prove that a subgroup of a topologi al group is dis rete, i it
ontains an isolated point. 20:18. Prove that a subgroup H of a topologi al group G is losed, i it is lo ally losed, i.e., there exists an open set U G su h that U \ H = U \ Cl H 6= ?. 20:19. Prove that if H is a non- losed subgroup of a topologi al group G then Cl H r H is dense in Cl H . 20:20. Prove that the losure of a subgroup of a topologi al group is a subgroup.
20 14.
20. TOPOLOGICAL GROUPS
94
: Is it true that the interior of a subgroup of a topologi al group is a subgroup?
20 21.
Re all that the smallest subgroup of a group G ontaining a set S is said to be generated by S . : The subgroup generated by S is the interse tion of all the subgroups whi h ontain S . On the other hand, this is the set of all the elements whi h an be obtained as produ ts of elements of S and elements inverse to elements of S . 20 22.
20:R. A onne ted topologi al group is generated by any neighborhood
of the unity.
Re all that for a subgroup H of a group G right osets are sets Ha =
fxa : x 2 H g with a 2 G. Analogously, sets aH are left osets of H in G.
20:23. Let H be a subgroup of a group G. De ne a relation: a b if ab 1 2 H . Prove that this is an equivalen e relation and the right
osets of H in G are the equivalen e lasses.
:
20 24.
What is the ounter-part of 20:23 for left osets?
The set of left osets of H in G is denoted by G=H , the set of right osets of H in G, by H r G. If G is a topologi al group and H is its subgroup then the sets G=H and H n G are provided with the quotient topology. Equipped with these topologies, they are alled spa es of osets. 20:S. For any topologi al group G and its subgroup H , the natural
proje tions G ! G=H and G ! H n G are open (i.e., the image of every open set is open).
20:25. The spa e of left (or right) osets of a losed subgroup in a topologi al group is regular.
Normal Subgroups Re all that a subgroup H of a group G is said to be normal if a 1 ha 2 H for all h 2 H and a 2 G. Normal subgroups are alled also normal divisors or invariant subgroups. : Prove that the losure of a normal subgroup of a topologi al group is a normal subgroup.
20 26.
: The onne ted omponent of the unity of a topologi al group is a losed normal subgroup.
20 27.
: The path- onne ted omponent of the unity of a topologi al group is a normal subgroup.
20 28.
20. TOPOLOGICAL GROUPS
95
Re all that for a normal subgroup left osets oin ide with right osets and the set of osets is a group with the multipli ation de ned by formula (aH )(bH ) = abH . The group of osets of H in G is alled the quotient group or fa tor group of G by H and denoted by G=H . 20:T. The quotient group of a topologi al group is a topologi al group (provided that it is onsidered with the quotient topology). 20:29. The natural proje tion of a topologi al group onto its quotient group is open. 20:30. A quotient group of a rst (or se ond) ountable group is rst (respe tively, se ond) ountable. 20:31. The quotient group G=H of a topologi al group G is regular, i H is losed. 20:32. Prove that if a normal subgroup H of a topologi al group G is open then the quotient group G=H is dis rete. 20:33. Let G be a nite topologi al group. Prove that there exists a normal subgroup H of G su h that a set U G is open, i it is a union of several osets of H in G.
Homomorphisms Re all that a map f of a group G to a group H is alled a (group) homomorphism if f (xy) = f (x)f (y) for all x; y 2 G. If G and H are topologi al groups then by a homomorphism G ! H one means a group homomorphism whi h is ontinuous. 20:U. A group homomorphism of a topologi al group to a topologi al group is ontinuous, i it is ontinuous at 1. Besides similar modi ations, whi h an be summarized by the following prin iple: everything is assumed to respe t the topologi al stru tures , the terminology of group theory passes over without hanges. In parti ular, the kernel Ker f of a homomorphism f : G ! H is de ned as the preimage of the unity of H . A homomorphism f is a monomorphism if it is inje tive. This is known to be equivalent to Ker f = 1. A homomorphism f : G ! H is an epimorphism if it is surje tive, i.e, its image Im f = f (G) is the whole H . In group theory, an isomorphism is an invertible homomorphism. Its inverse is a homomorphism (and hen e an isomorphism) automati ally. In theory of topologi al groups this must be in luded in the de nition of isomorphism: an isomorphism of topologi al groups is an invertible homomorphism whose inverse is also a homomorphism. In other words, an isomorphism of topologi al groups is a map whi h is both an algebrai homomorphism and a homeomorphism. Cf. Se tion 8. : An epimorphism f : G ! H is open, i its inje tive fa tor, f=S (f ) : G= Ker f ! H , is an isomorphism.
20 34.
20. TOPOLOGICAL GROUPS
96
: An epimorphism of a ompa t topologi al group onto a topologi al group with losed unity is open. 20:36. Prove that the quotient group R=Z of the additive group of real numbers by the subgroup of integers is isomorphi to the multipli ative group S 1 = fz 2 C : jz j = 1g of omplex numbers with absolute value 1.
20 35.
Lo al Isomorphisms Let G and H be topologi al groups. A lo al isomorphism of G to H is a homeomorphism f of a neighborhood U of the unity of G to a neighborhood V of the unity of H su h that f (xy) = f (x)f (y) for every x; y 2 U su h that xy 2 U , f 1(zt) = f 1(z)f 1(t) for every z; t 2 V su h that zt 2 V . Topologi al groups G, H are said to be lo ally isomorphi if there exists a lo al isomorphism of G to H . 20:V. Isomorphi topologi al groups are lo ally isomorphi . 20:W. Additive group R of real numbers and multipli ative group S 1 of
omplex numbers with absolute value 1 are lo ally isomorphi , but not isomorphi .
: Prove that the relation of being lo ally isomorphi is an equivalen e relation on the lass of topologi al groups. 20:38. Find neighborhoods of unities in R and S 1 and a homeomorphism between them, whi h satis es the rst ondition from the de nition of lo al isomorphism, but does not satisfy the se ond one. 20:39. Prove that for any homeomorphism between neighborhods of unities of two topologi al groups, whi h satis es the rst ondition from the de nition of lo al isomorphism, but does not satisfy the se ond one, there exists a submapping, whi h is a lo all isomorphsm between these topologi al groups.
20 37.
Dire t Produ ts Let G and H be topologi al groups. In group theory, the produ t G H is given a group stru ture,1 in topology it is given a topologi al stru ture (see Se ion 16). 20:X. These two stru tures are ompatible: the group operations in G H are ontinuous with respe t to the produ t topology. Thus, G H is a topologi al group. It is alled the dire t produ t of the topologi al groups G and H . There are anoni al homomorphisms related with this: the in lusions iG : G ! G H : x 7! (x; 1) and iH : H ! G H : x 7! (1; x), whi h are monomorphisms, and the proje tions pG : G H ! G : (x; y) 7! x and pH : G H ! H : (x; y) 7! y, whi h are epimorphisms. 1 Re all
that the multipli ation in G H is de ned by formula (x; u)(y; v) = (xy; uv).
21. ACTIONS OF TOPOLOGICAL GROUPS
97
: Prove that the topologi al groups G H=iH and G are isomorphi . 20:41. The produ t operation is both ommutative and asso iative: G H is ( anoni ally) isomorphi to H G and G (H K ) is anoni ally isomorphi to (G H ) K .
20 40.
A topologi al group G is said to de ompose into the dire t produ t of its subgroups A and B if the map A B ! G : (x; y) 7! xy is an isomorphism of topologi al groups. If this is the ase, the groups G and A B are usually identi ed via this isomorphism. Re all that a similar de nition exists in ordinary group theory. The only di eren e is that there the isomorphism is just an algebrai isomorphism. Moreover, in that theory, G de omposes into the dire t produ t of its subgroups A and B , i A and B generate G, are normal subgroups and A \ B = 1. Therefore, if these onditions are satis ed in the ase of topologi al groups, then (x; y) 7! xy : AB ! G is a group isomorphism. : Prove that in this situation the map (x; y) 7! xy : A B ! G is ontinuous. Find an example where the inverse group isomorphism is not ontinuous. 20:43. Prove that a ompa t Hausdor group whi h de omposes algebrai ally into the dire t produ t of two subgroups, de omposes also into the dire t produ t of these subgroups in the ategory of topologi al groups. 20:44. Prove that the multipli ative group R r 0 of real numbers is isomorphi (as a topologi al group) to the dire t produ t of the multipli ative group S 0 = f1; 1g and the multipli ative group R+ = fx 2 R : x > 0g. 20:45. Prove that the multipli ative group C r 0 of omplex numbers is isomorphi (as a topologi al group) to the dire t produ t of the multipli ative group S 1 = fz 2 C : jz j = 1g and the multipli ative group R+ . 20:46. Prove that the multipli ative group H r 0 of quaternions is isomorphi (as a topologi al group) to the dire t produ t of the multipli ative group S 3 = fz 2 H : jz j = 1g and the multipli ative group R+ . 20:47. Prove that the subgroup S 0 = f1; 1g of S 3 = fz 2 H : jz j = 1g is not a dire t fa tor. 20:48. Find a topologi al group homeomorphi to RP 3 (the threedimensional real proje tive spa e). 20 42.
21. A tions of Topologi al Groups A tions of Group in Set
Must be written!
22. SPACES OF CONTINUOUS MAPS
98
Continuous A tions
Must be written! Orbit Spa es
Must be written! Homogeneous Spa es
Must be written!
22. Spa es of Continuous Maps Sets of Continuous Mappings By C (X; Y ) we denote the set of all ontinuous mappings of a topologival spa e X to a topologi al spa e Y . :
22 1.
Prove that C (X; Y ) onsists of a single element i so does Y .
: Prove that there exists an inje tion Y ! C (X; Y ). In other words, the ardinality ard C (X; Y ) of C (X; Y ) is greater than or equal to ard Y . 22 2.
:
22 3. Riddle.
Find natural onditions implying C (X; Y ) = Y .
: Let Y = f0; 1g equipped with topology f?; f0g; Y g. Prove that there exists a bije tion between C (X; Y ) and the topologi al stru ture of X .
22 4.
:
22 5.
Let X be a set of n points with dis rete topology. Prove that
C (X; Y ) an be identi ed with Y : : : Y (n times).
22:6. Let Y be a set of k points with dis rete topology. Find ne essary and suÆ ient ondition for the set C (X; Y ) ontain k2 elements.
Topologi al Stru tures on Set of Continuous Mappings Let X , Y be topologi al spa es, A X , B Y . Denote by W (A; B ) the set ff 2 C (X; Y ) j f (A) B g. Denote by (pw) the set fW (a; U ) j a 2 X; U is open in Y g and by ( o) the set fW (C; U ) j C X is ompa t, U is open in Y g 22:A. (pw) is a subbase of a topologi al stru ture on C (X; Y ).
22. SPACES OF CONTINUOUS MAPS
99
The topologi al stru ture generated by (pw) is alled the topology of pointwise onvergen y. The set C (X; Y ) equipped with this stru ture is denoted by C (pw) (X; Y ). 22:B. ( o) is a subbase of a topologi al stru tures on C (X; Y ). The topologi al stru ture de ned by ( o) is alled the ompa t-open topology. Hereafter we denote by C (X; Y ) the spa e of all ontinuous mappings X ! Y with the ompa t-open topology, unless the ontrary is spe i ed expli itly. 22:C Compa t-Open Versus Pointwise. The ompa t-open topology is ner than the topology of pointwise onvergen e. :
22 7.
Prove that C (I; I ) is not homeomorphi to C (pw)(I; I ).
Denote by Const(X; Y ) the set of all onstant mappings f : X ! Y . 22:8. Prove that the topology of pointwise onvergen e and ompa topen topology of C (X; Y ) indu e the same topologi al stru ture on Const(X; Y ), whi h, with this topology, is homeomorphi Y . 22:9. Let X be a dis rete spa e of n points. Prove that C (pw ) (X; Y ) is homeomorphi Y : : : Y (n times). Is this true for C (X; Y )?
Topologi al Properties of Spa es of Continuous Mappings 22:D. Prove that if Y is Hausdor , then C (pw) (X; Y ) is Hausdor for
any topologi al spa e X . Is this true for C (X; Y )? :
22 10.
: pa t?
22 11.
Prove that C (I; X ) is path onne ted i X is path onne ted. Prove that C (pw)(I; I ) is not ompa t. Is the spa e C (I; I ) om-
Metri Case 22:E. If Y is metrizable and X is ompa t then C (X; Y ) is metrizable.
Let (Y; ) be a metri spa e and X a ompa t spa e. For ontinuous maps f; g : X ! Y put d(f; g) = maxf(f (x); g(x)) j x 2 X g: 22:F This is a Metri . If X is a ompa t spa e and Y a metri spa e, then d is a metri on the set C (X; Y ). Let X be a topologi al spa e and Y a metri spa e with metri . A sequen e fn of maps X ! Y is said to uniformly onverge to f : X ! Y if for any " > 0 there exists a natural N su h that (fn (x); f (x)) < " for any n > N and x 2 X . This is a straightforward generalization of the notion of uniform onvergen e whi h is known from Cal ulus.
22. SPACES OF CONTINUOUS MAPS
100
22:G Metri of Uniform Convergen e. Let X be a ompa t spa e
and Y a metri spa e. A sequen e fn of maps X ! Y onverges to f : X ! Y in the topology de ned by d, i fn uniformly onverges to f . 22:H Uniform Convergen e Versus Compa t-Open. Let X be a
ompa t spa e and Y a metri spa e. Then the topology de ned by d on C (X; Y ) oin ides with the ompa t-open topology.
: Prove that the spa e C (R; I ) is metrizable. 22:13. Let Y be a bounded metri spa e and X a topologi al spa e S whi h admits presentation X = 1 i=1 Xi , where Xi is ompa t and Xi Int Xi+1 for i = 1; 2; : : : . Prove that C (X; Y ) is metrizable. 22 12.
Denote by Cb (X; Y ) the set of all ontinuous bounded maps from a topologi al spa e X to a metri spa e Y . For maps f; g 2 Cb (X; Y ), put d1 (f; g) = supf(f (x); g(x)) j x 2 X g: 22:I Metri on Bounded Mappings. This is a metri in Cb (X; Y ). 22:J d1 and Uniform Convergen e. Let X be a topologi al spa e and Y a metri spa e. A sequen e fn of bounded maps X ! Y onverges to f : X ! Y in the topology de ned by d1 , i fn uniformly onverges to f . 22:K When Uniform Is Not Compa t-Open. Find X and Y su h that the topology de ned by d1 on Cb (X; Y ) does not oin ide with the
ompa t-open topology.
Intera tions With Other Constru tions 22:L Continuity of Restri ting. Let X , Y be topologi al spa es and A X . Prove that the map C (X; Y ) ! C (A; Y ) : f 7! f jA is ontinuous. 22:M Continuity of Composing. Let X be a topologi al spa e and Y a lo ally ompa t Hausdor spa e. Prove that the map C (X; Y ) C (Y; Z ) ! C (X; Z ) : (f; g) 7! g Æ f is ontinuous. :
22 14.
Is lo al ompa tness of Y ne essary in 22:M?
22:N Extending Target. For any topologi al spa es X , Y and B Y
the map C (X; B ) ! C (X; Y ) : f 7! iB Æ f is a topologi al embedding. 22:O Maps to Produ t. For any topologi al spa es X , Y and Z the spa e C (X; Y Z ) is anoni ally homeomorphi to C (X; Y ) C (X; Z ). 22:P Restri ting to Sets Covering Sour e. Let fX1 ; : : : ; Xn g be a fundumental over of X . Prove that for any topologi al spa e Y ,
C (X; Y ) !
n Y i=1
C (Xi ; Y ) : f 7! (f jX1 ; : : : ; f jXn )
is a topologi al embedding. What if the over is not fundamental?
22. SPACES OF CONTINUOUS MAPS
101
22:Q Fa torizing Sour e. Let S be a losed partition2 of a Hausdor
ompa t spa e X . Prove that for any topologi al spa e Y the mapping C (X=S; Y ) ! C (X; Y ) is a topologi al embedding. :
22 15.
Are the onditions imposed on S and X in 22:Q ne essary?
22:R The Evaluation Map. Let X , Y be topologi al spa es. Prove that if X is lo ally ompa t and Hausdor then the map C (X; Y ) X ! Y : (f; x) 7! f (x) is ontinuous. :
22 16.
Are the onditions imposed on X in 22:R ne essary?
Mappings X Y ! Z and X ! C (Y; Z )
Y ! Z be a
ontinuous map. Then the map F : X ! C (Y; Z ) : F (x) : y 7! f (x; y); is ontinuous. 22:T. Let X , Z be topologi al spa es and Y a Hausdor lo ally ompa t spa e. Let F : X ! C (Y; Z ) be a ontinuous mapping. Then the mapping f : X Y ! Z : (x; y) 7! F (x)(y) is ontinuous. 22:U. Let X , Y and Z be topologi al spa es. Let the mapping : C (X Y; Z ) ! C (X; C (Y; Z )) be de ned by the relation (f )(x) : y 7! f (x; y): Then (a) is ontinuous; (b) if Y is lo ally ompa t and Hausdor then is a homeomorphism. 22:S. Let X , Y and Z be topologi al spa es and f X
2 Re all
that a partition is alled losed, if the saturation of ea h losed set is losed.
Part 2
Algebrai Topology
CHAPTER 4
Fundamental Group and Covering Spa es This part of the book an be onsidered as an introdu tion to algebrai topology. This is a part of topology, whi h relates topologi al and algebrai problems. The relationship is used in both dire tions, but redu tion of topologi al problems to algebra is at rst stages more useful, sin e algebra is usually easier. The relation is established a
ording to the following s heme. One invents a onstru tion, whi h assigns to ea h topologi al spa e X under onsideration an algebrai obje t A(X ). The latter may be a group, or a ring, or a quadrati form, or algebra, et . Another onstru tion assigns to a ontinuous mapping f : X ! Y a homomorphism A(f ) : A(X ) ! A(Y ). The onstru tions should satisfy natural onditions (in parti ular, they form a fun tor), whi h make it possible to relate topologi al phenomena with their algebrai images obtained via the onstru tions. There are in nitely many useful onstru tions of this kind. In this part we deal mostly with one of them. This is the rst one, rst from both the viewpoints of history and its role in mathemati s. It was invented by Henri Poin are in the end of the nineteenth entury.
103
23. HOMOTOPY
104
23. Homotopy Continuous Deformations of Maps
23.A. Is it possible to deform ontinuously (a) The identity map id : R 2 ! R 2 to the onstant map R 2 ! R 2 : x 7! (b) ( ) (d) (e) (f)
0, The identity map id : S 1 ! S 1 to the symmetry S 1 ! S 1 : x 7! x (here x is onsidered as a omplex number, sin e the ir le S 1 is fx 2 C : jxj = 1g), The identity map id : S 1 ! S 1 to the onstant map S 1 ! S 1 : x 7! 1, The identity map id : S 1 ! S 1 to the two-fold wrapping S 1 ! S 1 : x 7! x2 , The in lusion S 1 ! R 2 to a onstant map, The in lusion S 1 ! R 2 r 0 to a onstant map?
23.B. Riddle. When you (tried to) solve the previous problem, what did you mean by \deform ontinuously"?
This se tion is devoted to the notion of homotopy formalizing the naive idea of the ontinuous deformation of a map.
Homotopy as Map and Family of Maps Let f , g be ontinuous maps of a topologi al spa e X to a topologi al spa e Y , and H : X I ! Y a ontinuous map su h that H (x; 0) = f (x) and H (x; 1) = g (x) for any x 2 X . Then f and g are said to be homotopi , and H is alled a homotopy between f and g . For x 2 X , t 2 I denote H (x; t) by ht (x). This hange of notation results in a hange of the point of view of H . Indeed, for a xed t the formula x 7! ht (x) de nes a map ht : X ! Y and H appears to be a family of maps ht enumerated by t 2 I .
23.C. Prove that ea h ht is ontinuous. 23.D. Does ontinuity of all ht imply ontinuity of H ?
The onditions H (x; 0) = f (x) and H (x; 1) = g (x) in the de nition of homotopy above an be reformulated as h0 = f and h1 = g . Thus a homotopy between f and g an be onsidered as a family of ontinuous maps, whi h onne ts f and g . Continuity of a homotopy allows one to say that it is a ontinuous family of ontinuous maps.
23. HOMOTOPY
105
Homotopy as Relation
23.E.
Homotopy of maps is an equivalen e relation. 23.E.1. If f : X ! Y is a ontinuous map then H : X I ! Y de ned by H (x; t) = f (x) is a homotopy between f and f . 23.E.2. If H is a homotopy between f and g then H 0 de ned by H 0 (x; t) = H (x; 1 t) is a homotopy between g and f . 23.E.3. If H is a homotopy between f and f 0 and H 0 is a homotopy between f 0 and f 00 then H 00 de ned by ( H (x; 2t) for t 1=2; H 00 (x; t) = 0 H (x; 2t 1) for t 1=2 is a homotopy between f and f 00 .
Homotopy, being an equivalen e relation by 23.E, divides the set C (X; Y ) of all ontinuous mappings of a spa e X to a spa e Y into equivalen e
lasses. The latter are alled homotopy lasses. The set of these lasses is denoted by (X; Y ). Prove that for any X , the set (X; I ) has a single element. 23.2. Prove that the number of elements of (I; Y ) oin ides with the number of path onne ted omponents of Y . 23.1.
Straight-Line Homotopy
23.F. Any two ontinuous maps of the same spa e to R n are homotopi . 23.G. Solve the pre eding problem by proving that for ontinuous maps f; g : X ! R n formula H (x; t) = (1 t)f (x) + tg (x) de nes a homotopy
between f and g .
The homotopy de ned in 23.G is alled a straight-line homotopy. 23.H. Prove that any two ontinuous maps of a spa e to a onvex subspa e of R n are homotopi . A set A Rn is said to be star onvex, if there exists a point b 2 A su h that for any x 2 A the whole segment [a; x℄ onne ting x to a is ontained in A. 23.3. Prove that any two ontinuous maps of a spa e to a star onvex subspa e of Rn are homotopi . 23.4. Prove that any ontinuous map of a onvex set C Rn to any spa e is homotopi to a onstant map. 23.5. Under what onditions (formulated in terms of known topologi al properties of a spa e X ) any two ontinuous maps of any onvex set to X are homotopi ?
23. HOMOTOPY
106
Prove that any non-surje tive map of an arbitrary topologi al spa e to S n is homotopi to a onstant map. n 23.7. Prove that any two maps of a one-point spa e to R r f0g with n > 1 are homotopi . 23.8. Find two non-homotopi maps of a one-point spa e to R r f0g. 23.9. For various m, n, k , al ulate the number of homotopy lasses of maps f1; 2; : : : ; mg ! Rn r fx1 ; x2 ; : : : ; xk g, where f1; 2; : : : ; mg is equipped with dis rete topology. 23.10. Let f; g be maps of a topologi al spa e X to C r 0. Prove that if jf (x) g(x)j < jf (x)j for any x 2 X then f and g are homotopi . 23.11. Prove that for any polynomials p and q over C of the same degree in one variable there exists r > 0 su h that for any R > r formulas z 7! p(z ) and z 7! q(z ) de ne maps of ir le fz 2 C : jz j = Rg to C r 0 and these maps are homotopi . 23.12. Let f , g be maps of an arbitrary topologi al spa e X to S n . Prove that if jf (a) g(a)j < 2 for any a 2 X then f is homotopi to g. 23.13. Let f : S n ! S n be a ontinuous map. Prove that if it is xed point free, i.e., f (x) 6= x for any x 2 S n , then f is homotopi to the symmetry x 7! x. 23.6.
Two Natural Properties of Homotopies
23.I. Let f; f 0 : X ! Y , g : Y ! B , h : A ! X be ontinuous maps and F : X I ! Y a homotopy between f and f 0 . Prove that then g Æ F Æ (h idI ) is a homotopy between g Æ f Æ h and g Æ f 0 Æ h. 23.J. Riddle. Under onditions of 23.I de ne a natural mapping (X; Y ) ! (A; B ):
How does it depend on g and h? Write down all the ni e properties of this onstru tion. 23.K. Prove that maps f0; f1 : X ! Y Z are homotopi i prY Æf0 is homotopi to prY Æ f1 and prZ Æf0 is homotopi to prZ Æ f1 .
Stationary Homotopy Let A be a subset of X . A homotopy H : X I ! Y is said to be xed or stationary on A, or, brie y, to be an A-homotopy, if H (x; t) = H (x; 0) for all x 2 A, t 2 I . Maps whi h an be onne ted by an A-homotopy are said to be A-homotopi . Of ourse, A-homotopi maps oin ide on A. If one wants to emphasize that a homotopy is not assumed to be xed, one says that it is free. If one wants to emphasize the opposite (that it is xed), one says that the homotopy is relative.
23. HOMOTOPY
107
Warning: there is a similar, but di erent kind of homotopy, whi h is also
alled relative. See below. 23.L. Prove that, like free homotopy, A-homotopy is an equivalen e relation. The lasses into whi h A-homotopy divides the set of ontinuous maps X ! Y that agree on A with a map f : A ! Y are alled A-homotopy
lasses of ontinuous extensions of f to X . 23.M. For what A is a straight-line homotopy xed on A?
Homotopies and Paths Re all that by a path in a spa e X we mean a ontinuous mapping of the interval I into X . (See Se tion 10.) 23.N. Riddle. In what sense is any path a homotopy? 23.O. Riddle. In what sense does any homotopy onsist of paths? 23.P. Riddle. In what sense is any homotopy a path? 23.Q. Riddle. Introdu e a topology in the set C (X; Y ) of all ontinuous mappings X ! Y in su h a way that for any homotopy ht : X ! Y the map I ! C (X; Y ) : t 7! ht would be ontinuous. Re all that the ompa t-open topology in C (X; Y ) is the topology generated by the sets f' 2 C (X; Y ) j '(A) B g for ompa t A X and open B Y. 23:A. Prove that any homotopy ht : X ! Y de nes (by the formula presented in 23.Q) a path in C (X; Y ) with ompa t-open topology. 23:B. Prove that if X is lo ally ompa t and regular then any path in C (X; Y ) with ompa t-open topology is de ned by a homotopy.
Homotopy of Paths
23.R. Prove that any two paths in the same spa e X are freely homo-
topi , i their images belong to the same pathwise onne ted omponent of X .
This shows that the notion of free homotopy in the ase of paths is not interesting. On the other hand, there is a sort of relative homotopy playing a very important role. This is (0 [ 1)-homotopy. This auses the following ommonly a
epted deviation from the terminology introdu ed above: homotopy of paths always means not a free homotopy, but a homotopy xed on the end points of I (i.e. on 0 [ 1).
Notation: a homotopy lass of a path s is denoted by [s℄.
24. HOMOTOPY PROPERTIES OF PATH MULTIPLICATION
108
24. Homotopy Properties of Path Multipli ation Multipli ation of Homotopy Classes of Paths Re all (see Se tion 10) that paths u and v in a spa e X an be multiplied, provided the initial point v (0) of v oin ides with the nal point u(1) of u. The produ t uv is de ned by ( u(2t); if t 1=2 uv (t) = v (2t 1); if t 1=2:
24.A. Prove that if a path u is homotopi to u0 and a path v is homotopi
to v 0 and there exists produ t uv , then u0 v 0 exists and is homotopi to uv .
De ne a produ t of homotopy lasses of paths u and v to be the homotopy
lass of uv . So, [u℄[v ℄ is de ned as [uv ℄, provided uv is de ned. This is a de nition whi h demands a proof. 24.B. Prove that the produ t of homotopy lasses of paths is wellde ned (of ourse, when the initial point of paths of the rst lass oin ides with the nal point of paths of the se ond lass).
Asso iativity
24.C. Is multipli ation of paths asso iative? Of ourse, this question might be formulated with more details: 24.D. Let u, v, w be paths in the same spa e su h that produ ts uv and vw are de ned (i.e., u(1) = v (0) and v (1) = w(0)). Is it true that (uv )w = u(vw)? 24.1. Prove that for paths in a metri spa e (uv )w = u(vw ) implies that u, v, w are onstant maps. 24.2. Riddle. Find non- onstant paths u, v , and w in an indis rete spa e su h that (uv)w = u(vw).
24.E. Find a map ' : I ! I su h that for any paths u, v, w with u(1) = v (0) and v (1) = w(0) ((uv )w) Æ ' = u(vw):
24.F.
Multipli ation of homotopy lasses of paths is asso iative.
If you are troubled by 24.F, onsider the following problem. 24.G. Reformulate Theorem 24.F in terms of paths and their homotopies.
24. HOMOTOPY PROPERTIES OF PATH MULTIPLICATION
109
If you want to understand the essen e of 24.F, you have to realize that paths (uv )w and u(vw) have the same traje tories and di ers by time spent in the fragments of the path. Therefore to nd a homotopy between them one has to nd a ontinuous way to hange one s hedule to the other. If there is still a trouble in a formal prove, re all 24.E and solve the following problem. 24.H. Prove that any path in I beginning in 0 and nishing in 1 is homotopi to id : I ! I . Also, it may be useful to take into a
ount 23.I.
Unit Let a be a point of a spa e X . Denote by ea the path I ! X : t 7! a. 24.I. Is ea a unit for multipli ation of paths? The same question in more detailed form: 24.J. For a path u with u(0) = a is ea u = u? For a path v with v(1) = a is vea = v ?
24.K.
Problems 24.I and 24.J are similar to 24.C and 24.D, respe tively. 24.3. Riddle. Extending this analogy, formulate and solve problems similar to 24.E. 24.4. Prove that ea u = u implies u = ea .
The homotopy lass of ea is a unit for multipli ation of homotopy
lasses of paths.
Inverse Re all that for a path u there is inverse path u 1 de ned by u 1 (t) = u(1 t) (see Se tion 10). 24.L. Is the inverse path inverse with respe t to multipli ation of paths? In other words: 24.M. For a path u beginning in a and nishing in b is uu u 1 u = eb ?
1
= ea and
Prove that for a path u with u(0) = a equality uu 1 = ea implies u = ea . 24.6. Find a map ' : I ! I su h that (uu 1 ) = u Æ ' for any path u.
24.5.
25. FUNDAMENTAL GROUP
24.N.
For any path u the homotopy lass of path u homotopy lass of u.
110
1
is inverse to the
We see that from the algebrai viewpoint multipli ation of paths is terrible, but it de nes multipli ation of homotopy lasses of paths, whi h has ni e algebrai properties. The only unfortunate property is that the multipli ation of homotopy lasses of paths is not de ned for any two
lasses. 24.O. Riddle. How to sele t a subset of the set of homotopy lasses of paths to obtain a group?
25. Fundamental Group De nition of Fundamental Group Let X be a topologi al spa e, x0 its point. A path in X whi h starts and ends at x0 is alled a loop in X at x0 . Denote by (X; x0 ) the set of loops in X at x0 . Denote by 1 (X; x0 ) the set of homotopy lasses of loops in X at x0 . Both (X; x0 ) and 1 (X; x0 ) are equipped with multipli ation. 25.A. For any topologi al spa e X and a point x0 2 X the set 1 (X; x0 ) of homotopy lasses of loops at x0 with multipli ation de ned above is a group.
1 (X; x0 ) is alled the fundamental group of the spa e X with base point x0 . It was introdu ed by Poin are and that is why it is alled also Poin are group. The letter in its notation is also due to Poin are.
Why Index 1? The index 1 in the notation 1 (X; x0 ) appeared later than the letter . It is related to one more name of the fundamental group: the rst (or one-dimensional) homotopy group. There is an in nite series of groups r (X; x0 ) with r = 1; 2; 3; : : : and the fundamental group is one of them. The higher-dimensional homotopy groups were de ned by Witold Hurewi z in 1935, thirty years after the fundamental group was de ned. There is even a zero-dimensional homotopy group 0 (X; x0 ), but it is not a group, as a rule. It is the set of path-wise onne ted omponents of X . Although there is no natural multipli ation in 0 (X; x0 ) , unless X is equipped with some spe ial additional stru tures, there is a natural unit in 0 (X; x0 ). This is the omponent ontaining x0 .
25. FUNDAMENTAL GROUP
111
Roughly speaking, the general de nition of r (X; x0 ) is obtained from the de nition of 1 (X; x0 ) by repla ing I with the ube I r . 25.B. Riddle. How to generalize problems of this se tion in su h a way that in ea h of them I would be repla ed by I r ? High Homotopy Groups Let X be a topologi al spa e and x0 its point. A ontinuous map I r ! X whi h maps the boundary I r of I r to x0 is alled a spheroid of dimension r of X at x0 . Two r-dimensional spheroids are said to be homotopi , if they are I r -homotopi . For spheroids u, v of X at x0 of dimension r 1 de ne their produ t uv by formula ( u(2t1 ; t2 ; : : : ; tr ); if t1 1=2 uv(t1 ; t2 ; : : : ; tr ) = v(2t1 1; t2; : : : ; tr ); if t1 1=2: The set of homotopy lasses of r-dimensional spheroids of a spa e X at x0 is the r-th (or r-dimensional) homotopy group r (X; x0 ) of X at x0 . Thus, r (X; x0 ) = (I r ; I r ; X; x0 ): Multipli ation of spheroids indu es multipli ation in r (X; x0 ), whi h makes r (X; x0 ) a group. 25.1. For any X and x0 the group r (X; x0 ) with r 2 is Abelian. 25.2. Riddle. For any X; x0 and r 2 present group r (X; x0 ) as the fundamental group of some spa e.
Cir ular loops Let X be a topologi al spa e, x0 its point. A ontinuous map l : S 1 ! X su h that1 l(1) = x0 is alled a ( ir ular) loop at x0 . Assign to ea h
ir ular loop l the omposition of l with the exponential map I ! S 1 : t 7! e2it . This is a usual loop at the same point. 25.C. Prove that any loop an be obtained in this way from a ir ular loop. Cir ular loops l1 , l2 are said to be homotopi if they are 1-homotopi . Homotopy of a ir ular loop not xed at x0 is alled a free homotopy. 25.D. Prove that ir ular loops are homotopi , i the orresponding loops are homotopi . 25.3. What kind of homotopy of loops orresponds to free homotopy of ir ular loops? 25.4. Des ribe the operation with ir ular loops orresponding to the multipli ation of paths.
1 Re all,
that S 1 is onsidered as a subset of the plane R2 , whi h is identi ed in a
anoni al way with C . Hen e 1 2 fz 2 C : jz j = 1g.
25. FUNDAMENTAL GROUP
25.5.
112
Outline a onstru tion of fundamental group based on ir ular loops.
Similarly, high-dimensional homotopy groups an be onstru ted not out of homotopy lasses of maps (I r ; I r ) ! (X; x0 ), but as (S r ; (1; 0; : : : ; 0); X; x0 ): Another, also quite a popular way, is to de ne r (X; x0 ) as (Dr ; Dr ; X; x0 ): 25.6. Establish natural bije tions (I r ; I r ; X; x0 ) ! (Dr ; Dr ; X; x0 ) ! (S r ; (1; 0; : : : ; 0); X; x0 )
The Very First Cal ulations
25.E. Prove that 1 (Rn ; 0) is a trivial group (i.e., onsists of one element).
25:A. What about r (Rn ; 0)?
25.F. Generalize 25.E to the ases suggested by 23.H and 23.3.
Cal ulate the fundamental group of an indis rete spa e. 25.8. Cal ulate the fundamental group of the quotient spa e of disk D 2 obtained by identi ation of ea h x 2 D2 with x.
25.7.
25.G. Prove that 1 (S n; (1; 0; : : : ; 0)) with n 2 is a trivial group.
Whether you have solved 25.G or not, we would re ommend you onsider problems .1, .3, .4, .5 and .6 designed to give an approa h to 25.G, warn about a natural mistake and prepare an important tool for further al ulations of fundamental groups. 25.G.1. Prove that any loop s : I ! S n , whi h does not ll the whole S n (i.e., s(I ) 6= S n ) is homotopi to the onstant loop, provided n 2. (Cf. Problem 23.6.) Warning: for any n there exists a loop lling S n . See 7:I 25.G.2. Is a loop lling S 2 homotopi to the onstant loop? 25.G.3 Corollary of Lebesgue Lemma 13.V. Let s : I ! X be a path, and be an open overing of a topologi al spa e X . There exists a sequen e of points a1 ; : : : ; aN 2 I with 0 = a1 < a2 < < aN 1 < aN = 1 su h that s([ai ; ai+1 ℄) is ontained in an element of for ea h i. 25.G.4. Prove that if n 2 then for any path s : I ! S n there exists a subdivision of I into a nite number of subintervals su h that the restri tion of s to ea h of the subintervals is homotopi , via a homotopy xed on the endpoints of the subinterval, to a map with nowhere dense image. 25.G.5. Prove that if n 2 then any loop in S n is homotopi to a loop whi h is not surje tive. 25.G.6. Dedu e 25.G from .1 and .5. Find all the points of the proof of 25.G obtained in this way, where the ondition n 2 is used.
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113
Fundamental Group of Produ t
25.H. The fundamental group of the produ t of topologi al spa es is
anoni ally isomorphi to the produ t of the fundamental groups of the fa tors: 25.9.
1 (X Y; (x0 ; y0 )) = 1 (X; x0 ) 1 (Y; y0 ) Prove that 1 (Rn r 0; (1; 0; : : : ; 0)) is trivial if n 3
25:B. Prove the following generalization of 25.H:
r (X Y; (x0 ; y0 )) = r (X; x0 ) r (Y; y0 ):
Simply-Conne tedness A non-empty topologi al spa e X is said to be simply onne ted or one onne ted if it is path- onne ted and any loop in it is homotopi to a
onstant map. 25.I. For a path- onne ted topologi al spa e X the following statements are equivalent: (a) X is simply onne ted, (b) any ontinuous map f : S 1 ! X is (freely) homotopi to a onstant map, ( ) any ontinuous map f : S 1 ! X an be extended to a ontinuous map D2 ! X , (d) any two paths s1 ; s2 : I ! X onne ting the same points x0 and x1 are homotopi .
The following theorem implies Theorem 25.I. However, sin e it treats a single loop, it an be applied to more situations. Anyway, proving 25.I, one proves 25.J in fa t. 25.J. Let X be a topologi al spa e and s : S 1 ! X be a ir ular loop. Then the following statements are equivalent: (a) s is homotopi to the onstant loop, (b) s is freely homotopi to a onstant map, ( ) s an be extended to a ontinuous map D2 ! X , (d) the paths s+ ; s : I ! X de ned by formula s" (t) = s(e"it ) are homotopi . 25.J.1. Riddle. Proving that 4 statements are equivalent one has to prove at least 4 impli ations. What impli ations would you hoose for the shortest proof of Theorem 25.J? Whi h of the following spa es are simply onne ted: (a) a dis rete spa e, (b) an indis rete spa e, ( ) Rn ,
25.10.
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114
Sn, a onvex set, a star onvex set, Rn r 0. 25.11. Prove that a topologi al spa e X , whi h is presented as the union of open simply onne ted sets U and V with simply onne ted U \ V , is simply
onne ted. 25.12. Show that the assumption that U and V are open is ne essary in 25.11. 25.13*. Let X be a topologi al spa e, U and V its open sets. Prove that if U [ V and U \ V are simply onne ted, then U and V are simply onne ted, too. (d) (e) (f) (g)
Fundamental Group of a Topologi al Group Let G be a topologi al group. Given loops u; v : I ! G starting at the unity 1 2 G, let us de ne a loop u v : I ! G by the formula u v(t) = u(t) v(t), where denotes the group operation in G. 25:C. Prove that the set (G; 1) of all the loops in G starting at 1 equipped with the operation is a group. 25:D. Prove that the operation on (G; 1) de nes a group operation on 1 (G; 1) and that this operation oin ides with the standard group operation (de ned by multipli ation of paths). 25:D:1. For loops u; v ! G starting at 1, nd (ue1 ) (e1 v). 25:E. The fundamental group of a topologi al group is abelian. 25:F. Formulate and prove the analogues of Problems 25:C and 25:D for high homotopy groups and 0 (G; 1).
26. The Role of Base Point Overview of the Role of Base Point Roughly, the role of base point may be des ribed as follows: While the base point hanges within the same path- onne ted omponent, the fundamental group remains in the same lass of isomorphi groups. However, if the group is not ommutative, it is impossible to nd a natural isomorphism between fundamental groups at di erent base points even in the same path- onne ted omponent. Fundamental groups of a spa e at base points belonging to di erent path- onne ted omponents have no relation to ea h other. In this se tion these will be demonstrated. Of ourse, with mu h more details.
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De nition of Translation Maps Let x0 and x1 be points of a topologi al spa e X , and let s be a path
onne ting x0 with x1 . Denote by the homotopy lass [s℄ of s. De ne a map Ts : 1 (X; x0 ) ! 1 (X; x1 ) by formula Ts ( ) = 1 . Prove that for any loop a : I ! X representing 2 1 (X; x0 ) and a path s : I ! X with s(0) = x0 there exists a free homotopy H : I I ! X between a and a loop representing Ts ( ) su h that H (0; t) = H (1; t) = s(t) for t 2 I . 26.2. Let a; b : I ! X be loops whi h are homotopi via a homotopy H : I I ! X su h that H (0; t) = H (1; t) (i.e., H is a free homotopy of loops: at ea h moment t 2 I it keeps the end points of the path oin iding). Set s(t) = H (0; t) (hen e s is the path run over by the initial point of the loop under the homotopy). Prove that the homotopy lass of b is the image of the homotopy lass of a under Ts : 1 (X; s(0)) ! 1 (X; s(1)). 26.1.
Properties of Ts
26.A. Ts
is a (group) homomorphism. (Re all that this means that Ts ( ) = Ts ( )Ts ( ).) 26.B. If u is a path onne ting x0 to x1 and v is a path onne ting x1 with x2 then Tuv = Tv Æ Tu . In other words the diagram
! 1(X; x1 ) ? ? Tuv & yTv
1 (X; x0 )
Tu
1 (X; x2 ) is ommutative. 26.C. If paths u and v are homotopi then Tu = Tv .
26.D. Tea = id : 1 (X; a) ! 1 (X; a) 26.E. Ts 1 = Ts 1. 26.F. Ts is an isomorphism for any path s. 26.G. For any points x0 and x1 lying in the same path- onne ted omponent of X groups 1 (X; x0 ) and 1 (X; x1 ) are isomorphi .
Role of Path
26.H. If s is a loop representing an element of fundamental group
1 (X; x0 ) then Ts is the internal automorphism of 1 (X; x0 ) de ned by 7! 1 . 26.I. Let x0 and x1 be points of a topologi al spa e X belonging to the same path- onne ted omponent. Isomorphisms Ts : 1 (X; x0 ) ! 1 (X; x1 ) do not depend on s, i 1 (X; x0 ) is ommutative.
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High Homotopy Groups 26.3. Riddle.
Guess how Ts is generalized to r (X; x0 ) with any r.
Here is another form of the same question. We put it sin e it ontains in its statement a greater pie e of an answer. 26.4. Riddle. Given a path s : I ! X with s(0) = x0 and a spheroid f : I r ! X at x0 , how to ook up a spheroid at x1 = s(1) out of these? r 26.5. Prove that for any path s : I ! X and a spheroid f : I ! X with r r f (Fr I ) = s(0) there exists a homotopy H : I I ! X of f su h that H (Fr I r t) = s(t) for any t 2 I and that the spheroid obtained by su h a homotopy is unique up to homotopy and de nes an element of r (X; s(1)) well-de ned by the homotopy lass of s and the element of r (X; s(0)) represented by f . Of ourse, a solution of 26.5 gives an answer to 26.4 and 26.3. The map r (X; s(0)) ! r (X; s(1)) de ned by 26.5 is denoted by Ts . By 26.2 this Ts generalizes Ts de ned in the beginning of the se tion for the ase r = 1. 26.6. Prove that the properties of Ts formulated in Problems 26.A { 26.G hold true in all dimensions.
In Topologi al Group In a topologi al group G there is another way to relate 1 (G; x0 ) with 1 (G; x1 ): there are homeomorphisms Lg : G ! G : x 7! xg and Rg : G ! G : x 7! gx, so that there are the indu ed isomorphisms (Lx0 1 x1 ) : 1 (G; x0 ) ! 1 (G; x1 ) and (Rx1 x0 1 ) : 1 (G; x0 ) ! 1 (G; x1 ). 26:A. Let G be a topologi al group, s I ! G be a path. Prove that Ts = (Ls(0) 1 s(1) ) = (Rs(1)s(0) 1 ) : 1 (G; s(0)) ! 1 (G; s(1)): 26:B. Dedu e from 26:A that the fundamental group of a topologi al group is abelian ( f. 25:E). 26:1. Prove that the fundamental groups of the following spa es are
ommutative: (a) the spa e of non-degenerate real n n matri es GL(n; R) = fA j det A 6= 0g; (b) the spa e of orthogonal real n n matri es O(n; R) = fA j A (t A) = 1g; ( ) the spa e of spe ial unitary omplex n n matri es SU (n) = fA j A (t A) = 1; det A = 1g (d) RP n ; (e) Vk;n = Hom(Rk ; Rn );
26:C. Generalize 26:A and 26:B to a homogeneous spa e G=H . 26:D. Riddle. What are the ounterparts for 26:A and 26:B and 26:C
for high homotopy groups?
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27. Covering Spa es De nition Let X , B topologi al spa es, p : X ! B a ontinuous map. Assume that p is surje tive and ea h point of B possesses a neighborhood U su h that the preimage p 1 (U ) of U is presented as a disjoint union of open sets V and p maps ea h V homeomorphi ally onto U . Then p : X ! B is
alled a overing, (of the spa e B ), the spa e B is alled the base of this
overing, X is alled the overing spa e for B and the total spa e of the
overing. Neighborhoods like U are said to be trivially overed. The map p is alled also a overing map, or a overing proje tion. 27.A. Let B be a topologi al spa e and F be a dis rete spa e. Prove that the proje tion prB : B F ! B is a overing. The following statement shows that in a sense lo ally any overing is organized as the overing of 27.A. 27.B. A ontinuous surje tive map p : X ! B is a overing, i for ea h point a of B the preimage p 1 (a) is dis rete and there exist a neighborhood U of a and a homeomorphism h : p 1 (U ) ! U p 1 (a) su h that pjp 1 (U ) = prU Æ h. However, the overings of 27.A are not interesting. They are said to be trivial. Here is the rst really interesting example. 27.C. Prove that R ! S 1 : x 7! e2ix is a overing. To distinguish the most interesting examples, a overing with a onne ted total spa e is alled a overing in narrow sense. Of ourse, the overing of 27.C is a overing in a narrow sense. 27.1.
Any overing is an open map.2
Lo al Homeomorphisms Versus Coverings A map f : X ! Y is said to be lo ally homeomorphi if ea h point of X has a neighborhood U su h that the image f (U ) is open in Y and the map U ! f (U ) de ned by f is a homeomorphism. 27.2. Any overing is lo ally homeomorphi . 27.3. Show that there exists a lo ally homeomorphi map whi h is not a
overing. 27.4. Prove that a restri tion of a lo ally homeomorphi map to an open set is lo ally homeomorphi . 2 Remind
that a map is said to be open if the image of any open set is open.
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27.5. For whi h subsets of R is the restri tion of the map of Problem 27.C a overing. 27.6. Find nontrivial overings X ! B with X homeomorphi to B and prove that they satisfy the de nition of overing.
Number of Sheets Let p : X ! B be a overing. The ardinality (i.e., number of points) of the preimage p 1 (a) of a point a 2 B is alled the multipli ity of the
overing at a or the number of sheets of the overing over a.
27.D.
If the base of a overing is onne ted then the multipli ity of the
overing at a point does not depend on the point.
In the ase of overing with onne ted base the multipli ity is alled the number of sheets of the overing. If the number of sheets is n then the
overing is said to be n-sheeted and we talk about n-fold overing. Of
ourse, unless the overing is trivial, it is impossible to distinguish the sheets of it, but this does not prevent us from speaking about the number of sheets.
More Examples
27.E. Prove that R 2 ! S 1 R : (x; y) 7! (e2ix; y) is a overing. 27.F. Prove that C ! C r 0 : z 7! ez is a overing. In what sense the overings of 27.E and 27.F are the same? De ne an appropriate equivalen e relation for overings.
27.7. Riddle.
27.G. Prove that R 2 ! S 1 S 1 : (x; y) 7! (e2ix; e2iy ) is a overing. 27.H. Prove that for any natural n the map S 1 ! S 1 : z 7! zn is an n-fold overing.
Prove that for any natural n the map C r 0 ! C r 0 : z 7! z n is an n-fold overing.
27.8.
27.I. Prove that for any natural p and q the map S 1 S 1 ! S 1 S 1 : (z; w) 7! (z p ; wq ) is a overing. Find its number of sheets.
Prove that if p : X ! B and p0 : X 0 ! B 0 are overings, then p p0 : X X 0 ! B B 0 is a overing. 27.10. Let p : X ! Y and q : Y ! Z be overings. Prove that if q is nitely-fold then q Æ p : x ! Y is a overing. 27.11*. Show that the assumption about the number of sheets in Problem 27.10 is ne essary. 27.12. Let X be a topologi al spa e, whi h an be presented as a union of open onne ted sets U and V . Prove that if U \ V is dis onne ted then X has a onne ted in nite-fold overing spa e 27.9.
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119
27.J. Prove that the natural proje tion S n ! R P n is a two-fold over-
ing.
27.K. Is (0; 3) ! S 1 : x 7! e2ix a overing? (Cf. 27.5.) 27.L. Is the proje tion R 2 ! R : (x; y) 7! x a overing? Indeed, why not take an open interval (a; b) R as a trivially overed neighborhood: its preimage (a; b) R is the union of open intervals (a; b) fy g whi h are proje ted homeomorphi ally by the proje tion (x; y ) 7! x onto (a; b)? 27.13. Find overings of M obius strip by ylinder. What numbers an you realize as the number of sheets for su h a overing? 27.14. Find non-trivial overings of M obius strip by itself. What numbers
an you realize as the number of sheets for su h a overing? 27.15. Find a two-fold overing of the Klein bottle by torus. Cf. Problem 18.14. 2 1 27.16. Find overings of the Klein bottle by plane R , ylinder S R and a non-trivial overing by itself. What numbers an you realize as the numbers of sheets for su h overings? 27.17. Constru t a overing of the Klein bottle by R2 . Des ribe expli itly the partition of R2 into preimages of points under this overing. 27.18. Constru t a d-fold overing of a sphere with p handles by a sphere with 1 + d(p 1) handles. 27.19. Find a overing of a sphere with any number of ross aps by a sphere with handles.
Universal Coverings A overing p : X ! B is said to be universal if X is simply onne ted. The appearan e of word universal in this ontext will be explained below in Se tion 30. 27.M. Whi h overings of the problems stated above in this se tion are universal?
Theorems on Path Lifting Let p : X ! B and f : A ! B be arbitrary maps. A map g : A ! X su h that p Æ g = f is said to over f or be a lifting of f . A lot of topologi al problems an be phrased in terms of nding a ontinuous lifting of some
ontinuous map. Problems of this sort are alled lifting problems. They may involve additional requirements. For example, the desired lifting has to oin ide with a lifting already given on some subspa e. 27.N. Prove that the identity map S 1 ! S 1 does not admit a ontinuous lifting with respe t to the overing R ! S 1 : x 7! e2ix . (In other words, there exists no ontinuous map g : S 1 ! R su h that e2ig(x) = x for x 2 S 1 .)
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27.O Path Lifting Theorem. Let p : X ! B be a overing, x0 2 X , b0 2 B be points su h that p(x0 ) = b0 . Then for any path s : I ! B starting at b0 there exists a unique path s~ : I ! X starting at x0 and being a lifting of s. (In other words, there exists a unique path s~ : I ! X with s~(0) = x0 and p Æ s~ = s.) 27.O.1 Lemma 1. Let p : X ! B be a trivial overing. Then for any
ontinuous map f of any spa e A to B there exists a ontinuous lifting f~ : A ! X . 27.O.2 Lemma 2. Let p : X ! B be a trivial overing and x0 2 X , b0 2 B be points su h that p(x0 ) = b0 . Then for any ontinuous map f of a spa e A to B mapping a point a0 to b0 , a ontinuous lifting f~ : A ! X with f~(a0 ) = x0 is unique. 27.O.3 Lemma 3. 3 Let p : X ! B be a overing, A a onne ted spa e. If f; g : A ! X are ontinuous maps oin iding in some point and pÆf = pÆg, then f = g. If in the Problem .2 one repla es x0 , b0 and a0 by pairs of points, then it may happen that the lifting problem has no solution f~ with f~(a0 ) = x0 . Formulate a ondition ne essary and suÆ ient for existen e of su h a solution. 27.21. What goes wrong with the Path Lifting Theorem 27.O for the lo al homeomorphism of Problem 27.K? 27.22. Consider the overing C ! C r 0 : z 7! ez : Find liftings of the paths u(t) = 2 t, v(t) = (1 + t)e2it , and their produ t uv. 27.23. Prove that any overing p : X ! B with simply onne ted B and path onne ted X is a homeomorphism.
27.20.
27.P Homotopy Lifting Theorem. Let p : X ! B be a overing, x0 2 X , b0 2 B be points su h that p(x0 ) = b0 . Let u; v : I ! B be paths starting at b0 and u~; v~ : I ! X be the lifting paths for u; v starting at x0 . If the paths u and v are homotopi then the overing paths u~ and v~ are homotopi . 27.Q Corollary. Under the assumptions of Theorem 27.P, the overing paths u~ and v~ have the same nal point (i.e., u~(1) = v~(1)). Noti e that in 27.P and 27.Q paths are assumed to share the initial point x0 . In the statement of 27.Q we emphasize that then they share also the nal point. 27.R Corollary of 27.Q. Let p : X ! B be a overing and s : I ! B be a loop. If there exists a lifting s~ : I ! X of s with s~(0) 6= s~(1) (i.e., there exists a overing path whi h is not a loop), then s is not homotopi to a onstant loop. 3 This
is rather a generalization of the uniqueness, than a ne essary step of the proof. But a good lemma should emphasize the real ontents of the proof, and a generalization is one of the best ways to do this.
28. CALCULATIONS OF FUNDAMENTAL GROUPS
121
Prove that if a pathwise onne ted spa e B has a non trivial pathwise
onne ted overing spa e, then the fundamental group of B is not trivial. 27.25. What orollaries an you dedu e from 27.24 and the examples of
overings presented above in this Se tion? 27.24.
High-Dimensional Homotopy Groups of Covering Spa e 27:A. Let p : X ! B be a overing. Then for any ontinuous map s : I n ! B and a lifting u : I n 1 ! X of the restri tion sjI n 1 there exists a unique lifting of s extending u. 27:B. For any overing p : X ! B and points x0 2 X , b0 2 B su h that p(x0 ) = b0 the homotopy groups r (X; x0 ) and r (B; b0 ) with r > 1 are
anoni ally isomorphi .
27:C. Prove that homotopy groups of dimensions greater than 1 of ir le,
torus, Klein bottle and Mobius strip are trivial.
28. Cal ulations of Fundamental Groups Using Universal Coverings Fundamental Group of Cir le For an integer n denote by sn the loop in S 1 de ned by formula sn (t) = e2int . The initial point of this loop is 1. Denote the homotopy lass of s1 by . Thus 2 1 (S 1 ; 1). Clearly, sn represents n . 28.A. What are the paths in R starting at 0 2 R and overing the loops sn with respe t to the universal overing R ! S 1 ? 28.B. The homomorphism Z ! 1(S 1 ; 1) de ned by formula n 7! n is an isomorphism. 28.B.1. Rephrase the statement that the homomorphism of Theorem 28.B is surje tive in terms of loops and loop homotopies. 28.B.2. Prove that a loop s : I ! S 1 starting at 1 is homotopi to sn if the path s~ : I ! R overing s and starting at 0 2 R nishes at n 2 R (i.e., s~(1) = n). 28.B.3. Rephrase the statement that the homomorphism of Theorem 28.B is inje tive in terms of loops and loop homotopies. 28.B.4. Prove that if loop sn is homotopi to onstant then n = 0. What is the image under2 the isomorphism of Theorem 28.B of the homotopy lass of loop t 7! e2it ?
28.1.
For a loop s : I ! S 1 starting at 1 take the overing path s~ : I ! R starting at 0. By Theorem 27.O su h a path exists and is unique. Its nal point belongs to the preimage of 1 under the universal overing proje tion
28. CALCULATIONS OF FUNDAMENTAL GROUPS
122
R ! S 1 : x 7! e2ix . Hen e, this nal point is an integer n. By 27.Q, it does
not hange if s is repla ed by a homotopi loop. Therefore, this onstru tion provides a well-de ned map 1 (S 1 ; 1) ! Z assigning n to [s℄. Denote this map by deg. 28.2. Prove that deg is an isomorphism inverse to the isomorphism of Theorem 28.B
28.C Corollary of Theorem 28.B.
The fundamental group of (S 1 )n is a free abelian group of rank n (i.e., isomorphi to Zn).
28.D. On torus S 1 S 1 nd two loops whose homotopy lasses generate the fundamental group of the torus.
28.E Corollary of Theorem 28.B. The fundamental group of pun tured plane R 2 r 0 is an in nite y li group. 28.3. Solve Problems 28.C { 28.E without referen e to Theorems 28.B and 25.H, but using expli it onstru tions of the orresponding universal overings.
Fundamental Group of Proje tive Spa e The fundamental group of the proje tive line is an in nite y li group. It is al ulated in the previous subse tion, sin e the proje tive line is a
ir le. The zero-dimensional proje tive spa e is a point, hen e its fundamental group is trivial. Here we al ulate the fundamental groups of proje tive spa es of all other dimensions. Let n 2 and l : I ! R P n be a loop overed by a path ~l : I ! S n whi h onne ts two antipodal points, say the poles P+ = (1; 0; : : : ; 0) and P = ( 1; 0; : : : ; 0), of S n . Denote by the homotopy lass of l. It is an element of 1 (R P n ; (1 : 0 : : 0)). 28.F. For any n 2 group 1 (R P n ; (1 : 0 : : 0)) is a y li group of order 2. It onsists of two elements: and 1. 28.F.1 Lemma. Any loop in RP n at (1 : 0 : : 0) is homotopi either to l or onstant. This depends on whether the overing path of the loop
onne ts the poles P+ and P , or is a loop. 28.4. Where in the proofs of Theorem 28.F and Lemma .1 the assumption n 2 is used?
Fundamental Groups of Bouquet of Cir les Consider a family of topologi al spa es fX g. In ea h of the spa es let a point x be marked. Take the sum q X and identify all the marked points. The resulting quotient spa e is alled the bouquet of fX g and denoted by _ X . Hen e bouquet of q ir les is a spa e whi h is a union of q opies of ir le. The opies meet in a single ommon point, and this
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123
is the only ommon point for any two of them. The ommon point is
alled the enter of the bouquet. Denote the bouquet of q ir les by Bq and its enter by . Let u1 , : : : , uq be loops in Bq starting at and parametrizing the q opies of ir le
omprising Bq . Denote the homotopy lass of ui by i . 28.G. 1(Bq ; ) is a free group freely generated by 1, : : : , q .
Algebrai Digression. Free Groups Re all that a group G is a free group freely generated by its elements a1 , : : : , aq if: ea h its element x 2 G an be expressed as a produ t of powers (with positive or negative integer exponents) of a1 , : : : , aq , i.e., x = aei11 aei22 : : : aeinn and this expression is unique up to the1 following trivial ambiguity: one r s may insert or delete fa tors ai ai and ai 1 ai or repla e am i by ai ai with r + s = m.
28.H.
A free group is de ned up to isomorphism by the number of its free generators.
The number of free generators is alled the rank of the free group. For a standard representative of the isomorphism lass of free groups of rank q one an take the group of words in alphabet of q letters a1 ; : : : ; aq and their inverses a1 1 ; : : : ; aq 1 . Two words represent the same element of the group, i they an be obtained from ea h other by a sequen e of insertions or deletions of fragments ai ai 1 and ai 1 ai . This group is denoted by F (a1 ; : : : ; aq ), or just F q , when the notations for the generators are not to be emphasized. 28.I. Ea h element of F (a1 ; : : : ; aq ) has a unique shortest representative. This is a word without fragments that ould have been deleted.
The number of letters in the shortest representative of x 2 F (a1 ; : : : ; aq ) is alled the length of x and denoted by l(x). Of ourse, this number is not well de ned, unless the generators are xed. Show that an automorphism of Fq an map x 2 Fq to an element with di erent length. For what value of q does su h an example not exist? Is it possible to hange the length in this way arbitrarily? 28.5.
28.J.
A group G is a free group freely generated by its elements a1 , : : : , aq if and only if every map of the set fa1 ; : : : ; aq g to any group X an be extended to a unique homomorphism G ! X .
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Sometimes Theorem 28.J is taken as a de nition of free group. (A definition of this sort emphasizes relations among di erent groups, rather than the internal stru ture of a single group. Of ourse, relations among groups an tell everything about internal a airs of ea h group.) Now we an reformulate Theorem 28.G as follows:
28.K.
The homomorphism
F (a1 ; : : : ; aq ) ! 1 (Bq ; )
taking ai to i for i = 1; : : : ; q is an isomorphism.
First, for the sake of simpli ity let us agree to restri t ourselves to the
ase of q = 2. It would allow us to avoid super uous ompli ations in notations and pi tures. This is the simplest ase, whi h really represents the general situation. The ase q = 1 is too spe ial. To take advantages of this, let us hange notations. Put B = B2 , u = u1 , v = u2 , = 1 , = 2 . Now Theorem 28.K looks as follows: The homomorphism F (a; b) isomorphism.
! (B; ) taking a to and b to
is an
This theorem an be proved like Theorems 28.B and 28.F, provided the universal overing of B is known.
Universal Covering for Bouquet of Cir les Denote by U and V the points antipodal to on the ir les of B . Cut B at these points, removing U and V and putting instead ea h of them two new points. Whatever this operation is, its result is a ross K , whi h is the union of four losed segments with a ommon end point . There appears a natural map P : K ! B , whi h takes the enter of the ross to the enter of B and maps homeomorphi ally the rays of the ross onto half- ir les of B . Sin e the ir les of B are parametrized by loops u and v , the halves of ea h of the ir les are ordered: the orresponding loop passes rst one of the halves and then the other one. Denote by U + the point of P 1 (U ), whi h belongs to the ray mapped by P onto the se ond half of the ir le, and by U the other point of P 1 (U ). Similarly denote points of P 1(V ) by V + and V . The restri tion of P to K r fU + ; U ; V + ; V g maps this set homeomorphi ally onto B rfU; V g. Therefore P provides a overing of B rfU; V g. But it fails to be a overing at U and V : ea h of this points has no trivially
overed neighborhood. Moreover, the preimage of ea h of these points
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onsists of 2 points (the end points of the ross), where P is not even a lo al homeomorphism. To re over, we may atta h a opy of K at ea h of the 4 end points of K and extend P in a natural way to the result. But then new 12 end points, where the map is not a lo al homeomorphism, appear. Well, we repeat the tri k and re over the property of being a lo al homeomorphism at ea h of the new 12 end points. Then we have to do this at ea h of the new 36 points, et . But if we repeat this in nitely many times, all the bad points are turned to ni e ones.4
28.L. Formalize the onstru tion of a overing for B des ribed above.
Consider F (a; b) as a dis rete topologi al spa e. Take K F (a; b). It an be thought of as a olle tion of opies of K enumerated by elements of F (a; b). Topologi ally this is a disjoint sum of the opies, sin e F (a; b) is equipped with dis rete topology. In K F (a; b) identify points (U ; g ) with (U + ; ga) and (V ; g ) with (V + ; gb) for ea h g 2 F (a; b). Denote the resulting quotient spa e by X .
28.M. The omposition of the natural proje tion K F (a; b) ! K and P : K ! B de nes a ontinuous quotient map p : X ! B . 28.N. p : X ! B is a overing. 28.O. X is path- onne ted. For any g 2 F (a; b) there exists a path
onne ting ( ; 1) with ( ; g ) and overing loop obtained from g by substituting a by u and b by v .
28.P. X is simply onne ted.
29. Fundamental Group and Continuous Maps Indu ed Homomorphisms Let f : X ! Y be a ontinuous map of a topologi al spa e X to a topologi al spa e Y . Let x0 2 X and y0 2 Y be points su h that f (x0 ) = 4 This
sounds like a story about a battle with a dragon, but the happy ending demonstrates that modern mathemati ians have a magi power of the sort that the heros of tales ould not dream of. Indeed, we meet a dragon K with 4 heads, ut o all the heads, but, a
ording to the old tradition of the genre, 3 new heads appear in pla e of ea h of the original heads. We ut o them, and the story repeats. We do not even try to prevent this multipli ation of heads. We just ght. But ontrary to the real heros of tales, we a t outside of Time and hen e have no time restri tions. Thus after in nite repetitions of the exer ise with an exponentially growing number of heads we su
eed! No heads left! This is a typi al story about an in nite onstru tion in mathemati s. Sometimes, as in our ase, su h a onstru tion an be repla ed by a nite one, but whi h deals with in nite obje ts. However, there are important
onstru tions, in whi h an in nite fragment is unavoidable.
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y0 . The latter property of f is expressed by saying that f maps pair (X; x0 ) to pair (Y; y0 ) and writing f : (X; x0 ) ! (Y; y0 ).
Denote by f℄ the map (X; x0 ) ! (Y; y0 ) de ned by formula f℄ (s) = f Æ s. This map assigns to a loop its omposition with f . 29.A. f℄ maps homotopi loops to homotopi loops. Therefore f℄ indu es a map 1 (X; x0 ) ! 1 (Y; y0). The latter is denoted by f . 29.B. f : (X; x0 ) ! 1(Y; y0) is a homomorphism for any ontinuous map f : (X; x0 ) ! (Y; y0).
f : (X; x0 ) ! 1 (Y; y0 ) is alled a homomorphism indu ed by f . 29.C. Let f : (X; x0 ) ! (Y; y0) and g : (Y; y0) ! (Z; z0) be ( ontinuous) maps. Then
(g Æ f ) = g Æ f : 1 (X; x0 ) ! 1 (Z; z0 ): 29.D. Let f; g : (X; x0 ) ! (Y; y0) be ontinuous maps homotopi via a homotopy xed at x0 . Then f = g . 29.E. Riddle. How to generalize Theorem 29.D to the ase of freely homotopi f and g ? 29.F. Let f : X ! Y be a ontinuous map, x0 and x1 points of X
onne ted by a path s : I ! X . Denote f (x0 ) by y0 and f (x1 ) by y1 . Then the diagram
1 (X; x0 ) ? ? Ts y
! 1 (Y;? y0)
f
?T y f Æs
1 (X; x1 ) f! 1 (Y; y1 ) is ommutative, i.e., Tf Æs Æ f = f Æ Ts . 29.1. Prove that the map C r 0 ! C r 0 : z 7! z 3 is not homotopi to the identity map C r 0 ! C r 0 : z 7! z . 29.2. Let X be a subset of Rn . Prove that a if a ontinuous map f : X ! Y is extentable to a ontinuous map Rn ! Y then f : 1 (X; x0 ) ! 1 (Y; f (x0 )) is the trivial homomorphism (i.e., maps everything to 1) for any x0 2 X . Prove that a Hausdor spa e, whi h ontains an open set homeomorphi to S 1 S 1 r (1; 1), has an in nite non- y li fundamental group.
29.3.
29.3.1. Prove that a spa e X satisfying the onditions of 29.3 an be ontinuously mapped to a spa e with in nite non- y li fundamental group in su h a way that the map would indu e an epimorphism of 1 (X ) onto this in nite group.
29.4. Prove that the fundamental group of the spa e GL(n; C ) of omplex n n-matri es with non-zero determinant is in nite.
29. FUNDAMENTAL GROUP AND CONTINUOUS MAPS
29.4.1. Constru t ontinuous maps S 1
omposition is the identity.
127
! GL(n; C ) ! S 1, whose
Fundamental Theorem of High Algebra Here our goal is to prove the following theorem, whi h at rst glan e has no relation to fundamental group.
29.G Fundamental Theorem of High Algebra.
Every polynomial of a positive degree in one variable with omplex oeÆ ients has a omplex root. With more details: Let p(z ) = z n + a1 z n 1 + + an be a polynomial of degree n > 0 in z with omplex oeÆ ients. Then there exists a omplex number w su h that p(w) = 0.
Although it is formulated in an algebrai way and alled \The Fundamental Theorem of High Algebra," it has no purely algebrai proof. Its proofs are based either on topologi al arguments or use omplex analysis. This is be ause the eld C of omplex numbers annot be des ribed in purely algebrai terms: all its des riptions involve a sort of ompletion
onstru tion, f. Se tion 15. 29.G.1 Redu tion to Problem on a Map. Dedu e Theorem 29.G from the following statement:
For any omplex polynomial p(z ) of a positive degree the zero belongs to the image of the map C ! C : z 7! p(z ). In other words, the formula z 7! p(z ) does not de ne a map C ! C r 0. 29.G.2 Estimate of Reminder. Let p(z ) = z n + a1 z n 1 + + an be a
omplex polynomial, q(z ) = z n and r(z ) = p(z ) q(z ). Then there exists a positive number R su h that jr(z )j < jq(z )j = Rn for any z with jz j = R 29.G.3 Lemma on Lady with Doggy. (Cf. 23.10.) A lady q(z ) and her dog p(z ) walk on pun tured plane C r 0 periodi ally (i.e., say, with z 2 S 1 ). Prove that if the lady does not let the dog to run further than by jq(z )j from her then the doggy loop S 1 ! C r 0 : z 7! p(z ) is homotopi to the lady loop S 1 ! C r 0 : z 7! q(z ). 29.G.4 Lemma for Dummies. (Cf. 23.11.) If f : X ! Y is a ontinuous map and s : S 1 ! X is a loop homotopi to the trivial one then f Æ s : S 1 ! Y is also homotopi to trivial.
Generalization of Intermediate Value Theorem
29.H. Riddle. How to generalize Intermediate Value Theorem 9.S to the ase of maps f : D2 ! R 2 ?
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128
Let f : D2 ! R2 be a ontinuous map whi h leaves xed ea h point of the bounding ir le S 1 . Then f (D2 ) D2 . 29.5.
29.I. Let f : D2 ! R 2 be a ontinuous map. If f (S 1) does not ontain a 2 R 2 and the ir ular loop f j : S 1 ! R 2 r a de nes a nontrivial element of 1 (R 2 r a) then there exists x 2 D2 su h that f (x) = a.
Let f : R2 ! R2 be a ontinuous map su h that jf (x) xj 1. Prove that f is a surje tion. 29.7. Let u; v : I ! I I be two paths su h that u(0) = (0; 0), u(1) = (1; 1) and v(0) = (0; 1), v(1) = (1; 0). Prove that u(I ) \ v(I ) 6= ?. 29.6.
29.7.1. Let u, v be as in 29.7. Denote by w the map I 2 ! R2 de ned by w(x; y) = u(x) v(y). Prove that 0 2 R2 is a value of w.
Let C be a smooth simple losed urve on the plane with two in e tion points. Prove that there is a line interse ting C in four points a, b, , d with segments [a; b℄, [b; ℄ and [ ; d℄ of the same length.
29.8.
Winding Number As we know (see 28.E), the fundamental group of the pun tured plane R 2 r 0 is Z. There are two isomorphisms whi h di er by multipli ation by 1. We hoose the one whi h maps the homotopy lass of the loop t 7! ( os 2t; sin 2t) to 1 2 Z. In terms of ir ular loops, the isomorphism means that to any loop f : S 1 ! R2 r 0 we asso iate an integer. It is the number of times the loop goes arround 0 in the ounter- lo kwise dire tion. Now we hange the viewpoint in this onsideration, and x the loop, but vary the point. Let f : S 1 ! R2 be a ir ular loop and x 2 R2 r f (S 1 ). Then f de nes an element of 1 (R2 r x) = Z (we hoose basi ally the same identi ation of 1 (R2 r x) with Z assigning 1 to the homotopy
lass of t 7! x + ( os 2t; sin 2t)). This number is denoted by ind(f; x) and alled winding number or index of x with respe t to f . 29:A. Let f : S 1 ! R2 be a loop and x; y 2 R2 r f (S 1 ). Prove that if ind(f; x) 6= ind(f; y) then any path onne ting x and y in R2 meets f (S 1 ). 29:B. Find a loop f : S 1 ! R2 su h that there exist x; y 2 R2 r f (S 1 ) with ind(f; x) = ind(f; y), but lying in di erent onne ted omponents of R2 r f (S 1 ). 29:C. Prove that for any ray R radiating from x the number of points in f 1 (R) is not less than j ind(f; x)j.
Borsuk-Ulam Theorem 29:D One-Dimensional Borsuk-Ulam. For ea h ontinuous map f :
S 1 ! R1 there exists x 2 S 1 su h that f (x) = f ( x).
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29:E Two-Dimensional Borsuk-Ulam. For ea h ontinuous map f :
S 2 ! R2 there exists x 2 S 2 su h that f (x) = f ( x).
29:E:1 Lemma. If there exists a ontinuous map f : S 2 ! R2 with f (x) 6= f ( x) for any x 2 S 2 then there exists a ontinuous map ' : RP 2 ! RP 1 indu ing a non-zero homomorphism 1 (RP 2 ) ! 1 (RP 1 ). : Prove that at ea h instant of time, there is a pair of antipodal points on the earth's surfa e where the pressures and also the temperatures are equal.
29 1.
Theorems 29:D and 29:E are spe ial ases of the following general theorem. We do not assume the reader to be ready to prove Theorem 29:F in the full generality, but is there another easy spe ial ase? 29:F Borsuk-Ulam Theorem. For ea h ontinuous map f : S n ! Rn
there exists x 2 S n su h that f (x) = f ( x).
30. Covering Spa es via Fundamental Groups Homomorphisms Indu ed by Covering Proje tions
30.A. Let p : X ! B be a overing, x0 2 X , b0 = p(x0 ). Then p : 1 (X; x0 ) ! 1 (B; b0 ) is a monomorphism. Cf. 27.P. The image of the monomorphism p : 1 (X; x0 ) ! 1 (B; b0 ) indu ed by a overing proje tion p : X ! B is alled the group of overing p with base point x0 .
30.B. Riddle on Lifting Loops. Des ribe loops in the base spa e of
a overing, whose homotopy lasses belong to the group of the overing, in terms provided by Path Lifting Theorem 27.O.
30.C. Let p : X ! B be a overing, let x0 ; x1 2 X belong to the same path- omponent of X , and b0 = p(x0 ) = p(x1 ). Then p (1 (X; x0 )) and p (1 (X; x1 )) are onjugate subgroups of 1 (B; b0 ) (i.e. there exists an element of 1 (B; b0 ) su h that p (1 (X; x1 )) = 1 p (1 (X; x0 )) ). 30.D. Let p : X ! B be a overing, x0 2 X , b0 = p(x0 ). Let 2 1 (B; b0 ). Then there exists x1 2 p 1 (b0 ) su h that p (1 (X; x1 )) = 1 p (1 (X; x0 )) .
30.E. Let p : X ! B be a overing in a narrow sense and G 1 (B; b0 ) be the group of this overing with base point x0 . A subgroup H
1 (B; b0 ) is a group of the same overing, i it is onjugate to G.
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130
Number of Sheets
30.F Number of Sheets and Index of Subgroup. Let p : X ! B
be a overing in narrow sense with nite number of sheets. Then the number of sheets is equal to the index of the group of this overing. 30.G Sheets and Right Cosets. Let p : X ! B be a overing in narrow sense, b0 2 B , x0 2 p 1 (b0 ). Constru t a natural bije tion of p 1 (b0 ) and the set p (1 (X; x0 ))n1 (B; b0 ) of right osets of the group of the overing in the fundamental group of the base spa e. The number of sheets of a universal overing equals the order of the fundamental group of the base spa e. 30.2 Covering Means Non-Trivial 1 . Any topologi al spa e, whi h has a nontrivial path- onne ted overing spa e, has a nontrivial fundamental group. 30.1 Number of Sheets in Universal Covering.
30:A A tion of 1 in Fiber. Let p : X
! B be a overing, b0 2 B .
Constru t a natural right a tion of 1 (B; b0 ) in p 1 (b0 ). 30:B. When the a tion in 30:A is transitive?
Hierar hy of Coverings Let p : X ! B and q : Y ! B be overings, x0 2 X , y0 2 Y and p(x0 ) = q (y0 ) = b0 . One says that q with base point y0 is subordinate to p with base point x0 if there exists a map ' : X ! Y su h that q Æ ' = p and '(x0 ) = y0 . In this ase the map ' is alled a subordination. 30.H. A subordination is a overing map. 30.I. If a subordination exists, then it is unique. Cf. 27.O. Coverings p : X ! B and q : Y ! B are said to be equivalent if there exists a homeomorphism h : X ! Y su h that p = q Æ h. In this ase h and h 1 are alled equivalen ies 30.J. If two overings are mutually subordinate, then the orresponding subordinations are equivalen ies. 30.K. Let p : X ! B and q : Y ! B be overings, x0 2 X , y0 2 Y and p(x0 ) = q (y0) = b0 . If q with base point y0 is subordinate to p with base point x0 then the group of overing p is ontained in the group of
overing q , i.e. p (1 (X; x0 )) q (1 (Y; y0)). A topologi al spa e X is said to be lo ally path- onne ted if for ea h point a 2 X and ea h neighborhood U of a there is a neighborhood V U whi h is path- onne ted.
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30.L. Let B be a lo ally path- onne ted spa e, p : X ! B and q : Y ! B be overings in narrow sense, x0 2 X , y0 2 Y and p(x0 ) = q (y0 ) = b0 . If p (1 (X; x0 )) q (1 (Y; y0)) then q is subordinate to p.
30.L.1. Under the onditions of 30.L, if paths u; v : I ! X have the same initial point x0 and a ommon nal point, then the paths whi h over p Æ u and p Æ v and have the same initial point y0 also have the same nal point. 30.L.2. Under the onditions of 30.L, the map X ! Y de ned by .1 (guess, what is this map!) is ontinuous.
30.M. Two overings, p : X ! B and q : Y ! B , with a ommon lo ally path- onne ted base are equivalent, i for some x0 2 X and y0 2 Y with p(x0 ) = q (y0 ) = b0 the groups p (1 (X; x0 )) and q (1 (Y; y0 )) are onjugate in 1 (B; b0 ).
To be nished Automorphisms of Covering Regular Coverings Existen e of Coverings Lifting Maps
CHAPTER 5
More Appli ations and Cal ulations 31. Retra tions and Fixed Points Retra tions and Retra ts A ontinuous map of a topologi al spa e onto a subspa e is alled a retra tion if the restri tion of the map to the subspa e is the identity mapping. In other words, if X is a topologi al spa e, A X then : X ! A is a retra tion if it is ontinuous and jA = idA .
31.A. Let be a ontinuous map of a spa e X onto its subspa e A. Then the following statements are equivalent: (a) is a retra tion, (b) (a) = a for any a 2 A, ( ) Æ in = idA , (d) : X ! A is an extension of the identity mapping A ! A.
A subspa e A of a spa e X is said to be a retra t of X if there exists a retra tion X ! A. 31.1.
Any one-point subset is a retra t.
Two-point set may be a non-retra t. 31.2. Any subset of R onsisting of two points is not a retra t of R. 31.3. If A is a retra t of X and B is a retra t of A then B is a retra t of X . 31.4. If A is a retra t of X and B is a retra t of Y then A B is a retra t of X Y . 31.5. A losed interval [a; b℄ is a retra t of R. 31.6. An open interval (a; b) is not a retra t of R. 31.7. What topologi al properties of ambient spa e are inherited by a retra t? 31.8. Prove that a retra t of a Hausdor spa e is losed. 31.9. Prove that the union of Y -axis and the set f(x; y ) 2 R2 : x > 0; y = sin x1 g is not a retra t of R2 and moreover is not a retra t of any of its neighborhoods. 132
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133
The role of the notion of retra t is lari ed by the following theorem.
31.B.
A subset A of a topologi al spa e X is a retra t of X , i any
ontinuous map A ! Y to any spa e Y an be extended to a ontinuous map X ! Y .
Fundamental Group and Retra tions
31.C. If : X ! A is a retra tion, i : A ! X is the in lusion and x0 2 A, then : 1 (X; x0 ) ! 1 (A; x0 ) is an epimorphism and i : 1 (A; x0 ) ! 1 (X; x0 ) is a monomorphism. 31.D. Riddle. Whi h of the two statements of Theorem 31.C (about or i ) is easier to use for proving that a set A X is not a retra t of X?
31.E Borsuk Theorem in Dimension 2. S 1 is not a retra t of D2. 31.10.
Is the proje tive line a retra t of the proje tive plane?
The following problem is more diÆ ult than 31.E in the sense that its solution is not a straightforward onsequen e of Theorem 31.C, but rather demands to reexamine the arguments used in proof of 31.C. 31.11. Prove that the boundary ir le of M obius band is not a retra t of Mobius band. 31.12.
handle.
Prove that the boundary ir le of a handle is not a retra t of the
The Borsuk Theorem in its whole generality annot be dedu ed like Theorem 31.E from Theorem 31.C. However, it an be proven using a generalization of 31.C to higher homotopy groups. Although we do not assume that you an su
essfully prove it now relying only on the tools provided above, we formulate it here.
31.F Borsuk Theorem.
Sphere S n
1
is not a retra t of ball Dn .
At rst glan e this theorem seems to be useless. Why ould it be interesting to know that a map with a very spe ial property of being retra tion does not exists in this situation? However in mathemati s non-existen e theorems may be losely related to theorems, whi h may seem to be more attra tive. For instan e, Borsuk Theorem implies Brower Theorem dis ussed below. But prior to this we have to introdu e an important notion related to Brower Theorem.
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134
Fixed-Point Property. Let f : X ! X be a ontinuous map. A point a 2 X is alled a xed point of f if f (a) = a. A spa e X is said to have the xed-point property if any ontinuous map X ! X has a xed point. Fixed point property means solvability of a wide lass of equations. Prove that the xed point property is a topologi al property. 31.14. A losed interval [a; b℄ has the xed point property. 31.15. Prove that if a topologi al spa e has xed point property then ea h its retra t also has the xed-point property. 31.16. Prove that if topologi al spa es X and Y have xed point property, x0 2 X and y0 2 Y , then X q Y=x0 y0 also has the xed point property. 31.17. Prove that Rn with n > 0 does not have the xed point property. 31.18. Prove that S n does not have the xed point property. 31.19. Prove that RP n with odd n does not have the xed point property. 31.20*. Prove that C P n with odd n does not have the xed point property. 31.13.
Information. R P n and C P n with any even n have the xed point property.
31.G Brower Theorem. Dn has the xed point property. 31.H. Dedu e from Borsuk Theorem in dimension n (i.e., from the state-
ment that S n 1 is not a retra t of Dn ) Brower Theorem in dimension n (i.e., the statement that any ontinuous map Dn ! Dn has a xed point).
32. Homotopy Equivalen es Homotopy Equivalen e as Map Let X and Y be topologi al spa es, f : X ! Y and g : Y ! X ontinuous maps. Consider ompositions f Æ g : Y ! Y and g Æ f : X ! X . They would be equal to the orresponding identity maps, if f and g were homeomorphisms inverse to ea h other. If f Æ g and g Æ f are only homotopi to the identity maps then f and g are said to be homotopy inverse to ea h other. If a ontinuous map possesses a homotopy inverse map then it is alled homotopy invertible or a homotopy equivalen e. 32.A. Prove the following properties of homotopy equivalen es: (a) any homeomorphism is a homotopy equivalen e, (b) a map homotopy inverse to a homotopy equivalen e is a homotopy equivalen e, ( ) the omposition of homotopy equivalen es is a homotopy equivalen e. 32.1.
Find a homotopy equivalen e that is not a homeomorphism.
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Homotopy Equivalen e as Relation Topologi al spa es X and Y are said to be homotopy equivalent if there exists a homotopy equivalen e X ! Y . 32.B. Homotopy equivalen e of topologi al spa es is an equivalen e relation. The lasses of homotopy equivalent spa es are alled homotopy types. Thus homotopy equivalent spa es are said to be of the same homotopy type. Prove that homotopy equivalent spa es have the same number of path onne ted omponents. 32.3. Prove that homotopy equivalent spa es have the same number of onne ted omponents. 32.4. Find in nite series of topologi al spa es belonging to the same homotopy type, but pairwise non-homeomorphi .
32.2.
Deformation Retra tion A retra tion , whi h is homotopy inverse to the in lusion, is alled a deformation retra tion. Sin e is a retra tion, one of the two onditions from the de nition of homotopy inverse maps is satis ed automati ally: its omposition with the in lusion Æ in is equal to the identity idA . The other ondition says that in Æ is homotopi to the identity idX . If X admits a deformation retra tion onto A, then A is alled a deformation retra t of X .
Examples
32.C. Cir le S 1 is a deformation retra t of R 2 r 0 Prove that Mobius strip is homotopy equivalent to ir le. 32.6. Prove that a handle is homotopy equivalent to a union of two ir les interse ting in a single point. 32.7. Prove that a handle is homotopy equivalent to a union of three ar s with ommon end points (i.e., letter ). 32.8. Classify letters of Latin alphabet up to homotopy equivalen e. 32.5.
32.D. Prove that a plane with s points deleted is homotopy equivalent
to a union of s ir les interse ting in a single point. 32.E. Prove that the union of a diagonal of a square and the ontour of the same square is homotopy equivalent to a union of two ir les interse ting in a single point.
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Prove that the spa e obtained from S 2 by identi ation of a two (distin t) points is homotopy equivalent to the union of a two-dimensional sphere and a ir le interse ting in a single point. 2 32.10. Prove that the spa e f(p; q ) 2 C : z + pz + q has two distin t rootsg of quadrati omplex polynomials with distin t roots is homotopy equivalent to the ir le. 32.11. Prove that the spa e GL(n; R) of invertible n n real matri es is homotopy equivalent to the subspa e O(n) onsisting of orthogonal matri es. 32.9.
Deformation Retra tion Versus Homotopy Equivalen e
32.F. Spa es of Problem 32.E annot be embedded one to another. On the other hand, they an be embedded as deformation retra ts to plane with two points removed. Deformation retra tions omprise a spe ial type of homotopy equivalen es. They are easier to visualize. However, as follows from 32.F, homotopy equivalent spa es may be su h that none of them an be embedded to the other one, and hen e none of them is homeomorphi to a deformation retra t of the other one. Therefore deformation retra tions seem to be not suÆ ient for establishing homotopy equivalen es. Though it is not the ase: 32.12*. Prove that any two homotopy equivalent spa es an be embedded as deformation retra ts to the same topologi al spa e.
Contra tible Spa es A topologi al spa e X is said to be ontra tible if the identity map id : X ! X is homotopi to a onstant map. 32.13. Show that R and I are ontra tible. 32.14. Prove that any ontra tible spa e is path- onne ted. 32.15. Prove that the following three statements about a topologi al spa e X are equivalent: (a) X is ontra tible, (b) X is homotopy equivalent to a point, ( ) there exists a deformation retra tion of X onto a point, (d) any point a of X is a deformation retra t of X , (e) any ontinuous map of any topologi al spa e Y to X is homotopi to a
onstant map, (f) any ontinuous map of X to any topologi al spa e Y is homotopi to a
onstant map. 32.16. Is it right that if X is a ontra tible spa e then for any topologi al spa e Y (a) any two ontinuous maps X ! Y are homotopi ?
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(b) any two ontinuous maps Y ! X are homotopi ? 32.17. Che k if spa es of the following list are ontra tible: (a) Rn , (b) a onvex subset of Rn , ( ) a star onvex subset of Rn , (d) f(x; y) 2 R2 : x2 y2 1g, (e) a nite tree (i.e., a onne ted spa e obtained from a nite olle tion of losed intervals by some identifying of their end points su h that deleting of an internal point of ea h of the segments makes the spa e dis onne ted.) 32.18. Prove that X Y is ontra tible, i both X and Y are ontra tible.
Fundamental Group and Homotopy Equivalen es
32.G. Let f : X ! Y and g : Y ! X be homotopy inverse maps, 2 X and y0 2 Y be points su h that f (x0 ) = y0 and g(y0) = x0 and, moreover, the homotopies relating f Æ g to idY and g Æ f to idX are
x0
xed at y0 and x0 , respe tively. Then f and g are inverse to ea h other isomorphisms between groups 1 (X; x0 ) and 1 (Y; y0).
32.H Corollary. If : X ! A is a strong deformation retra tion, x0 2 A, then : 1 (X; x0 ) ! 1 (A; x0 ) and in : 1 (A; x0 ) ! 1 (X; x0 )
are isomorphisms inverse to ea h other.
Cal ulate the fundamental group of the following spa es: (a) Mobius strip, (b) R3 r R1 , ( ) RN r Rn , (d) R3 r S 1 , (e) RN r S n, (f) S 3 r S 1 , (g) S N r S k , (h) RP 3 r RP 1 , (i) handle, (j) sphere with s holes, (k) Klein bottle with a point removed, (l) Mobius strip with s holes. 32.20. Prove that the boundary of the M obius band standardly embedded in R3 (see 18.18) ould not be the boundary of a disk embedded in R3 in su h a way that its interior does not interse t the band. 32.21. Cal ulate the fundamental group of the spa e of all the omplex polynomials ax2 + bx + with distin t roots. Cal ulate the fundamental group of the subspa e of this spa e onsisting of polynomials with a = 1. 32.22. Riddle. Can you solve 32.21 along deriving of the formular for roots of quadrati trinomial? 32.19.
32.I. What if the hypothesis of Theorem 32.G were weakened as follows: g (y0 ) = 6 x0 and/or the homotopies relating f Æ g to idY and g Æ f to idX
33. CELLULAR SPACES
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are not xed at y0 and x0 , respe tively? How would f and g be related? Would 1 (X; x0 ) and 1 (Y; y0 ) be isomorphi ?
33. Cellular Spa es De nition of Cellular Spa es In this se tion we study a lass of topologi al spa es, whi h play an important role in algebrai topology. Their role in the ontext of this book is more restri ted: this is the lass of spa es for whi h we learn how to al ulate the fundamental group. This lass of spa es was introdu ed by J.H.C.Whitehead. He alled these spa es CW - omplexes, and they are known under this name. However, for many reasons it is not a good name. For very rare ex eptions (one of whi h is CW - omplex, other is simpli ial omplex), the word omplex is used nowadays for various algebrai notions, but not for spa es. We have de ided to usethe term ellular spa e instead of CW - omplexes, following D. B. Fu hs and V. A. Rokhlin, Beginner's Course in Topology: Geometri Chapters. Berlin; New York: Springer-Verlag, 1984. A zero-dimensional ellular spa e is just a dis rete spa e. Points of a 0-dimensional ellular spa e are also alled (zero-dimensional) ells or 0- ells. A one-dimensional ellular spa e is a spa e, whi h an be obtained as follows. Take any 0-dimensional ellular spa e X0 . Take a family of maps ' : S 0 ! X0 . Atta h to X0 by ' the sum of a family of opies of D1 (indexed by the same indi es as the maps ' ):
X0 [' (q D1 ): The images of the interior parts of opies of D1 are alled (open) 1dimensional ells, or 1- ells, or edges . The subsets obtained out of D1 are alled losed 1- ells. The ells of X0 (i.e., points of X0 ) are also
alled verti es. Open 1- ells and 0- ells omprise a partition of a onedimensional ellular spa e. This partition is in luded in the notion of
ellular spa e, i.e., a one-dimensional ellular spa e is a topologi al spa e equipped with a partition, whi h an be obtained in this way. One-dimensional ellular spa es are asso iated also with the term graph. However, rather often this term is used for one-dimensional ellular spa es either equipped with additional stru tures (like orientations on edges), or satisfying to additional restri tions (su h as inje tivity of ' ).
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A two-dimensional ellular spa e is a spa e, whi h an be obtained as follows. Take any ellular spa e X1 of dimension 1. Take a family of
ontinuous1 maps ' : S 1 ! X1 . Atta h to X1 by ' the sum of a family of opies of D2 : X1 [' (q D2 ): The images of the interior parts of opies of D2 are alled open 2dimensional ells, or 2- ells, or fa es . The ells of X1 are also onsidered as ells of the 2-dimensional ellular spa e. A set obtained out of a opy of D2 is alled a losed 2- ell. Open ells of both kinds omprise a partition of a 2-dimensional ellular spa e. This partition is in luded in the notion of ellular spa e, i.e., a two-dimensional ellular spa e is a topologi al spa e equipped with a partition, whi h an be obtained in the way des ribed above. A ellular spa e of dimension n is de ned in a similar way: This is a spa e equipped with a partition. It an be obtained from a ellular spa e Xn 1 of dimension < n by atta hing a family of opies of ball Dn by a family of ontinuous maps of their boundary spheres: Xn 1 [' (q Dn ): The images of interior parts of the atta hed n-dimensional balls are
alled (open) n-dimensional ells, or n- ells . The images of the whole n-dimensional balls are alled losed n- ells. Cells of Xn 1 are also
onsidered as ells of the n-dimensional ellular spa e. A ellular spa e is obtained as a union of in reasing sequen e of ellular spa es X0 X1 Xn : : : obtained in this way from ea h other. The sequen e may be nite or in nite. In the latter ase topologi al stru ture is introdu ed by saying that the over of the union by Xn 's is fundamental, i.e., that a set U [1 n=0 Xn is open, i its interse tion U \ Xn with ea h Xn is open in Xn . The union of all ells of dimension n of a ellular spa e X is alled the n-dimensional skeleton of X . This term may be misleading, sin e ndimensional skeleton may be without ells of dimension n, hen e it may
oin ide with (n 1)-dimensional skeleton. Thus n-dimensional skeleton may have dimension < n. Therefore it is better to speak about n-th skeleton or n-skeleton. Cells of dimension n are alled also n- ells. A
ellular spa e is said to be nite if it ontains a nite number of ells. A
ellular spa e is said to be lo ally nite if any its point has a neighborhood whi h interse ts a nite number of ells. A ellular spa e is said to be
ountable if it ontains a ountable number of ells. Let X be a ellular spa e. A subspa e A X , whi h an be presented both as a union 1 Above,
in the de nition of 1-dimensional ellular spa e, the restri tion of ontinuity for ' also ould be stated, but it would be empty, sin e any map of S 0 to any spa e is ontinuous.
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of losed ells and a union of open ells, is alled a ellular subspa e of X . Of ourse, it is provided with a partition into the open ells of X ontained in A. Obviously, the k-skeleton of a ellular spa e X is a
ellular subspa e of X .
33.A. Prove that a ellular subspa e of a ellular spa e is a ellular spa e.
First Examples
33.B. A ellular spa e onsisting of two ells, one 0-dimensional and one n-dimensional, is homeomorphi to S n .
33.C. Present Dn with n > 0 as a ellular spa e made of three ells. 33.D. A ellular spa e onsisting of a single zero-dimensional ell and q one-dimensional ells is a bouquet of q ir les.
33.E. Present torus S 1 S 1 as a ellular spa e with one 0- ell, two 1- ells, and one 2- ell.
33.F. How to obtain a presentation of torus S 1 S 1 as a ellular spa e with 4 ells from a presentation of S 1 as a ellular spa e with 2 ells?
33.1. Prove that if X and Y are nite ellular spa es then X Y an be equipped in a natural way with a stru ture of nite ellular spa e.
Does the statement of 33.1 remain true if one skips the niteness
ondition in it? If yes, prove; if no, nd an example when the produ t is not a ellular spa e. 33.2*.
33.G. Present sphere S n as a ellular spa e su h that spheres S 0 S 1 S 2 S n 1 are its skeletons. 33.H. Present R P n as a ellular spa e with n + 1 ells. Des ribe the
atta hing maps of its ells. 33.3. Present C P n as a ellular spa e with n +1 ells. Des ribe the atta hing maps of its ells.
Present the following topologi al spa es as ellular ones handle, Mobius strip, S1 I , sphere with p handles, sphere with p ross aps.
33.4.
(a) (b) ( ) (d) (e)
33.5. What is the minimal number of ells in a ellular spa e homeomorphi to (a) Mobius strip, (b) sphere with p handles, ( ) sphere with p ross aps?
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Find a ellular spa e, in whi h a losure of a ell is not equal to a union of other ells. What is the minimal number of ells in a spa e ontaining a
ell of this sort? 33.7. Consider a disjoint sum of a ountable olle tion of opies of losed interval I and identify the opies of 0 in all of them. Present the result (whi h is the bouquet of the ountable family of intervals) as a ountable
ellular spa e. Prove that this spa e is not rst ountable. 33.6.
33.I. Present R 1 as a ellular spa e.
Prove that for any two ellular spa es homeomorphi to R1 there exists a homeomorphism between them mapping ea h ell of one of them homeomorphi ally onto a ell of the other one. 33.8.
33.J. Present R n as a ellular spa e.
Denote by R 1 the union of the sequen e of Eu lidean spa es R 0 R 1 R n anoni ally in luded to ea h other: R n = fx 2 R n+1 : xn+1 = 0g. Equip R 1 with the topologi al stru ture, for whi h the spa es R n omprise a fundamental over. 33.K. Present R 1 as a ellular spa e.
More Two-Dimensional Examples Let us onsider a lass of 2-dimensional ellular spa es, whi h admit a simple ombinatorial des ription. Ea h spa e of this lass an be presented as a quotient spa e of a nite family of onvex polygons by identi ation of sides via aÆne homeomorphisms. The identi ation of verti es is de ned by the identi ation of the sides. The quotient spa e is naturally equipped with de omposition into 0- ells, whi h are the images of verti es, 1- ells, whi h are the images of sides, and fa es, the images of the interior parts of the polygons. To des ribe su h a spa e, one needs, rst, to show, what sides are to be identi ed. Usually this is indi ated by writing the same letters at the sides that are to be identi ed. There are only two aÆne homeomorphisms between two losed intervals. To spe ify one of them, it is enough to show orientations of the intervals whi h are identi ed by the homeomorphism. Usually this is done by drawing arrows on the sides. Here is a des ription of this sort for the standard presentation of torus S 1 S 1 as the quotient spa e of square:
It is possible to avoid a pi ture by a des ription. To do this, go around the polygons ounter- lo kwise writing down the letters, whi h stay at
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the sides of polygon along the ontour. The letters orresponding to the sides, whose orientation is opposite to the ounter- lo kwise dire tion, put with exponent 1. This gives rise to a olle tion of words, whi h
ontains a suÆ ient information about the family of polygons and the partition. For instan e, the presentation of torus shown above is en oded by the word ab 1 a 1 b. Prove that: word a 1 a des ribes a ellular spa e homeomorphi to S 2 , word aa des ribes a ellular spa e homeomorphi to RP 2 , word aba 1b 1 des ribes a handle, word ab b 1 des ribes ylinder S 1 I , ea h of the words aab and aba des ribe Mobius strip, word abab des ribes a ellular spa e homeomorphi to RP 2 , ea h of the words aabb and ab 1 ab des ribe Klein bottle, word a1 b1a1 1 b1 1a2 b2 a2 1 b2 1 : : : ag bg ag 1 bg 1 : des ribes sphere with g handles, word a1 a1 a2 a2 : : : ag ag des ribes sphere with g ross aps.
33.9.
(a) (b) ( ) (d) (e) (f) (g) (h)
(i)
Topologi al Properties of Cellular Spa es 33:A. Closed ells omprise a fundamental over of a ellular spa e. 33:B. If A is ellular subspa e of a ellular spa e X then A is losed in
X.
33:C. Prove that any ompa t subset of a ellular spa e interse ts a
nite number of ells. 33:D Corollary. A ellular spa e is ompa t, i it is nite. 33:E. Any ell of a ellular spa e is ontained in a nite ellular subspa e of this spa e. 33:F. Any ompa t subset of a ellular spa e is ontained in a nite
ellular subspa e. 33:G. A ellular spa e is separable, i it is ountable. 33:H. Any path- onne ted omponent of a ellular spa e is a ellular subspa e. 33:I. Any path- onne ted omponent of a ellular spa e is both open and losed. It is a onne ted omponent. In parti ular, a ellular spa e is path- onne ted, i it is onne ted. 33:J. Any onne ted lo ally nite ellular spa e is ountable. 33:K. A ellular spa e is onne ted, i its 1-skeleton is onne ted. 33:L. Any ellular spa e is normal.
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Embedding to Eu lidean Spa e
33.L. Any ountable 0-dimensional ellular spa e an be embedded into R. 33.M. Any ountable lo ally nite 1-dimensional ellular spa e an be embedded into R 3 . 33.10.
Find a 1-dimensional ellular spa e, whi h you annot embed into
R2 . (We do not ask to prove that it is impossible to embed.) 33.N. Any nite dimensional ountable lo ally nite ellular spa e an be embedded into Eu lidean spa e of suÆ iently high dimension. 33.N.1. Let X and Y be topologi al spa es su h that X an be embedded into Rp and Y an be embedded into Rq . Let A be a losed subset of Y . Assume that A has a neighborhood U in Y su h that there exists a homeomorphism h : Cl U ! A I mapping A to A 0. Let ' : A ! X be any ontinuous map. Then there exists an embedding of X [' Y into R p+q+1 . 33.N.2. Let X be a lo ally nite ountable k-dimensional ellular spa e and A be its (k 1)-skeleton. Prove that if A an be embedded to Rp then X an be embedded into Rp+k+1 . 33.O. Any ountable lo ally nite ellular spa e an be embedded into R1 . 33.P. Any ountable lo ally nite ellular spa e is metrizable.
One-Dimensional Cellular Spa es
33.Q.
Any onne ted nite 1-dimensional ellular spa e is homotopy equivalent to a bouquet of ir les. 33.Q.1 Lemma. Let X be a 1-dimensional ellular spa e, and e its 1- ell, whi h is atta hed by an inje tive map S 0 ! X0 (i.e., it has two distin t end points). Prove that the natural proje tion X ! X=e is a homotopy equivalen e. Des ribe the homotopy inverse map expli itly.
A 1-dimensional ellular spa e is alled a tree if it is onne ted and the
omplement of any its 1- ell is not onne ted. 33.R. A ellular spa e X is a tree, i there is no an embedding S 1 ! X . 33.S. Prove that any point of a tree is a deformation retra t of the tree. 33.11.
Prove that any nite tree has xed point property.
Cf. 31.14, 31.15 and 31.16. 33.12. Does the same hold true for any tree, for a nite graph?
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A ellular subspa e A of a ellular spa e X is alled a maximal tree of X if A is a tree and is not ontained in any other ellular subspa e B X , whi h is a tree.
33.T. Prove that any nite onne ted 1-dimensional ellular spa e ontains a maximal tree.
33.U. Prove that a ellular subspa e A of a ellular spa e X is a maximal tree, i it is a tree and the quotient spa e X=A is a bouquet of ir les.
33.V. Let X be a 1-dimensional ellular spa e and A its ellular subspa e. Prove that if A is a tree then the natural proje tion X ! X=A is a homotopy equivalen e.
Problems 33.T, 33.V and 33.U provide a proof of Theorem 33.Q. 33:M. Prove that any 1-dimensional onne ted ellular spa e has a max-
imal tree.
33:N. Any onne ted one-dimensional ellular spa e is homotopy equivalent to a bouquet of ir les. 33:O. Prove that if T is a tree and a ellular subspa e of a ellular spa e
X then the natural proje tion X ! X=T is a homotopy equivalen e.
33:P. Any onne ted ellular spa e is homotopy equivalent to a ellular spa e with 0-skeleton onsisting of one point.
Euler Chara teristi Let X be a nite ellular spa e. Let i (X ) denote the number of its ells of dimension i. Euler hara teristi of X is the alternating sum of i (X ):
(X ) = 0 (X ) 1 (X ) + 2 (X )
+(
1)i i (X ) + : : :
. 33:Q. Prove that Euler hara teristi is additive in the following sense:
for any ellular spa e X and its nite ellular subspa es A and B
(A [ B ) = (A) + (B ) (A \ B ): 33:R. Prove that Euler hara teristi is multipli ative in the following sense: for any nite ellular spa es X and Y the Euler hara teristi of their produ t X Y is (X ) (Y ).
33.W.
A nite onne ted ellular spa e X of dimension one is homotopy equivalent to the bouquet of 1 (X ) ir les.
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145
34. Fundamental Group of a Cellular Spa e One-Dimensional Cellular Spa es
34.A. If X is a nite 1-dimensional ellular spa e, then 1 (X ) is a free group of rank 1
(X ).
34.B. Homotopy Classi ation of Finite 1-Dimensional Cellular Spa es. Two nite 1-dimensional ellular spa es are homotopy equivalent, i their Euler hara teristi s are equal.
Prove that the fundamental group of 2-dimensional sphere with n points removed is a free group of rank n 1. 3 34.2 Euler Theorem. For any bounded onvex polyhedron in R the number of edges plus 2 is equal to the sum of the numbers of verti es and fa es.
34.1.
34.C. Let X be a nite 1-dimensional ellular spa e, T a maximal tree of X and x0 2 T . For ea h ell e X r T hoose a loop se, whi h starts
at x0 , goes inside T to e, then goes on e along e and then omes ba k to x0 in T . Prove that 1 (X; x0 ) is freely generated by homotopy lasses of se .
Generators
34.D. Let A be a topologi al spa e, x0 2 A. Let ' : S k 1 ! A be a
ontinuous map, X = A [' Dk . Prove that if k > 1 then the in lusion homomorphism 1 (A; x0 ) ! 1 (X; x0 ) is surje tive. Cf. .5, .4. 34.E. Let X be a ellular spa e, x0 its 0- ell and X1 the 1-skeleton of X . Then the in lusion homomorphism 1 (X1 ; x0 ) ! 1 (X; x0 )
is surje tive.
34.F. Let X be a nite ellular spa e, T a maximal tree of X1 and x0 2 T . For ea h ell e X1 r T hoose a loop se , whi h starts at x0 ,
goes inside T to e, then goes on e along e and then omes ba k to x0 in T . Prove that 1 (X; x0 ) is generated by homotopy lasses of se . 34.3. 34.4.
Dedu e Theorem 25.G from Theorem 34.E. Find 1 (C P n ).
Relators Let X be a ellular spa e, x0 its 0- ell. Denote by Xn the n-skeleton of X . Re all that X2 is obtained from X1 by atta hing opies of disk D2 by
ontinuous maps ' : S 1 ! X1 . The atta hing maps are ir ular loops in X1 . For ea h hoose a path s : I ! X1 onne ting ' (1) with x0 .
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146
Denote by N the normal subgroup of 1 (X; x0 ) generated (as a normal subgroup2 .) by elements Ts [' ℄ 2 1 (X1 ; x0 ): 34.G. Prove that N does not depend on the hoi e of paths s . 34.H. N oin ides with the kernel of the in lusion homomorphism i : 1 (X1 ; x0 ) ! 1 (X; x0 ): 34.H.1 Lemma 1. N Ker i , f. 25.J ( ). 34.H.2 Lemma 2. Let p1 : Y1 ! X1 be a overing with overing group N . Then for any and a point y 2 p1 1 (' (1)) there exists a lifting 'e : S 1 ! Y1 of ' with 'e (1) = y. 34.H.3 Lemma 3. Let Y2 be a ellular spa e obtained by atta hing opies of disk to Y1 by all liftings of atta hing maps ' . Then there exists a map p2 : Y2 ! X2 extending p1 and this is a overing. 34.H.4 Lemma 4. Any loop s : I ! X1 realizing an element of the kernel of the in lusion homomorphism 1 (X1 ; x0 ) ! 1 (X2 ; x0 ) (i.e., homotopi to onstant in X2 ) is overed by a loop of Y2 . The overing loop is ontained in Y1 . 34.H.5 Lemma 5. N oin ides with the kernel of the in lusion homomorphism 1 (X1 ; x0 ) ! 1 (X2 ; x0 ). 34.H.6 Lemma 6. Atta hing maps of n- ells with n 3 are lifted to any
overing spa e. Cf. 27:A, 27:B. 34.H.7 Lemma 7. Covering p2 : Y2 ! X2 an be extended to a overing of the whole X . 34.H.8 Lemma 8. Any loop s : I ! X1 realizing an element of Ker i (i.e., homotopi to onstant in X ) is overed by a loop of Y . The overing loop is ontained in Y1 .
Writing Down Generators and Relators Theorems 34.F and 34.H imply the following pres ription for writing down presentation for the fundamental group of a nite dimensional ellular spa e by generators and relators: Let X be a nite ellular spa e, x0 its 0- ell. Let T a maximal tree of 1-skeleton of X . For ea h 1- ell e 6 T of X hoose a loop se , whi h starts at x0 , goes inside T to e, then goes on e along e and then omes ba k to x0 in T . Let g1 , : : : , gm be the homotopy lasses of these loops. 2 Re all that a subgroup is said to be normal
if it oin ides with onjugate subgroups. The normal subgroup generated by a set A is the minimal normal subgroup ontaining A. As a subgroup, it is generated by elements of A and elements onjugate to them. This means that ea h element of this normal subgroup is a produ t of elements
onjugate to elements of A
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Let '1 ; : : : ; 'n : S 1 ! X1 be atta hing maps of 2- ells of X . For ea h 'i
hoose a path si onne ting 'i (1) with x0 in 1-skeleton of X . Express the homotopy lass of the loop si 1 'i si as a produ t of powers of generators gj . Let r1 , : : : , rn are the words in letters g1 , : : : , gm obtained in this way. The fundamental group of X is generated by g1 , : : : , gm , whi h are subje t to de ning relators r1 = 1, : : : , rn = 1. 34.I. Che k that this rule gives orre t answers in the ases of R P n and S 1 S 1 for the ellular presentations of these spa es provided in Problems 33.H and 33.E.
Fundamental Groups of Basi Surfa es
34.J.
tation
The fundamental group of a sphere with g handles admits presen-
fa1; b1 ; a2; b2 ; : : : ag ; bg 34.K.
:
a1 b1 a1 1 b1 1 a2 b2 a2 1 b2 1 : : : ag bg ag 1 bg 1 = 1g: The fundamental group of a sphere with g ross aps admits pre-
sentation
a1 ; a2 ; : : : ag : a21 a22 : : : a2g = 1 : 34.L. Prove that fundamental groups of spheres with di erent number of handles are not isomorphi . When one needs to prove that two nitely presented groups are not isomorphi , one of the rst natural moves is to abelianize the groups. Re all that to abelianize a group G means to quotient it out by the ommutator subgroup. The ommutator subgroup [G; G℄ is the normal subgroup generated by ommutators a 1 b 1 ab for all a; b 2 G. Abelianization means adding relations that ab = ba for any a; b 2 G. Abelian nitely generated groups are well known. Any nitely generated abelian group is isomorphi to a produ t of a nite number of y li groups. If the abelianized groups are not isomorphi then the original groups are not isomorphi as well. 34.L.1. Abelianized fundamental group of a sphere with g handles is a free abelian group of rank 2g (i.e., is isomorphi to Z2g). 34.L.2. Prove that fundamental groups of spheres with di erent number of ross aps are not isomorphi . 34.L.3. Abelianized fundamental group of a sphere with g ross aps is isomorphi to Zg 1 Z2.
34.M. Spheres with di erent numbers of handles are not homotopy equivalent.
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34.N.
Spheres with di erent numbers of ross aps are not homotopy equivalent.
34.O.
A sphere with handles is not homotopy equivalent to a sphere with ross aps.
If X is a path- onne ted spa e then the abelianized fundamental group of X is alled the 1-dimensional (or rst) homology group of X and denoted by H1 (X ). If X is not path- onne ted then H1 (X ) is the dire t sum of the rst homology groups of all path- onne ted omponents of X . Thus .1 an be rephrased as follows: if Fg is a sphere with g handles then H1 (Fg ) = Z2g. Seifert - van Kampen Theorem Let X be a onne ted ellular spa e, A and B its ellular subspa es whi h
over X . Denote A \ B by C . 34:A. How fundamental groups of X , A, B and C are related? 34:B Seifert - van Kampen Theorem. Suppose A, B , and C are
onne ted. Let x0 2 C , 1 (A; x0 ) = f 1 ; : : : ; p : 1 = 1; : : : ; r = 1g; 1 (B; x0 ) = f 1 ; : : : ; q : 1 = 1; : : : ; s = 1g; and 1 (C; x0 ) be generated by 1 , : : : t . Let the images of i under the in lusion homomorphisms 1 (C; x0 ) ! 1 (A; x0 ) and 1 (C; x0 ) ! 1 (B; x0 ) be expressed as i ( 1 ; : : : ; p ) and i ( 1 ; : : : ; q ), respe tively. Then
1 (X ) = f 1 ; : : : ; p ; 1 ; : : : ; q : 1 = 1; : : : ; r = 1; 1 = 1; : : : ; s = 1; 1 = 1 ; : : : ; t = t g: 34:C. Let X , A, B and C be as above. Suppose A; B are simply onne ted and C onsists of two path onne ted omponents. Prove that 1 (X ) is isomorphi to Z.
To write details: van Kampen published mu h more general theorem!
35. One-Dimensional Homology and Cohomology Sometimes the fundamental group ontains too mu h information to deal with, and it is more onvinient to ignore a part of this information. A regular way to do his is to use some of the natural quotient groups of the fundamental group. One of the quotients, the abelianized fundamental
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group, was introdu ed and used in Se tion 34 to prove, in parti ular, that spheres with di erent numbers of handles are not homotopy equivalent, see Problems 34.L, .1-.3 and 34.M. Re all that for a path- onne ted spa e X the abelianized fundamental group of X is alled its one-dimensional homology group and denoted by H1 (X ). If X is an arbitrary topologi al spa e then H1 (X ) is the dire t sum of the one-dimensional homology groups of all the onne ted
omponents of X . In this Se tion we will study the one-dimensional homology and its losest relatives. Usually they are studied in the framework of homology theory together with high-dimensional generalizations. This general theory requires mu h more algebra and takes more time and e orts. On the other hand, one-dimensional ase is useful on its own, involves a lot of spe i details and provides a geometri intuition, whi h is useful, in parti ular, for studying high-dimensional homology. First, few new words. Elements of a homology group is alled homology
lasses. They really admit several interpretations as equivalen e lasses of obje ts of various nature. For example, a
ording to the de nition we start with, a homology lass is a oset onsisting of elements of the fundamental group. In turn, ea h element of the fundamental group
onsists of loops. Thus, we an think of a homology lass as of a set of loops. A loop whi h belongs to the zero homology lass is said to be zero-homologous. Loops, whi h belong to the same homology lass, are said to be homologous to ea h other. 35:A Zero-Homologous Loop. Let X be a topologi al spa e. A ir ular loop s : S 1 ! X is zero-homologous, i there exist a ontinuous map f of a disk D with handles (i.e., a sphere with a hole and handles) to X and a homeomorphism h of S 1 onto the boundary ir le of D su h that f Æ h = s. 35:A:1. In the fundamental group of a disk with handles, a loop, whose homotopy lass generates the fundamental group of the boundary ir le, is homotopi to a produ t of ommutators of meridian and longitude loops of the handles. A homotopy between a loop and a produ t of ommutators of loops
an be thought of as an extension of the loop to a ontinuous map of a sphere with handles and a hole.
Des ription of H1 (X ) in Terms of Free Cir ular Loops Fa torization by the ommutator subgroup kills the di eren e between translation maps de ned by di erent paths. Therefore the abelianized fundamental groups of a path- onne ted spa e an be naturally identi ed. Hen e ea h free loop de nes a homology lass. This suggests that H1 (X )
an be de ned starting with free loops, rather than loops at a base point.
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35:B. On the sphere with two handles and three holes shown in Figure
1 the sum of the homology lasses of the three loops, whi h go ounter lo kwise arround the three holes, is zero.
Sphere with two handles and three holes. The boundary ir les of the holes are equipped with arrows showing the ounter- lo kwise orientation. Figure 1.
35:C Zero-Homologous Colle tions of Loops. Let X be a pathwise
onne ted spa e and s1 ; : : : ; sn : S 1 ! X be a olle tion of n free loops. Prove that the sum of homology lasses of s1 , : : : , sn is equal to zero, i there exist a ontinuous map f : F ! X , where F is a sphere with handles and n holes, and embeddings i1 ; : : : ; in : S 1 ! F parametrizing the boundary ir les of the holes in the ounter- lo kwise dire tion (as in Figure 1) su h that sk = f Æ ik for k = 1; : : : ; n. 35:D Homologous Colle tions of Loops. In a topologi al spa e X any lass 2 H1 (X ) an be represented by a nite olle tion of free
ir ular loops. Colle tions fu1 ; : : : ; up g and fv1 ; : : : ; vq g of free ir ular loops in X de ne the same homology lass, i there exist a ontinuous map f : F ! X , where F is a disjoint sum of several spheres with handles and holes with the total number of holes equal p + q, and embeddings i1 ; : : : ; ip+q : S 1 ! F parametrizing the boundary ir les of all the holes of F in the ounter- lo kwise dire tion su h that uk = f Æ ik for k = 1; : : : ; p and vk 1 = f Æ ik+p for k = 1; : : : ; q. Find H1 (X ) for the following spa es Mobius strip, handle, sphere with p handles and r holes, sphere with p ross aps and r holes, the omplement in R3 of the ir les f(x; y; z ) 2 R3 j z = 0; x2 + y2 = 1g and f(x; y; z ) 2 R3 j x = 0; z 2 + (y 1)2 = 1g, (f) the omplement in R3 of the ir les f(x; y; z ) 2 R3 j z = 0; x2 + y2 = 1g and f(x; y; z ) 2 R3 j z = 1; x2 + y2 = 1g,
: (a) (b) ( ) (d) (e)
35 1.
One-Dimensional Cohomology Let X be a path- onne ted topologi al spa e and G a ommutative group. 35:E. The homomorphisms 1 (X; x0 ) ! G omprise a ommutative group in whi h the group operation is the pointwise addition.
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The group Hom(1 (X; x0 ); G) of all the homomorphisms 1 (X; x0 ) ! G is alled one-dimensional ohomology group of X with oeÆ ients in G and denoted by H 1 (X ; G). For an arbitrary topologi al spa e X , the one-dimensional ohomology group of X with oeÆ ients in G is de ned as the dire t produ t of one-dimensional ohomology group with oeÆ ients in G of all the path onne ted omponents of X . 35:F Cohomology via Homology. H 1 (X ; G) = Hom(H1 (X ); G).
The following subse tion is to be rewritten when the se tion on lassi ation of overings will be done! Cohomology and Classi ation of Regular Coverings Re all that a overing p : X ! B is a regular G- overing if X is a G-spa e, in whi h the orbits of the a tion of G are the bers of p and G a ts e e tively on ea h of them. Regular G- overing may be with dis onne ted total spa e. For example, X G ! X is a regular G overing. For any loop s : I ! B in the base B of a regular G- overing p : X ! B there is a map Ms : p 1 (s(0)) ! p 1 (s(0)) assigning to x 2 p 1 (s(0)) the nal point of the path overing s 1 and beginning at x. This map is alled the monodromy transformation of p 1 (s(0)) de ned by s. It oin ides with a tion of one of the elements of G. In this way a homomorphism 1 (B ) ! G is de ned. It is alled the monodromy representaion of the fundamental group. Thus any regular G- overing of X de nes a
ohomology lass belonging to H 1 (X ; G). 35:G Cohomology and Regular Coverings. This map is a bije tion of the set of all the regular G- overings of X onto H 1 (X ; G). 35:2 Addition of G-Coverings. What operation on the set of regular G- overings orresponds to addition of ohomology lasses?
Integer Cohomology and Maps to S 1 Let X be a topologi al spa e and f : X ! S 1 a ontinuous map. It indu es a homomorphism f : H1 (X ) ! H1 (S 1 ) = Z. Therefore it de nes an element of H 1 (X ; Z). 35:H. This onstru tion de nes a bije tion of the set of all the homotopy
lasses of maps X ! S 1 onto H 1 (X ; Z).
35:I Addition of Maps to Cir le. What operation on the set of ho-
motopy lasses of maps to S 1 orresponds to the addition in H 1 (X ; Z)?
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35:J. What regular Z- overing of X orresponds to a homotopy lass of
mappings X ! S 1 under the ompositions of the bije tions des ribed in 35:H and 35:G
One-Dimensional Homology Modulo 2 Here we de ne yet another natural quotient group of the fundamental group. It is even simpler than H1 (X ). For a path- onne ted X , onsider the quotient group of 1 (X ) by the normal subgroup generated by squares of all the elements of (X ). It is denoted by H1 (X ; Z2) and alled one-dimensional homology group of X with oeÆ ients in Z2. For an arbitrary X , the group H1 (X ; Z2) is de ned as the sum of one-dimensional homology group with oeÆ ients in Z2 of all the path- onne ted omponents of X . Elements of H1 (X ; Z2) are alled one-dimensional homology lasses modulo 2 or one-dimensional homology lasses with oeÆ ients in Z2. They
an be thought of as lasses of elements of the fundamental groups or
lasses of loops. A loop de ning the zero homology lass modulo 2 is said to be zero-homologous modulo 2. 35:K. In a disk with ross aps the boundary loop is zero-homologous
modulo 2.
35:L Loops Zero-Homologous Modulo 2. Prove that a ir ular loop s : S 1 ! X is zero-homologous modulo 2, i there exist a ontinuous map f of a disk with ross aps D to X and a homeomorphism h of S 1 onto the boundary ir le of D su h that f Æ h = s. 35:M. If a loop is zero-homologous then it is zero-homologous modulo
2.
35:N Homology and Mod 2 Homology. H1 (X ; Z2) is ommutative for any X , and an be obtained as the quotient group of H1 (X ) by the subgroup of all even homology lasses, i.e. elements of H1 (X ) of the form 2 with 2 H1 (X ). Ea h element of is of order 2 and H1 (X ; Z2) is a ve tor spa e over the eld of two elements Z2. Find H1 (X ; Z2) for the following spa es Mobius strip, handle, sphere with p handles, sphere with p ross aps, sphere with p handles and r holes, sphere with p ross aps and r holes, the omplement in R3 of the ir les f(x; y; z ) 2 R3 j z = 0; x2 + y2 = 1g and f(x; y; z ) 2 R3 j x = 0; z 2 + (y 1)2 = 1g, (h) the omplement in R3 of the ir les f(x; y; z ) 2 R3 j z = 0; x2 + y2 = 1g and f(x; y; z ) 2 R3 j z = 1; x2 + y2 = 1g, : (a) (b) ( ) (d) (e) (f) (g)
35 3.
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: Z2-Homology of Cellular Spa e. Dedu e from the al ulation of the fundamental group of a ellular spa e (see Se tion 34) an algorithm for al ulation of the one-dimensional homology group with Z2
oeÆ ients of a ellular spa e. 35 4
35:O Colle tions of Loops Homologous Mod 2. Let X be a topo-
logi al spa e. Any lass 2 H1 (X ; Z2) an be represented by a nite olle tion of free ir ular loops in X . Colle tions fu1 ; : : : ; up g and fv1 ; : : : ; vq g of free ir ular loops in X de ne the same homology lass modulo 2, i there exist a ontinuous map f : F ! X , where F is a disjoint sum of several spheres with ross aps and holes with the total number of holes equal p+q, and embeddings i1 ; : : : ; ip+q : S 1 ! F parametrizing the boundary ir les of all the holes of F su h that uk = f Æ ik for k = 1; : : : ; p and vk = f Æ ik+p for k = 1; : : : ; q. : Compare 35:O with 35:D. Why in 35:O the ounter- lo kwise dire tion does not appear? In what other aspe ts 35:O is simpler than 35:D and why?
35 5.
35:P Duality Between Mod 2 Homology and Cohomology.
H 1 (X ; Z2) = Hom(H1 (X ; Z2); Z2) = HomZ2(H1 (X ; Z2); Z2) for any spa e X . If H1 (X ; Z2) is nite then H1 (X ; Z2) and H 1 (X ; Z2) are nite-dimensional ve tor spa es over Z2 dual to ea h other. 35:6. A loop is zero-homologous modulo 2 in X , i it is overed by a loop in any two-fold overing spa e of X .
35:Q. Riddle. Homology Modulo n? Generalize all the theory above
about Z2-homology to de ne and study Zn-homology for any natural n.
Part 3
Manifolds
This part is devoted to study of the most important topologi al spa es. These spa es provide a s ene for most of geometri bran hes of mathemati s.
CHAPTER 6
Bare Manifolds 36. Lo ally Eu lidean Spa es De nition of Lo ally Eu lidean Spa e Let n be a non-negative integer. A topologi al spa e X is alled a lo ally Eu lidean spa e of dimension n if ea h point of X has a neighborhood homeomorphi either to R n or R n+ . Re all that R n+ = fx 2 R n : x1 0g, it is de ned for n 1.
36.A. The notion of 0-dimensional lo ally Eu lidean spa e oin ides with the notion of dis rete topologi al spa e. 36.B. Prove that the following spa es are lo ally Eu lidean: (a) R n , (b) any open subset of R n ,
( ) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n)
S n, RP n , C P n, R n+ , any open subset of R n+ , Dn , torus S 1 S 1 , handle, sphere with handles, sphere with holes, Klein bottle, sphere with ross aps.
Prove that an open subspa e of a lo ally Eu lidean spa e of dimension n is a lo ally Eu lidean spa e of dimension n.
36.1.
36.2.
Prove that a bouquet of two ir les is not lo ally Eu lidean.
36.C. If X is a lo ally Eu lidean spa e of dimension p and Y is a lo ally Eu lidean spa e of dimension q then X Y is a lo ally Eu lidean spa e
of dimension p + q .
156
36. LOCALLY EUCLIDEAN SPACES
157
Dimension
36.D. Can a topologi al spa e be simultaneously a lo ally Eu lidean spa e of dimension both 0 and n > 0? 36.E. Can a topologi al spa e be simultaneously a lo ally Eu lidean spa e of dimension both 1 and n > 1? Prove that any nonempty open onne ted subset of a lo ally Eu lidean spa e of dimension 1 an be made dis onne ted by removing two points. 36.4. Prove that any nonempty lo ally Eu lidean spa e of dimension n > 1 ontains a nonempty open set, whi h annot be made dis onne ted by removing any two points.
36.3.
36.F. Can a topologi al spa e be simultaneously a lo ally Eu lidean spa e of dimension both 2 and n > 2? 36.G. Let U be an open subset of R 2 and a p 2 U . Prove that 1 (U r fpg) admits an epimorphism onto Z. 36.H. Dedu e from 36.G that a topologi al spa e annot be simultaneously a lo ally Eu lidean spa e of dimension both 2 and n > 2. We see that dimension of lo ally Eu lidean topologi al spa e is a topologi al invariant at least for the ases when it is not greater than 2. It is orre ted without this restri tion. However, one needs some te hnique to prove this. One possibility is provided by dimension theory, see, e.g., W. Hurewi z and H. Wallman, Dimension Theory Prin eton, NJ, 1941. Other possibility is to generalize the arguments used in 36.H to higher dimensions. However, this demands a knowledge of high-dimensional homotopy groups. 36.5. Dedu e that a topologi al spa e annot be simultaneously a lo ally Eu lidean spa e of dimension both n and p > n from the fa t that n 1 (S n 1 ) = Z. Cf. 36.H
Interior and Boundary A point a of a lo ally Eu lidean spa e X is said to be an interior point of X if a has a neighborhood (in X ) homeomorphi to R n . A point a 2 X , whi h is not interior, is alled a boundary point of X . 36.6.
Whi h points of Rn+ have a neighborhood homeomorphi to Rn+ ?
36.I. Formulate a de nition of boundary point independent of a de nition for interior point. Let X be a lo ally Eu lidean spa e of dimension n. The set of all interior points of X is alled the interior of X and denoted by int X . The set
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of all boundary points of X is alled the boundary of X and denoted by X . These terms (interior and boundary) are used also with di erent meaning. The notions of boundary and interior points of a set in a topologi al spa e and the interior part and boundary of a set in a topologi al spa e are introdu ed in general topology, see Se tion 5. They have almost nothing to do with the notions dis ussed here. In both senses the terminology is
lassi al, whi h is impossible to hange. This does not reate usually a danger of onfusion. Notations are not as ommonly a
epted as words. We take an easy opportunity to sele t unambiguous notations: we denote the interior part of a set A in a topologi al spa e X by IntX A or Int A, while the interior of a lo ally Eu lidean spa e X is denoted by int X ; the boundary of a set in a topologi al spa e is denoted by symbol Fr, while the boundary of lo ally Eu lidean spa e is denoted by symbol . 36.J. For a lo ally Eu lidean spa e X the interior int X is an open dense subset of X , the boundary X is a losed nowhere dense subset of X . 36.K. The interior of a lo ally Eu lidean spa e of dimension n is a lo ally Eu lidean spa e of dimension n without boundary (i.e., with empty boundary; in symbols: (int X ) = ?). 36.L. The boundary of a lo ally Eu lidean spa e of dimension n is a lo ally Eu lidean spa e of dimension n 1 without boundary (i.e., with empty boundary; in symbols: (X ) = ?). 36.M. int R n+ fx 2 R n : x1 > 0g and R n+ fx 2 R n : x1 = 0g: 36.7. For any x; y 2 fx 2 Rn : x1 = 0g, there exists a homeomorphism f : Rn+ ! Rn+ with f (x) = y. 36.N. Either R n+ = ? (and then X = ? for any lo ally Eu lidean spa e X of dimension n), or R n+ = fx 2 R n : x1 = 0g: In fa t, the se ond alternative holds true. However, this is not easy to prove for any dimension. 36.O. Prove that R 1+ = f0g. 36.P. Prove that R 2+ = fx 2 R 2 : x1 = 0g. (Cf. 36.G.) n n : x = 0g from n 1 ) = Z. (Cf. 36.8. Dedu e that a R+ = fx 2 R 1 n 1 (S 36.P, 36.5)
36.Q. Dedu e from R n+ = fx 2 R n : x1 = 0g for all n 1 that int(X Y ) = int X int Y
37. MANIFOLDS
and
159
(X Y ) = ( (X ) Y ) [ (X Y ):
The last formula resembles Leibniz formula for derivative of a produ t.
36.R. Riddle. Can this be a matter of han e? 36.S. Prove that (a) (I I ) = (I I ) [ (I I ),
(b) Dn = S n 1 , ( ) (S 1 I ) = S 1 I = S 1 q S 1 , (d) the boundary of Mobius strip is homeomorphi to ir le.
36.T Corollary. Mobius strip is not homeomorphi to ylinder S 1 I .
37. Manifolds De nition of Manifold A topologi al spa e is alled a manifold of dimension n if it is lo ally Eu lidean of dimension n, se ond ountable, Hausdor .
37.A. Prove that the three onditions of the de nition are independent
(i.e., there exist spa es not satisfying any one of the three onditions and satisfying the other two.) 37.A.1. Prove that R [i R, where i : fx 2 R : x < 0g ! R is the in lusion, is a non-Hausdor lo ally Eu lidean spa e of dimension one.
37.B. Che k whether the spa es listed in Problem 36.B are manifolds. A ompa t manifold without boundary is said to be losed. As in the ase of interior and boundary, this term oin ides with one of the basi terms of general topology. Of ourse, the image of a losed manifold under embedding into a Hausdor spa e is a losed subset of this Hausdor spa e (as any ompa t subset of a Hausdor spa e). However absen e of boundary does not work here, and even non- ompa t manifolds may be losed subsets. They are losed in themselves, as any spa e. Here we meet again an ambiguity of lassi al terminology. In the ontext of manifolds the term losed relates rather to the idea of a losed surfa e.
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160
Components of Manifold
37.C. A onne ted omponent of a manifold is a manifold. 37.D. A onne ted omponent of a manifold is path- onne ted. 37.E. A onne ted omponent of a manifold is open in the manifold. 37.F. A manifold is the sum of its onne ted omponents. 37.G. The set of onne ted omponents of any manifold is ountable.
If the manifold is ompa t, then the number of the omponents is nite. 37.1.
Prove that a manifold is onne ted, i its interior is onne ted.
37.H. The fundamental group of a manifold is ountable. Making New Manifolds out of Old Ones
37.I. Prove that an open subspa e of a manifold of dimension n is a
manifold of dimension n.
37.J. The interior of a manifold of dimension n is a manifold of dimension n without boundary. 37.K. The boundary of a manifold of dimension n is a manifold of dimension n 1 without boundary. 37.2. The boundary of a ompa t manifold of dimension n is a losed manifold of dimension n 1.
37.L. If X is a manifold of dimension p and Y is a manifold of dimension q then X Y is a manifold of dimension p + q . 37.M. Prove that a overing spa e (in narrow sense) of a manifold is a manifold of the same dimension.
37.N. Prove that if the total spa e of a overing is a manifold then the base is a manifold of the same dimension. 37.O. Let X and Y be manifolds of dimension n, A and B omponents of X and Y respe tively. Then for any homeomorphism h : B ! A the spa e X [h Y is a manifold of dimension n. 37.O.1. Prove that the result of gluing of two opy of Rn+ by the identity map of the boundary hyperplane is homeomorphi to Rn .
37.P. Let X and Y be manifolds of dimension n, A and B losed subsets
of X and Y respe tively. If A and B are manifolds of dimension n 1 then for any homeomorphism h : B ! A the spa e X [h Y is a manifold of dimension n.
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Double
37.Q. Can a manifold be embedded into a manifold of the same dimension without boundary? Let X be a manifold. Denote by DX the spa e X [idX X obtained by gluing of two opies of X by the identity mapping idX : X ! X of the boundary. 37.R. Prove that DX is a manifold without boundary of the same dimension as X .
DX is alled the double of X . 37.S. Prove that a double of a manifold is ompa t, i the original manifold is ompa t. Collars and Bites Let X be a manifold. An embedding : X I ! X su h that (x; 0) = x for ea h x 2 X is alled a ollar of X . A ollar an be thought of as a neighborhood of the boundary presented as a ylinder over boundary. 37:A. Every manifold has a ollar. Let U be an open set in the boundary of a manifold X . For a ontinuous fun tion ' : X ! R+ with ' 1 (0; 1) = U set B' = f(x; t) 2 X R+ : t '(x)g: A bite on X at U is an embedding b : B' ! X with some ' : X ! R+ su h that b(x; 0) = x for ea h x 2 X . This is a generalization of ollar. Indeed, a ollar is a bite at U = X with ' = 1. 37:A:1. Prove that if U X is ontained in an open subset of X homeomorphi to Rn+ , then there exists a bite of X at U . 37:A:2. Prove that for any bite b : B ! X of a manifold X the
losure of X r b(B ) is a manifold. 37:A:3. Let b1 : B1 ! X be a bite of X and b2 : B2 ! Cl(X r b1 (B1 )) be a bite of Cl(X r b1 (B1 )). Constru t a bite b : B ! X of X with b(B ) = b1 (B1 ) [ b2 (B2 ). 37:A:4. Prove that if there exists a bite of X at X then there exists a ollar of X . 37:B. For any two ollars 1 ; 2 : X I ! X there exists a homeomorphism h : X ! X with h(x) = x for x 2 X su h that h Æ 1 = 2 . This means that a ollar is unique up to homeomorphism.
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37:B:1. For any ollar : X I ! X there exists a ollar 0 : X I ! X su h that (x; t) = 0 (x; t=2). 37:B:2. For any ollar : X I ! X there exists a homeomorphism h : X ! X [x7!(x;1) X I with h( (x; t)) = (x; t).
38. Isotopy Isotopy of Homeomorphisms Let X and Y be topologi al spa es, h; h0 : X ! Y homeomorphisms. A homotopy ht : X ! Y , t 2 [0; 1℄ onne ting h and h0 (i.e., with h0 = h, h1 = h0 ) is alled an isotopy between h and h0 if ht is a homeomorphism for ea h t 2 [0; 1℄. Homeomorphisms h, h0 are said to be isotopi if there exists an isotopy between h and h0 . 38.A. Being isotopi is an equivalen e relation on the set of homeomorphisms X ! Y . 38.B. Find a topologi al spa e X su h that homotopy between homeomorphisms X ! X does not imply isotopy. This means that isotopy lassi ation of homeomorphisms an be more re ned than homotopy lassi ation of them. 38.1. 38.2.
Classify homeomorphisms of ir le S 1 to itself up to isotopy. Classify homeomorphisms of line R1 to itself up to isotopy.
The set of isotopy lasses of homeomorphisms X ! X (i.e. the quotient of the set of self-homeomorphisms of X by isotopy relation) is alled the mapping lass group or homeotopy group of X . 38.C. For any topologi al spa e X , the mapping lass group of X is a group under the operation indu ed by omposition of homeomorphisms. 38.3. Find the mapping lass group of the union of the oordinate lines in the plane. 38.4. Find the mapping lass group of the union of bouquet of two ir les.
Isotopy of Embeddings and Sets Homeomorphisms are topologi al embeddings of spe ial kind. The notion of isotopy of homeomorphism is extended in an obvious way to the ase of embeddings. Let X and Y be topologi al spa es, h; h0 : X ! Y topologi al embeddings. A homotopy ht : X ! Y , t 2 [0; 1℄ onne ting
38. ISOTOPY
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h and h0 (i.e., with h0 = h, h1 = h0 ) is alled an (embedding) isotopy between h and h0 if ht is an embedding for ea h t 2 [0; 1℄. Embeddings h, h0 are said to be isotopi if there exists an isotopy between h and h0 . 38.D. Being isotopi is an equivalen e relation on the set of embeddings X ! Y. A family At , t 2 I of subsets of a topologi al spa e X is alled an isotopy of the set A = A0 , if the graph = f(x; t) 2 X I j x 2 At g of the family is brewise homeomorphi to the ylinder A I , i. e. there exists a homeomorphism A I ! mapping A ftg to \ X ftg for any t 2 I . Su h a homeomorphism gives rise to an isotopy of embeddings t : A ! X , t 2 I with 0 = in, t (A) = At . An isotopy of a subset is also alled a subset isotopy. Subsets A and A0 of the same topologi al spa e X are said to be isotopi in X , if there exists a subset isotopy At of A with A0 = A1 . 38.E. It is easy to see that this is an equivalen e relation on the set of subsets of X . As it follows immediately from the de nitions, any embedding isotopy determines an isotopy of the image of the initial embedding and any subset isotopy is a
ompanied with an embedding isotopy. However the relation between the notions of subset isotopy and embedding isotopy is not too lose be ause of the following two reasons: (a) an isotopy t a
ompanying a subset isotopy At starts with the in lusion of A0 (while arbitrary isotopy may start with any embedding); (b) an isotopy a
ompanying a subset isotopy is determined by the subset isotopy only up to omposition with an isotopy of the identity homeomorphism A ! A (an isotopy of a homeomorphism is a spe ial
ase of embedding isotopies, sin e homeomorphisms an be onsidered as a sort of embeddings). An isotopy of a subset A in X is said to be ambient, if it may be a
ompanied with an embedding isotopy t : A ! X extendible to an isotopy ~ t : X ! X of the identity homeomorphism of the spa e X . The isotopy ~ t is said to be ambient for t . This gives rise to obvious re nements of the equivalen e relations for subsets and embeddings introdu ed above. 38.F. Find isotopi , but not ambiently isotopi sets in [0; 1℄. 38.G. If sets A1 ; A2 X are ambiently isotopi then the omplements X r A1 and X r A2 are homeomorphi and hen e homotopy equivalent. 38.5. Find isotopi , but not ambiently isotopi sets in R. 38.6. Prove that any isotopi ompa t subsets of R are ambiently isotopi . 38.7.
Find isotopi , but not ambiently isotopi ompa t sets in R3 .
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Prove that any two embeddings S 1 ! R3 are isotopi . Find embeddings S 1 ! R3 that are not ambiently isotopi . 38.8.
Isotopies and Atta hing 38:A. Any isotopy ht : X
! X extends to an isotopy Ht : X ! X .
38:B. Let X and Y be manifolds of dimension n, A and B omponents of X and Y respe tively. Then for any isotopi homeomorphisms f; g : B ! A the manifolds X [f Y and X [g Y are homeomorphi . 38:C. Let X and Y be manifolds of dimension n, let B be a ompa t subset of Y . If B is a manifold of dimension n 1 then for any embeddings f; g : B ! X ambiently isotopi in X the manifolds X [f Y and X [g Y are homeomorphi .
Conne ted Sums
38.H. Let X and Y be manifolds of dimension n, and ' : R n ! X , : R n ! Y be embeddings. Then X r '(Int Dn ) [ (S n )!X r'(Int Dn ): (a)7!'(a) Y r (Int Dn) is a manifold of dimension n.
This manifold is alled a onne ted sum of X and Y . 38.I. Show that the topologi al type of the onne ted sum of X and Y depends not only on the topologi al types of X and Y . 38.J. Let X and Y be manifolds of dimension n, and ' : R n ! X , : R n ! Y be embeddings. Let h : X ! X be a homeomorphism. Then the onne ted sums of X and Y de ned via and ', on one hand, and via and h Æ ', on the other hand, are homeomorphi . 38.9. Find pairs of manifolds onne ted sums of whi h are homeomorphi to (a) S 1 , (b) Klein bottle, ( ) sphere with three ross aps. 38.10. Find a dis onne ted onne ted sum of onne ted manifolds. Des ribe, under what ir umstan es this an happen.
39. One-Dimensional Manifolds Zero-Dimensional Manifolds This se tion is devoted to topologi al lassi ation of manifolds of dimension one. We skip the ase of 0-dimensional manifolds due to triviality
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of the problem. Indeed, any 0-dimensional manifold is just a ountable dis rete topologi al spa e, and the only topologi al invariant needed for topologi al lassi ation of 0-manifolds is the number of points: two 0dimensional manifolds are homeomorphi , i they have the same number of points. The ase of 1-dimensional manifolds is also simple, but it requires more detailed onsideration.
Redu tion to Conne ted Manifolds Sin e ea h manifold is the sum of its onne ted omponents, two manifolds are homeomorphi if and only if there exists a one-to-one orresponden e between their omponents su h that the orresponding omponents are homeomorphi . Therefore for topologi al lassi ation of n-manifolds it suÆ es to lassify only onne ted n-manifolds.
Examples
39.A. What onne ted 1-manifolds do you know? (a) (b) ( ) (d)
Do you know any losed onne ted 1-manifold? Do you know a onne ted ompa t 1-manifold, whi h is not losed? What non- ompa t onne ted 1-manifolds do you know? Is there a non- ompa t onne ted 1-manifolds with boundary? 39.B. Fill the following table with pluses and minuses. Manifold X Is X ompa t? Is X empty?
S1
R1 I
R 1+ Statements of Main Theorems
39.C.
Any onne ted manifold of dimension 1 is homeomorphi to one of the following for manifolds: ir le S 1, line R 1 , interval I , half-line R 1+ .
This theorem may be splitted into the following four theorems:
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39.D.
Any losed onne ted manifold of dimension 1 is homeomorphi to ir le S 1 .
39.E.
Any non- ompa t onne ted manifold of dimension 1 without boundary is homeomorphi to line R 1 .
39.F.
Any ompa t onne ted manifold of dimension 1 with nonempty boundary is homeomorphi to interval I .
39.G.
Any non- ompa t onne ted manifold of dimension one with nonempty boundary is homeomorphi to half-line R 1+ .
Lemma on 1-Manifold Covered with Two Lines
39.H Lemma. Any onne ted manifold of dimension 1 overed with two open sets homeomorphi to R 1 is homeomorphi either to R 1 , or S 1 . Let X be a onne ted manifold of dimension 1 and U; V X be its open subsets homeomorphi to R. Denote by W the interse tion U \ V . Let ' : U ! R and : V ! R be homeomorphisms. 39.H.1. Prove that ea h onne ted omponent of '(W ) is either an open interval, or an open ray, or the whole R. 39.H.2. Prove that a homeomorphism between two open onne ted subsets of R is a (stri tly) monotone ontinuous fun tion. 39.H.3. Prove that if a sequen e xn of points of W onverges to a point a 2 U r W then it does not onverge in V . 39.H.4. Prove that if there exists a bounded onne ted omponent C of '(W ) then C = '(W ), V = W , X = U and hen e X is homeomorphi to R. 39.H.5. In the ase of onne ted W and U 6= V , onstru t a homeomorphism X ! R whi h takes: W to (0; 1), U to (0; +1), and V to ( 1; 1). 39.H.6. In the ase of W onsisting of two onne ted omponents, onstru t a homeomorphism X ! S 1 , whi h takes: p p W to fz 2 S 1 : 1=p 2 < Im(z) < 1= 2g, U to fz 2 S 1 : 1= 2 < Im( p z)g, and V to fz 2 S 1 : Im(z) < 1= 2g.
Without Boundary 39.D.1. Dedu e Theorem 39.D from Lemma 39.G.
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39.E.1. Dedu e from Lemma 39.G that for any onne ted non- ompa t one-dimensional manifold X without a boundary there exists an embedding X ! R with open image. 39.E.2. Dedu e Theorem 39.E from .1.
With Boundary 39.F.1. Prove that any ompa t onne ted manifold of dimension 1 an be embedded into S 1 . 39.F.2. List all onne ted subsets of S 1 . 39.F.3. Dedu e Theorem 39.F from .2, and .1. 39.G.1. Prove that any non- ompa t onne ted manifold of dimension 1
an be embedded into R1 . 39.G.2. Dedu e Theorem 39.G from .1.
Consequen es of Classi ation
39.I. Prove that onne ted sum of losed 1-manifolds is de ned up homeomorphism by topologi al types of summands. 39.J. Whi h 0-manifolds bound a ompa t 1-manifold? Mapping Class Groups
39.K. Find the mapping lass groups of (a) S 1 , (b) R 1 , ( ) R 1+ , (d) [0; 1℄.
Find the mapping lass group of an arbitrary 1-manifold with nite number of omponents.
39.1.
40. Two-Dimensional Manifolds Examples
40.A. What onne ted 2-manifolds do you know? (a) (b) ( ) (d)
List losed onne ted 2-manifold that you know. Do you know a onne ted ompa t 2-manifold, whi h is not losed? What non- ompa t onne ted 2-manifolds do you know? Is there a non- ompa t onne ted 2-manifolds with boundary?
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Constru t non-homeomorphi non- ompa t onne ted manifolds of dimension two without boundary and with isomorphi in nitely generated fundamental group. 40.1.
Ends and Odds Let X be a non- ompa t Hausdor topologi al spa e, whi h is a union of an in reasing sequen e of its ompa t subspa es C1 C2 Cn X: Ea h onne ted omponent U of X r Cn is ontained in some onne ted
omponent of X r Cn 1 . A de reasing sequen e U1 U2 Un : : : of onne ted omponents of (X r C1 ) (X r C2 ) (X r Cn ) : : : respe tively is alled an end of X with respe t to C1 Cn : : : . 40:A. Let X and Cn be as above, D be a ompa t set in X and V a
onne ted omponent of X r D. Prove that there exists n su h that D Cn . 40:B. Let X and Cn be as above, Dn be an in reasing sequen e of
ompa t sets of X with X = [1 n=1 Dn . Prove that for any end U1 Un : : : of X with respe t to Cn there exists a unique end V1 Vn : : : of X with respe t to Dn su h that for any p there exists q su h that Vq Up . 40:C. Let X , Cn and Dn be as above. Then the map of the set of ends of X with respe t to Cn to the set of ends of X with respe t to Dn de ned by the statement of 40:B is a bije tion. Theorem 40:C allows one to speak about ends of X without spe ifying a system of ompa t sets C1 C2 Cn X with X = [1 n=1 Cn . Indeed, 40:B and 40:C establish a anoni al one-toone orresponden e between ends of X with respe t to any two systems of this kind. 40:D. Prove that R1 has two ends, Rn with n > 1 has one end. 40:E. Find the number of ends for the universal overing spa e of the bouquet of two ir les. 40:F. Does there exist a 2-manifold with a nite number of ends whi h
annot be embedded into a ompa t 2-manifold? 40:G. Prove that for any ompa t set K S 2 with onne ted omplement S 2 r K there is a natural map of the set of ends of S 2 r K to the set of onne ted omponents of K . Let W be an open set of X . The set of ends U1 Un : : : of X su h that Un W for suÆ iently large n is said to be open.
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40:H. Prove that this de nes a topologi al stru ture in the set of ends
of X .
The set of ends of X equipped with this topologi al stru ture is alled the spa e of ends of X . Denote this spa e by E (X ). 40.1:1. Constru t non- ompa t onne ted manifolds of dimension two without boundary and with isomorphi in nitely generated fundamental group, but with non-homeomorphi spa es of ends. 40.1:2. Constru t non- ompa t onne ted manifolds of dimension two without boundary and with isomorphi in nitely generated fundamental group, but with di erent number of ends. 40.1:3. Constru t non- ompa t onne ted manifolds of dimension two without boundary with isomorphi in nitely generated fundamental group and the same number of ends, but with di erent topology in the spa e of ends. 40.1:4. Let K be a ompletely dis onne ted losed set in S 2 . Prove that the map E (S 2 r K ) ! K de ned in 40:G is ontinuous. 40.1:5. Constru t a ompletely dis onne ted losed set K S 2 su h that this map is a homeomorphism.
40.B. Prove that there exists an un ountable family of pairwise non-
homeomorphi onne ted 2-manifolds without boundary.
The examples of non- ompa t manifolds dimension 2 presented above show that there are too many non- ompa t onne ted 2-manifolds. This makes impossible any useful topologi al lassi ation of non- ompa t 2-manifolds. Theorems redu ing the homeomorphism problem for 2manifolds of this type to the homeomorphism problem for their spa es of ends do not seem to be really useful: spa es of ends look not mu h simpler than the surfa es themselves. However, there is a spe ial lass of non- ompa t 2-manifolds, whi h admits a simple and useful lassi ation theorem. This is the lass of simply
onne ted non- ompa t 2-manifolds without boundary. We postpone its
onsideration to the end of this se tion. Now we turn to the ase, whi h is the simplest and most useful for appli ations.
Closed Surfa es
40.C.
Any onne ted losed manifold of dimension two is homeomorphi either to sphere S 2 , or sphere with handles, or sphere with ross aps.
Re all that a
ording to Theorem 34.M the basi surfa es represent pairwise distin t topologi al (and even homotopy) types. Therefore, 34.M
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and 40.C together give topologi al and homotopy lassi ations of losed 2-dimensional manifolds. We do not re ommend to prove Theorem 40.C immediately and, espe ially, in the formulation given here. All known proofs of 40.C an be de omposed into two main stages: rstly, a manifold under onsideration is equipped with some additional stru ture (like triangulation or smooth stru ture); then using this stru ture a required homeomorphism is onstru ted. Although the rst stage appears in the proof ne essarily and is rather diÆ ult, it is not useful outside the proof. Indeed, any losed 2-manifold, whi h we meet in a on rete mathemati al ontext, is either equipped, or an be easily equipped with the additional stru ture. The methods of imposing the additional stru ture are mu h easier, than a general proof of existen e for this stru ture in arbitrary 2-manifold. Therefore, we suggest for the rst ase to restri t ourselves to the se ond stage of the proof of Theorem 40.C, prefa ing it with general notions related to the most lassi al additional stru ture, whi h an be used for this purpose.
Triangulations of Surfa es By an Eu lidean triangle we mean the onvex hall of three non- ollinear points of Eu lidean spa e. Of ourse, it is homeomorphi to disk D2 , but not only the topologi al stru ture is relevant for us now. The boundary of a triangle ontains three distinguished points, its verti es, whi h separates the boundary into three pie es, its sides. A topologi al triangle in a topologi al spa e X is an embedding of an Eu lidean triangle into X . A vertex (respe tively, side ) of a topologi al triangle T ! X is the image of a vertex ( respe tively, side) of T in X . A set of topologi al triangles in a 2-manifold X is a triangulation of X provided the images of these triangles omprise a fundamental over of X and any two of the images either are disjoint or interse t in a ommon side or in a ommon vertex. 40.D. Prove that in the ase of ompa t X the former ondition (about fundamental over) means that the number of triangles is nite. 40.E. Prove that the ondition about fundamental over means that the
over is lo ally nite.
Two Properties of Triangulations of Surfa es Triangulations of surfa es are not rami ed
40.F. Let E be a side of a triangle involved into a triangulation of a 2-manifold X . Prove that there exist at most two triangles of this triangulation for whi h E is a side. Cf. 36.G, 36.H and 36.P.
40. TWO-DIMENSIONAL MANIFOLDS
Lo al strong
onne tedness
Triangulations present a surfa e
ombinatorially.
171
40.G. Let V be a vertex of a triangle involved into a triangulation of
a 2-manifold X and T , T 0 be two triangles of the triangulation adja ent to V . Prove that there exisits a sequen e T = T1 ; T2 ; : : : ; Tn = T 0 of triangles of the triangulation su h that V is a vertex of ea h of them and triangles Ti , Ti+1 have ommon side for ea h i = 1; : : : ; n 1. S heme of Triangulation Let X be a 2-manifold and T a triangulation of X . Denote the set of verti es of T by V . Denote by 2 the set of triples of verti es, whi h are verti es of a triangle of T . Denote by 1 the set of pairs of verti es, whi h are verti es of a side of T . Put 0 = S . This is the set of verti es of T . Put = 2 [ 1 [ 0 . The pair (V; ) is alled the ( ombinatorial) s heme of T . 40:I. Prove that the ombinatorial s heme (V; ) of a triangulation of a 2-manifold has the following properties: (a) is a set onsisting of subsets of V , (b) ea h element of onsists of at most 3 elements of V , ( ) three-element elements of over V , (d) any subset of an element of belongs to , (e) interse tion of any olle tion of elements of belongs to , (f) for any two-element element of there exist exa tly two threeelement elements of ontaining it. Let V be a set and is a set of nite subsets of V . The pair (V; ) is
alled a triangulation s heme if any subset of an element of belongs to , interse tion of any olle tion of elements of belongs to , any one element subset of V belongs to . There is a natural way to asso iate a topologi al spa e (in fa t, a ellular spa e) to any triangulation s heme. Namely, for a triangulation s heme (V; ) onsider the set S (V; ) of all fun tions : V ! I (= [0; 1℄) su h that Supp( ) = fv 2 V : (v) 6= 0g P belongs to and v2V (v) = 1. Equip S (V; ) with the ompa t open topology. 40:J. Prove that S (V; ) is a ellular spa e with ells f 2 S : Supp( ) = g with 2 . 40:K. Prove that if (V; ) is the ombinatorial s heme of a triangulation of a 2-manifold X then S (V; ) is homeomorphi to X . 40:L. Let (V; ) be a triangulation s heme su h that (a) V is ountable, (b) ea h element of onsists of at most 3 elements of V , ( ) three-element elements of over V ,
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(d) for any two-element element of there exist exa tly two threeelement elements of ontaining it Prove that (V; ) is a ombinatorial s heme of a triangulation of a 2manifold.
Examples 40.2. Consider the over of torus obtained in the obvious way from the
over of the square by its halves separated by a diagonal of the square. Is it a triangulation of torus? Why not?
Figure 1
Prove that the simplest triangulation of S 2 onsists of 4 triangles. 40.4*. Prove that a triangulation of torus S 1 S 1 ontains at least 14 A lot! triangles, and a triangulation of the proje tive plane ontains at least 10 Just say NO to triangles. triangulations.
40.3.
Families of Polygons The problems onsidered above show that triangulations provide a ombinatorial des ription of 2-dimensional manifolds, but this des ription is usually too bulky. Here we will study other, more pra ti al way to present 2-dimensional manifolds ombinatorially. The main idea is to use larger building blo ks. Let F be a olle tion of onvex polygons P1 ; P2 ; : : : . Let the sides of these polygons be oriented and paired o . Then we say that this is a family of polygons. There is a natural quotient spa e of the sum of polygons involved in a family: one identi es ea h side with its pair-mate by a homeomorphism, whi h respe ts the orientations of the sides. This quotient spa e is alled just the quotient of the family. 40.H. Prove that the quotient of the family of polygons is a 2-manifold without boundary. 40.I. Prove that the topologi al type of the quotient of a family does not
hange when the homeomorphism between the sides of a distinguished pair is repla ed by other homeomorphism whi h respe ts the orientations. 40.J. Prove that any triangulation of a 2-manifold gives rise to a family of polygon whose quotient is homeomorphi to the 2-manifold.
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A family of polygons an be des ribed ombinatorially: Assign a letter to ea h distinguished pair of sides. Go around the polygons writing down the letters assigned to the sides and equipping a letter with exponent 1 if the side is oriented against the dire tion in whi h we go around the polygon. At ea h polygon we write a word. The word depends on the side from whi h we started and on the dire tion of going around the polygon. Therefore it is de ned up to y li permutation and inversion. The olle tion of words assigned to all the polygons of the family is alled a phrase asso iated with the family of polygons. It des ribes the family to the extend suÆ ient to re overing the topologi al type of the quotient. 40.5. Prove that the quotient of the family of polygons asso iated with phrase aba 1 b 1 is homeomorphi to S 1 S 1 . 40.6. Identify the topologi al type of the quotient of the family of polygons asso iated with phrases (a) aa 1 ; (b) ab, ab; ( ) aa; (d) abab 1; (e) abab; (f) ab ab ; (g) aabb; (h) a1 b1 a1 1 b1 1 a2 b2 a2 1 b2 1 : : : ag bg ag 1 bg 1 ; (i) a1 a1 a2 a2 : : : ag ag .
40.K. A olle tion of words is a phrase asso iated with a family of polygons, i ea h letter appears twi e in the words.
A family of polygons is alled irredu ible if the quotient is onne ted. 40.L. A family of polygons is irredu ible, i a phrase asso iated with it does not admit a division into two olle tions of words su h that there is no letter involved in both olle tions.
Operations on Family of Polygons Although any family of polygons de nes a 2-manifold, there are many families de ning the same 2-manifold. There are simple operations whi h
hange a family, but do not hange the topologi al type of the quotient of the family. Here are the most obvious and elementary of these operations. (a) Simultaneous reversing orientations of sides belonging to one of the pairs. (b) Sele t a pair of sides and subdivide ea h side in the pair into two sides. The orientations of the original sides de ne the orderings of the halves. Unite the rst halves into one new pair of sides, and the se ond halves into the other new pair. The orientations of the original sides de ne in an obvious way orientations of their halves.
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This operation is alled 1-subdivision. In the quotient it e e ts in subdivision of a 1- ell (whi h is the image of the sele ted pair of sides) into two 1- ells. This 1- ells is repla ed by two 1- ells and one 0- ell. ( ) The inverse operation to 1-subdivision. It is alled 1- onsolidation. (d) Cut one of the polygons along its diagonal into two polygons. The sides of the ut omprise a new pair. They are equipped with an orientation su h that gluing the polygons by a homeomorphism respe ting these orientations re overs the original polygon. This operation is alled 2-subdivision. In the quotient it e e ts in subdivision of a 2- ell into two new 2- ells along an ar whose end-points are 0- ells (may be oin iding). The original 2- ell is repla ed by two 2- ells and one 1- ell. (e) The inverse operation to 2-subdivision. It is alled 2- onsolidation.
Topologi al and Homotopy Classi ation of Closed Surfa es
40.M Redu tion Theorem. Any nite irredu ible family of polygons
an be redu ed by the ve elementary operations to one of the following standard families: (a) aa 1 (b) a1 b1 a1 1 b1 1 a2 b2 a2 1 b2 1 : : : ag bg ag 1 bg 1 ( ) a1 a1 a2 a2 : : : ag ag for some natural g . 40.N Corollary. Any triangulated losed onne ted manifold of dimen-
sion 2 is homeomorphi to either sphere, or sphere with handles, or sphere with ross aps.
Theorems 40.N and 34.M provide lassi ations of triangulated losed
onne ted 2-manifolds up to homeomorphisms and homotopy equivalen e. 40.M.1 Redu tion to Single Polygon. Any nite irredu ible family of polygons an be redu ed by elementary operations to a family onsisting of a single polygon. 40.M.2 Can ellation. A family of polygons orresponding to a phrase
ontaining a fragment aa 1 or a 1 a, where a is any letter, an be transformed by elementary operations to a family orresponding to the phrase obtained from the original one by erasing this fragment, unless the latter is the whole original phrase. 40.M.3 Redu tion to Single Vertex. An irredu ible family of polygons
an be turned by elementary transformations to a polygon su h that all its verti es are proje ted to a single point of the quotient.
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40.M.4 Separation of Cross ap. A family orresponding to a phrase
onsisting of a word XaY a, where X and Y are words and a is a letter,
an be transformed to the family orresponding to the phrase bbY 1 X . 40.M.5. If a family, whose quotient has a single vertex in the natural ell de omposition, orresponds to a phrase onsisting of a word XaY a 1 , where X and Y are nonempty words and a is a letter, then X = UbU 0 and Y = V b 1 V 0 . 40.M.6 Separation of Handle. A family orresponding to a phrase onsisting of a word UbU 0 aV b 1 V 0 a 1 , where U , U 0 , V , and V 0 are words and a, b are letters, an be transformed to the family presented by phrase d d 1 1 UV 0 V U 0 . 40.M.7 Handle plus Cross ap Equals 3 Cross aps. A family orresponding to phrase aba 1 b 1
X an be transformed by elementary transformations to the family orresponding to phrase abdbadX .
Re ognizing Closed Surfa es
40.O. What is the topologi al type of the 2-manifold, whi h an be
obtained as follows: Take two disjoint opies of disk. Atta h three parallel strips onne ting the disks and twisted by . The resulting surfa e S has a onne ted boundary. Atta h a opy of disk along its boundary by a homeomorphism onto the boundary of the S . This is the spa e to re ognize.
40.P. Euler hara teristi of the ellular spa e obtained as quotient of a family of polygons is invariant under homotopy equivalen es. How an 40.P help to solve 40.O? 40.8. Let X be a losed onne ted surfa e. What values of (X ) allow to re over the topologi al type of X ? What ambiguity is left for other values of (X )? 40.7.
Orientations By an orientation of a segment one means an ordering of its end points (whi h one of them is initial and whi h one is nal). By an orientation of a polygon one means orientation of all its sides su h that ea h vertex is the nal end point for one of the adja ent sides and initial for the other one. Thus an orientation of a polygon in ludes orientation of all its sides. Ea h segment an be oriented in two ways, and ea h polygon
an be oriented in two ways. An orientation of a family of polygons is a olle tion of orientations of all the polygons omprising the family su h that for ea h pair of sides one of the pair-mates has the orientation inherited from the orientation of the polygon ontaining it while the other pair-mate has the orientation
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opposite to the inherited orientation. A family of polygons is said to be orientable if it admits an orientation. Whi h of the families of polygons from Problem 40.6 are orientable? 40.10. Prove that a family of polygons asso iated with a word is orientable i ea h letter appear in the word on e with exponent 1 and on e with exponent 1.
40.9.
40.Q.
Orientability of a family of polygons is preserved by the elementary operations.
A surfa e is said to be orientable if it an be presented as the quotient of an orientable family of polygons. 40.R. A surfa e S is orientable, i any family of polygons whose quotient is homeomorphi to S is orientable. 40.S. Spheres with handles are orientable. Spheres with ross aps are not.
More About Re ognizing Closed Surfa es 40.11.
How an the notion of orientability and 40.Q help to solve 40.O?
40.T.
Two losed onne ted manifolds of dimension two are homeomorphi i they have the same Euler hara teristi and either are both orientable or both nonorientable.
Compa t Surfa es with Boundary As in the ase of one-dimensional manifolds, lassi ation of ompa t two-dimensional manifolds with boundary an be easily redu ed to the
lassi ation of losed manifolds. In the ase of one-dimensional manifolds it was very useful to double a manifold. In two-dimensional ase there is a onstru tion providing a losed manifold related to a ompa t manifold with boundary even loser than the double.
40.U. Contra ting to a point ea h onne ted omponent of the boundary of a two-dimensional ompa t manifold with boundary gives rise to a losed two-dimensional manifold.
40.12. A spa e homeomorphi to the quotient spa e of 40.U an be onstru ted by atta hing opies of D2 one to ea h onne ted omponent of the boundary.
40.V.
Any onne ted ompa t manifold of dimension 2 with nonempty boundary is homeomorphi either to sphere with holes, or sphere with handles and holes, or sphere with ross aps and holes.
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40.W. Riddle. Generalize orientabilty to the ase of non losed manifolds of dimension two. (Give as many generalization as you an and prove that they are equivalent. The main riterium of su
ess is that the generalized orientability should help to re ognize the topologi al type.)
40.X.
Two ompa t onne ted manifolds of dimension two are homeomorphi i they have the same Euler hara teristi , are both orientable or both nonorientable and their boundaries have the same number of
onne ted omponents.
Simply Conne ted Surfa es 40:M Theorem. Any simply onne ted non- ompa t manifold of dimension two without boundary is homeomorphi to R2 .
41. One-Dimensional
mod2-Homology
of Surfa es
Polygonal Paths on Surfa e Let F be a triagulated surfa e. A path s : I ! F is said to be polygonal if s(I ) is ontained in the one-dimensional skeleton of the triangulation of F , the preimage of any vertex of the triangulation is nite, and the restri tion of s to a segment between any two onsequitive points whi h are mapped to verti es is an aÆne homeomorphism onto an edge of the triangulation. In terms of kinemati s, a polygonal path represents a moving point, whi h goes only along edges, does not stay anywhere, and, whenever it appears on an edge, it goes along the edge with a onstant speed to the opposite end-point. A ir ular loop l : S 1 ! F is said to be t7!exp(2it) 1 l polygonal if the orresponding path I ! S !F is polygonal. 41:A. Let F be a triagulated surfa e. Any path s : I ! F onne ting verti es of the triangulation is homotopi to a polygonal path. Any
ir ular loop l : S 1 ! F is freely homotopi to a polygonal one. A polygonal path is a ombinatorial obje t: 41:B. To des ribe a polygonal path up to homotopy, it is enough to spe ify the order in whi h it passes through verti es. On the other hand, pushing a path to the one-dimensional skeleton an
reate new double points. Some edges may appear several time in the same edge. 41:1. Let F be a triangulated surfa e and be an element of 1 (F ) di erent from 1. Prove that there exists a natural N su h that for any n N ea h polygonal loop representing n passes through some edge of the triangulation more than on e.
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Subdivisions of Triangulation To avoid a ongestion of paths on edges, one an add new edges, i.e., subdivide the triangulation. Although an elementary operation on families of polygons applied to a triangulation, gives rise to a family, whi h is not a triangulation, making several elementary operations, one an get a new triangulation with more edges. One triangulation of a surfa e is alled a re nement of another one if ea h triangle of the former is ontained in a triagle of the latter. There are several standard ways to onstru t a re nement of a triangulation. For example, add a new vertex, whi h is lo ated inside of a triangle of a given triangulation, onne t it with the verti es of this triangle with segments, whi h are three new edges. The triangle is subdivided into three new triangles. The other triangles of the original triangulations are kept inta t. This is alled the star subdivision entered at . See Figure 2.
τ
Figure 2.
Star subdivision entered at triangle
Another kind of lo al subdivision: add a new vertex lo ated on an edge " of a given triangulation, onne t by new edges this vertex to the verti es opposite to " of the triangles adja ent to ". Ea h of the adja ent triangles is subdived into two new triangles. Leave the other triangles inta t. This is a star subdivision entered at ". See Figure 3.
ε
Figure 3.
Star subdivision entered at edge "
: Constru t a triangulation and its subdivision whi h annot be obtained as a omposition of star subdivisions entered at edges and triangles.
41 2.
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: Prove that a subdivision of a triangulation of a ompa t surfa e
an be presented as a result of a nite sequen e of star sudivisions
entered at triangles and edges and operations inverse to operations of these types.
41 3.
Bringing Loops to General Position 41:C. Let F be a triangulated and u, v polygonal ir ular loops on F .
Then there exist a subdivision of the triangulation of F and polygonal loops u0 , v0 homotopi to u and v, respe tively, su h that u0 (I ) \ v0 (I ) is nite. 41:D. Let F be a triangulated and u a polygonal ir ular loop on F .
Then there exist a subdivision of the triangulation of F and a polygonal loop v homotopi to u su h that v maps the preimage v 1 (") of any edge " v(I ) homeomorphi ally onto ". (In other words, v passes along ea h edge at most on e). Let u, v be polygonal ir ular loops on a triangulated surfa e F and a be an isolated point of u(I ) \ v(I ). Suppose u 1 (a) and v 1 (a) are one point sets. One says that u interse ts v translversally at a if there exist a neighborhood U of a in F and a homeomorphism U ! R2 whi h maps u(I ) \ U onto the x-axes and v(I ) \ U to y-axes. Polygonal ir ular loops u, v on a triangulated surfa e are said to be in general position to with respe t ea h other, if u(I ) \ v(i) is nite, for ea h point a 2 u(i) \ v(I ) ea h of the sets u 1 (a) and v 1 (a) ontains a single point and u, v are transversal at a. 41:E. Any two ir ular loops on a triangulated surfa e are homotopi to ir ular loops, whi h are polygonal with respe t to some subdivision of the triangulation and in general position with respe t to ea h other.
For a map f : X ! Y denote by Sk (f ) the set fa 2 X j f 1f (a) onsists of k elementsg and put S (f ) = fa 2 X j f 1 f (a) onsists of more than 1 elementg: A polygonal ir ular loop l on a triangulated surfa e F is said to be generi if (a) S (l) is nite, (b) S (l) = S2 (l), ( ) at ea h a 2 l(S2 (l)) the two bran hes of s(I ) interse ting at a are transversal, that is a has a neighborhood U in F su h that there exists a homeomorphism U ! R2 mapping the images under s of the onne ted omponents of s 1 (U ) to the oordinate axis.
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41:F. Any ir ular loop on a triangulated surfa e is homotopi to a
ir ular loop, whi h is polygonal with respe t to some subdivision of the triangulation and generi . Generi ir ular loops are espe ially suitable for graphi representation, be ause the image of a ir ular loop de nes it to a great extend: 41:G. Let l be a generi polygonal loop on a triangulated surfa e. Then any generi polygonal loop k with k(S 1 ) = l(S 1 ) is homotopi in l(S 1 ) to either l or l 1 . Thus, to des ribe a generi ir ular loop up to a reparametrization homotopi to identity, it is suÆ ient to draw the image of the loop on the surfa e and spe ify the dire tion in whi h the loop runs along the image. The image of a generi polygonal loop is alled a generi (polygonal)
losed onne ted urve. A union of a nite olle tion of generi losed
onne ted polygonal urves is alled a generi (polygonal) losed urve. A generi losed onne ted urve without double points (i.e., an embedded oriented ir le ontained in the one-dimensional skeleton of a triangulated surfa e) is alled a simple polygonal losed urve. The adje tive losed in the de nitions above appears be ause there is a version of the de nitions with (non- losed) paths instead of loops. 41:H. Riddle. What modi ations in Problems 41:C { 41:G and orresponding de nitions should be done to repla e loops by paths everywhere? By a generi polygonal urve we will mean a union of a nite olle tion of pairwise disjoint images of generi polygonal loops and paths.
Cutting Surfa e Along Curve 41:I Cutting Surfa e Along Curve. Let F be a triangulated twodimensional manifold and C F a one-dimensional manifold ontained in the 1-skeleton of the triangulation of F . Assume that C = F \ C . Prove that there exists a two-dimensional manifold T and a surje tive
ontinuous map p : T ! F su h that: (a) pj : T r p 1 (C ) ! F r C is a homeomorphism, (b) pj : p 1 (C ) ! C is a two-fold overing. Su h T and p are unique up to a homeomorphism: if T~ and p~ are other manifold and mapping satisfying the same onditions then there exists a homeomorphism h : T~ ! T su h that p Æ h = p~.
The surfa e T des ribed in 41:I is alled the result of utting F along C . It is denoted by F C . This is not the omplement F r C , though a
opy of F r C is ontained in F C as a dense subset, whi h is homotopy equivalent to the whole F C .
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41:J Triangulation of F
C . There exists a unique triangulation of F C su h that the natural map F C ! F maps edges onto edges and triangles onto triangles homeomorphi ally. : Des ribe the topologi al type of F C for the following F and C: (a) F is Mobius band, C its ore ir le (deformation retra t); (b) F = S 1 S 1 , C = S 1 1; ( ) F is S 1 S 1 standardly embedded into R3 , C the trefoil knot on F , that is f(z; w) 2 S 1 S 1 j z 2 = w3 g; (d) F is Mobius band, C is a segment: show that there are two possible pla ements of C in F and des ribe F C for both of them; (e) F = RP 2 , C = RP 1 . (f) F = RP 2 , C is homeomorphi to ir le: show that there are two possible pla ements of C in F and des ribe F C for both of them. 41:5 Euler Chara teristi and Cutting. Find the Euler hara teristi of F C when C = ?. What if C 6= ?? 41 4.
Curves on Surfa es and Two-Fold Coverings Let F be a two-dimensional triangulated surfa e and C F a manifold of dimension one ontained in the 1-skeleton of the triangulation of F . Let C = F \ C . Sin e the preimage C~ of C under the natural proje tion F C ! F is a two-fold overing spa e of C , there is an involution : C~ ! C~ whi h is the only nontrivial automorphism of this overing. Take two opies of F C and identify ea h x 2 C~ in one of them with (x) in the other opy. The resulting spa e is denoted by F C . 41:K. The natural proje tion F C ! F de nes a ontinuous map F C ! F . This is a two-fold overing. Its restri tion over F r C is trivial.
One-Dimensional Z2-Cohomology of Surfa e By 35:G, a two-fold overing of F an be thought of as an element of H 1 (F ; Z2). Thus any one-dimensional manifold C ontained in the 1skeleton of F and su h that C = F \ C de nes a ohomology lass of F with oeÆ ients in Z2. This lass is said to be realized by C . 41:L. The ohomology lass with oeÆ ients in Z2 realized by C in a
ompa t surfa e F is zero, i C divides F , that is, F = G [ H , where G and H are ompa t two-dimensional manifolds with G \ H = C . Re all that the ohomology group of a path- onne ted spa e X with
oeÆ ients in Z2 is de ned above in Se tion 35 as Hom(1 (X ); Z2). 41:M. Let F be a triangulated onne ted surfa e, let C F be a manifold of dimension one with C = F \ C ontained in the 1-skeleton of F . Let l be a polygonal loop on F whi h is in general position with respe t
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182
to C . Then the value whi h the ohomology lass with oeÆ ients in Z2 de ned by C takes on the element of 1 (F ) realized by l equals the number of points of l \ C redu ed modulo 2.
One-Dimensional Z2-Homology of Surfa e 41:N Z2-Classes via Simple Closed Curves. Let F be a triangu-
lated onne ted two-dimensional manifold. Every homology lass 2 H1 (F ; Z2) an be represented by a polygonal simple losed urve. 41:O. A Z2-homology lass of a triangulated two-dimensional manifold F represented by a polygonal simple losed urve A F is zero, i there exists a ompa t two-dimensional manifold G F su h that A = G. Of ourse, the \if" part of 41:O follows straightforwardly from 35:L. The \only if" part requires tri kier arguments. 41:O:1. If A is a polygonal simple losed urve on F , whi h does not bound in F a ompa t 2-manifold, then there exists a onne ted
ompa t 1-manifold C F with C = F \ C , whi h interse ts A in a single point transversally.
41:O:2. Let F be a two-dimensional triangulated surfa e and C F a manifold of dimension one ontained in the 1-skeleton of the triangulation of F . Let C = F \C . Any polygonal loop f : S 1 ! F , whi h interse ts C in an odd number of points and transversally at ea h of them, is overed in F C by a path with distin t end-points. 41:O:3. See 35:6.
Poin are Duality
To be written! One-Sided and Two-Sided Simple Closed Curves on Surfa es
To be written! Orientation Covering and First Stiefel-Whitney Class
To be written! Relative Homology
To be written!
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183
42. Surfa es Beyond Classi ation To be written! Genus of Surfa e
To be written! Systems of disjoint urves on a surfa e
To be written! Polygonal Jordan and S hon ies Theorems
To be written! Polygonal Annulus Theorem
To be written! Dehn Twists
To be written! Coverings of Surfa es
To be written! Bran hed Coverings
To be written! Mapping Class Group of Torus
To be written! Lifting homeomorphisms to the universal overing spa e. Nielsen and Baer Theorems for torus. GL(2; Z). Dehn twists along meridian and longitude and relation between them. Center of the mapping lass group. Braid Groups
To be written!
43. THREE-DIMENSIONAL MANIFOLDS
43. Three-Dimensional Manifolds To be written! Poin are Conje ture Lens Spa es Seifert Manifolds Fibrations over Cir le Heegaard Splitting and Diagrams
184
CHAPTER 7
Smooth Manifolds Although manifolds provide a s ene for almost all geometri bran hes of Mathemati s, the topologi al stru ture of a manifold does not de orate this s ene enough. It is not suÆ ient to dis uss most of phenomena of Analysis and Geometry. Usually in appli ations, manifolds arise equipped with various additional stru tures. One of them, smooth or di erential stru ture, appears more often than others. The goal of this Chapter is to introdu e the smooth stru ture and develope the basi theory. While topologi al stru tures provide a basis for dis ussing phenomena related to ontinuity, smooth stru tures provide a basis for dis ussing phenomena related to di erentiability. The traditional de nition of smooth stru tures is quite long and di erent from de nitions of similar, and, in fa t, losely related stru tures whi h are studied in algebrai geometry and topology. Furthermore, smooth stru tures are traditionally de ned only on manifolds. This deprives us of exibility that we enjoy in general topology, where any set-theoreti
onstru tion has a topologi al ounter-part: a subset turns into a subspa e, a quotient set turns into a quotient spa e, et . The image of a di erential manifold under a di erentiable map may happen to be not a manifold, and hen e not eligible to bear any tra e of a di erential stru ture. Therefore we dare to hange the very basi de nitions of the di erential topology. The notion of di erential manifold be omes a spe ial ase of more general notions of di erential spa e and di erential variety. Of
ourse, spe ialists are aware about the possibility of these generalizations. However as far as we know, nobody did a serious attempt to rely on the generalizations in a textbook written for beginners. We try to over ome the phobia about singularities, whi h was a hara teristi property of texts on di erential topology. We believe, this makes the subje t simpler, although introdu es possibilty to speak about pathologi al obje ts. As we laimed above, we think on tea hing the elementary topology as about tea hing a language. This is a great language, one of the main parts of the language of Mathemati s. It is not our goal to tea h only 185
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\politi ally orre t" words: we do not want to ex lude a single word just be ause it an be used in a des ription of \bad", \pathologi al" obje ts. Of ourse, the standard approa h to smooth manifolds is also presented, right after the new one. But rst, we must refresh the ba kground from Multivariable Cal ulus.
44. Analyti Digression: Di erentiable Fun tions in Eu lidean Spa e Di erentiability and Di erentials Re all that a fun tion f : R ! R is alled di erentiable at a 2 R if there exists a number f 0 (a) su h that f (a + x) f (a) lim (21) = f 0 (a): x!0 x This de nition does not admit immediate generalization to the ase of a map R n ! R k , but an be reformulated in a way that does. Namely, denote by L the linear map R ! R : x 7! f 0 (a)x. Then (21) is equivalent to f (a + x) f (a) L(x) lim = 0: x!0 x
Let f be a map de ned in a neighborhood of a point a 2 R n and taking values in R k . One says that f is di erentiable at a if there exists a linear map L : R n ! R k su h that jf (a + x) f (a) L(x)j = 0: lim x!0 jxj In this ase L is alled the di erential of f at a. 44.A. If f is di erentiable at a, then its di erential at a is unique. The di erential of f at a is denoted by da f . Prove that for any linear map L : Rn ! Rk di erent from da f there exists a neighborhood U of a su h that jf (x) f (a) da f (x a)j < jf (x) f (a) L(x a)j for x 2 U r a. 44.1.
Theorem 44.1 means that the aÆne map x 7! f (a)+ da f (x a) approximates f in a neighborhood of a better than any other aÆne map. Prove that if the dimensions of both sour e and target are equal to 1 df then da f is multipli ation by (a). dx 44.2.
44. ANALYTIC DIGRESSION ON DIFFERENTIAL FUNCTIONS
187
Derivative Along Ve tor The image of a ve tor v under da f , i.e., da f (v ) is denoted also by Dv f (a) and alled the derivative of f at a in dire tion v . 44.3.
Prove that Dv f (a) = limt!0
f (a + tv) f (a) . t
Thus, Dv f (a) is the velo ity of hanging of f when a moves with velo ity v. 44.4. Prove that if v is the i-th standard base ve tor (i.e., all the omponents of v, but i-th, equal 0, and the i-th omponent is 1), then Dv f (a) is equal to f2 fk 1 ( f xi (a); xi (a); : : : xi (a)), where fj is the j -th omponent of f . The di erential da f of a map f : Rn Ja obian matrix of f at a).
44.5.
! Rk
j has matrix f xi (the
Main Properties of Di erential
44.B. Let U R n , V R k , W R m . If a map f : U ! V is di erentiable at a 2 U and g : V ! W is di erentiable at f (a) then g Æ f : U ! W is di erentiable at a and da (g Æ f ) = df (a) g Æ da f . In other words, the di erential of omposition is the omposition of di erentials. 44.6.
Re ognize Theorem 44.B as a reformulation of the Chain Rule.
44.C. The di erential of the identity map is the identity map. 44.7 Generalization of 44.C.
oinsides with L.
Di erential of a linear map L at ea h point
Higher Order Derivatives If U R n , V R k and a map f : U ! V is di erentiable at ea h point of U , the di erentials da f give rise to map U ! Hom(R n ; R k ) : a 7! da f . This generalizes the notion of derivative fun tion. There is a map losely related to this one, and more onvinient for generalizations. It is de ned as follows: U R n ! R k : (a; v ) 7! Dv f (a). The relation is provided by the de nition Dv f (a) = da f (v ). 44.D. Prove that U ! Hom(R n ; Rk ) : a 7! da f is ontnuous i U R n ! R k : (a; v) 7! Dv f (a) is ontinuous. 44.E. Prove that U ! Hom(R n ; Rk ) : a 7! daf is di erentiable at a i U R n ! R k : (a; v ) 7! Dv f (a) is di erentiable at (a; v ) for ea h v 2 R n . 44.8. Riddle.
1?
How does this look like in the ase of n = 1 or even n = k =
Is it possible to redu e in 44.E the set of v, for whi h U Rn Dv f (a) is di erentiable at (a; v)?
! Rk : (a; v) 7!
44. ANALYTIC DIGRESSION ON DIFFERENTIAL FUNCTIONS
188
The map U R n ! R k : (a; v ) 7! Dv f (a) is alled the derivative map for f and denoted by Df . Sin e it is also a map of the same kind, one an iterate the onstru tion and de ne the se ond derivative D2 f : U (R n )2 ! R k , third derivative D3 f : U (R n )3 ! R k and r-th derivative Dr f : U (R n )r ! R k . 44.9.
: : : vr .
Prove that Dr f (a; v1 ; : : : vr ) does not hange when one inter hange v1 ,
Express Dr f (a; v1 ; : : : vr ) in \ lassi al" terms, i.e., write down an expression for Dr f (a; v1 ; : : : vr ) in terms of partial derivatives of omponents of f and oordinates of v1 , : : : vr . 44.10.
Let e1 , : : : en be the standard basis of Rn . Prove that for x = i ei x i=1
44.11.
Pn
lim
x!0
jf (a + x) f (a)
Ps
1 Pn ir i1 r r=1 r! i1 ;:::;ir =1 D f (a; ei1 ; : : : ; eir )x : : : x j jxjs
=0
C r -Maps Let U be an open subset of R n and r a non-negative integer or 1. A map f : U ! R k is said to be of lass C r or a C r -map if at ea h point of U it has all the derivatives of order r and all of them are ontinuous. A map is of lass C 1 if it is of lass C r for all nite r.
44.F. A map f : U ! R k is of lass C r , i its omponents f1 , : : : fk have all the partial derivatives of order r and these partial derivatives are ontinuous.
Constru t a map whi h has all the partial derivatives of order r at ea h point, but is not of lass C r .
44.12.
Let U be an open subset of Rn . A map f : U ! R k is said to be real analyti or of lass C a at x0 2 U if there exists a neighborhood V of x0 in U su h that the Taylor series 1 X r=0
n 1 X Dr f (x0 ; ei1 ; : : : ; eir )xi1 : : : xir r! i1 ;:::;ir =1
onverges to f (x0 + x) for x0 + x 2 V .
44.G. A real analyti map is of lass C 1. A map of lass C 0 is just a ontinuous map. It is onvinient to assume a > 1 and speak about lasses C r with 0 r a.
44. ANALYTIC DIGRESSION ON DIFFERENTIAL FUNCTIONS
189
Di eomorphisms Let U , V be open subsets of R n and r be a natural number, or 1 or a. A map f : U ! V is alled a di eomorphism of lass C r , or C r di eomorphism or just di eomorphism (of U to V ) if f is of lass C r at ea h point of U , invertible, and f 1 is of lass C r at ea h point of V . 44.H. The di erential of a di eomorphism at any point is an isomorphism. 44.I. Composition of C r -di eomorphisms is a C r -di eomorphism. The map inverse to a C r -di eomorphism is a C r -di eomorphism. 44.13. Whi h of the following maps are di eomorphisms and what are the
lasses of the di eomorphisms: (a) R ! R : x 7! x2 , (b) R ! R : x 7! x3 , ( ) (0; 1) ! (0; 1) : x 7! x3 , (d) (0; 1) ! (0; 1) : x 7! x2 , (e) C ! C : x 7! x3 , (f) R ! R : x 7! x + x3 , (g) R ! R : x 7! ( x + x2 , x + x2 ; if x 0 (h) R ! R : x 7! , x x2 ; if x < 0 ( x + x5 ; if x 0 (i) R ! R : x 7! , x x44 ; if x < 0 (j) R ! R : x 7! x5=3 , (k) R ! R : x 7! x + x101=3 ?
Inverse Fun tion Theorem The following important and famous theorem is a sort of inverse to 44.H. 44.J Inverse Fun tion Theorem. If f : U ! R n is a C r -map with r 1 de ned in a neighborhood U of a 2 R n and da f : R n ! R n is invertible then there exist neighborhoods V U of a and W R n of f (a) su h that f j : V ! W is a C r -di eomorphism. 44.K Corollary. Let U and V be open sets in R n . A map f : U ! V is a C r -di eomorphism i it is a bije tive C r -map and da f is an isomorphism for any a 2 U .
Impli it Fun tion Theorem
44.L Impli it Fun tion Theorem. Let U R n be an open set and f : U ! R k a C r -di erentiable map. If da f is surje tive then the point a has a neighborhood V U su h that V an be presented (by a renumeration of oordinates) as A B with A R n k , B R k and f 1 f (a) \ V
44. ANALYTIC DIGRESSION ON DIFFERENTIAL FUNCTIONS
190
is the graph of some C r -map ' : A ! B , i.e., f 1 f (a) \ V = f(x; y ) 2 A B j y = '(x)g.
C r -Fun tions The set of all the C r -fun tions U ! R is denoted by C r (U ). 44.M. If f 2 C r (U ) then f jV 2 C r (V ) for any open V U . In other words, for open sets V U R n formula f 7! f jV de nes a map C r (U ) ! C r (V ).
Useful C 1 -Fun tion
44.N Bell-Shape Fun tion. There exists a C 1-fun tion : R n ! R whi h takes value 1 on the unit ball Dn R n , takes value 0 on the
omplement of the ball B of radius 2 entered at 0 takes values in (0; 1) on Int B r Dn. A C a -fun tion with these properties does not exists. 44.N.1. The fun tion
1 : x 7!
(
exp( (x 0;
1 1)(x 2) );
if x 2 (1; 2); if x 6=2 (1; 2)
is a C 1-fun tion on R. 44.N.2 Lemma on Smooth Step Fun tion. The fun tion Rx (t) dt 2 (x) = R02 1 1 1 (t) dt 1 is a C -fun tion on R. It takes value 0 on [0; 1℄ and 1 on [2; 1). A C a -fun tion f : R ! R with f [0; 1℄ = 0 and f [2; 1) = 1 does not exists.
Appli ations of Bell-Shape Fun tion
44.O Retreat Ensures Expansion. Let U R n be an open set and f : U ! R a C r -fun tion with r 1. Prove that any point a 2 U has a neighborhood V U su h that f jV is a resti tion of a C r -fun tion g : Rn ! R. 44.P C r (R n ) Knows All the C r (U ). A fun tion f : U ! R is of lass C r i any point a 2 U has a neighborhood V U su h that f jV is a resti tion of a C r -fun tion R n ! R . C r -Maps Consider open sets U; V R n . Any map f : U ! V de nes a map of the set of all the fun tions V ! R to the set of fun tions U ! R : (' : V ! R ) 7! (f Æ ' : U ! R ): If f is a C r -map then this maps C r -fun tions to C r -fun tions.
45. DIFFERENTIAL SPACES
191
44.Q. A map f : U ! V is a C r -di eomorphism, i it de nes a bije tion C r (V ) ! C r (U ).
45. Di erential Spa es Motivation: Topologi al Stru ture via Continuous Fun tions Let X be a topologi al spa e. Consider the set of all ontinuous fun tions X ! R . It is denoted by C (X ). 45.A. C (X ) is an algebra over R with respe t to the pointwise addition of fun tions, multipli ation of fun tion by numbers and multipli ation of fun tions. In other words, if f; g 2 C (X ); then (x 7! f (x) + g (x)) 2 C (X ) and (x 7! f (x)g (x)) 2 C (X ) and these operations satisfy the axioms of algebra over R . Besides these linear operations, there are other operations, with respe t to whi h C (X ) is losed. 45.B. Let f1 ; : : : ; fn 2 C (X ) and f : X ! R n be a map de ned by f1 , : : : ,fn . Let f (X ) A and let g : A ! R be a ontinuous fun tion. Then g Æ f j 2 C (X ). 45.C. For any topologi al spa e X there is a minimal topologi al stru ture ' on X su h that C (X ) = C (X; '). Prove that if X is a metrizable spa e then ' oin ides with the original topology of X . Find a topologi al spa e su h that these topologi al stru tures are di erent. Find a non-metrizable spa e su h that these topologi al stru tures oin ide. Metrizable topologi al spa es omprise a large and important lass of topologi al spa es. The lass of topologi al spa es, whi h are re overable from algebras of ontinuous fun tions on them, is even larger. For spa es of this lass the whole theory ould be rebuilt on the basis of C (X ), whi h would be a repla ement for the topologi al stru ture (i.e., the set of open sets) of X . We des ribe this opportunity be ause of its similarity with our approa h to di erential stru tures. Exa tly as the notion of topologi al stru ture extends the notion of ontinuous fun tion to a more general situation, the notion of di erential stru ture is to extend the notion of di erentiable fun tion. However, di erential stru tures were never de ned in a generality omparable to the generality of topologi al spa es. Maybe this is why the approa h via distinguishing an algebra of \good" real valued fun tions, whi h in the ase of ontinuity looks more restri tive than the standard
45. DIFFERENTIAL SPACES
192
approa h, ts so well in the ase of di erentiability: it is applied to the situations, in whi h general topology ould be perfe tly based on algebras of ontinuous fun tions. 45.D. Prove that a topologi al spa e X an be embedded to R n , i : (a) the topologi al stru ture of X is de ned by C (X ), (b) the algebra C (X ) ontains n fun tions f1 , : : : , fn su h that any f 2 C (X ) an be obtained from f1 , : : : fn by an operation des ribed in 45.B and ( ) for any di erent a; b 2 X there exists fi with fi (a) 6= fi (b).
Di erential Spa es Let X be a set and r be a natural number or in nity. A di erential stru ture of lass C r on X is an algebra C r (X ) of fun tions X ! R satisfying the following two onditions: (a) For any f1 ; : : : ; fn 2 C r (X ) su h that the image of the map f : X ! R n de ned by f1, : : : ,fn is ontained in an open set A R n and any C r map g : A ! R the omposition g Æ (f j) : X ! R belongs to C r (X ). (Cf. 45.B above.) (b) A fun tion f : X ! R belongs to C r (X ) if for ea h a 2 X there exist g; h 2 C r (X ) su h that h(a) > 0 and f (x) = g (x) for ea h x with h(x) > 0. A set equipped with a di erential stru ture of lass C r is alled a differential spa e of lass C r or C r -spa e. Elements of C r (X ) are alled C r -fun tions on X . Any di erential spa e has a natural topologi al stru ture: the smallest one with respe t to whi h all the fun tions belonging to C r (X ) are ontinuous. It is alled the underlying topologi al stru ture and X equipped with this stru ture is alled the underlying topologi al spa e of the C r spa e. The terms from general topology applied to a C r -spa e are understood as being applied to the underlying topologi al spa e. For example, \Hausdor C r -spa e" means \C r -spa e whose underlying topologi al spa e is Hausdor ". 45.E. The underlying topologi al stru ture has a basis onsisting of the sets whi h are de ned by nite systems of inequalities f (x) > 0 with f 2 C r (X ). 45.1. Let X be a C r -spa e with r 1, let f1 ; : : : ; fr 2 C r (X ), and Ui = fi 1 (0; +1) for i = 1; : : : ; r. Constru t f; g 2 C r (X ) su h that \ri=1 Ui = f 1 (0; +1) and [ri=1 Ui = g 1 (0; +1). 45.2. The underlying topologi al stru ture of a C r -spa e with r 1 has the basis onsisting of the sets ea h of whi h is de ned by an inequality f (x) > 0 with f 2 C r (X ).
45. DIFFERENTIAL SPACES
193
45.F. In terms of the underlying topology, the se ond ondition in the
de nition of di erential stru ture is formulated as follows: the property of belonging to C r (X ) is lo al, i.e., a fun tion f : X ! R belongs to C r (X ), provided in a neighborhood of ea h point of X it oin ides with some g 2 C r (X ).
For a given set X , what is a di erential C r -stru ture on X with the indis rete (underlying) topology? Does it exist? Is it unique? 45.4. For a given set X , what is a di erential C r -stru ture on X with the dis rete topology? Does it exist? For whi h X is it unique? r 45.5. Prove that any C -spa e satisfying the rst separation axiom is Hausdor . r 45.6. Prove that any C -spa e satisfying the rst separation axiom is regular. r r 45.7. Let X be a C -spa e with r 1. Let f; g 2 C (X ) and A = 1 1 f [0; +1), B = g [0; +1). Prove that if A \ B = ? then there exists a fun tion h 2 C r (X ) su h that h(X ) [0; 1℄, h 1 (0) = A and h 1 (1) = B . 45.3.
45.G. The set of all the fun tions of lass C r on an open subset U R n
is a di erential stru ture of lass C r on U . This C r -stru ture is the minimal one whi h ontains all the n oordinate proje tions U ! R . Di erential Stru ture of a Metri Spa e Let X be a metri spa e with metri : X X ! R+ . A fun tion f : X ! R is said to be di erentiable at a 2 X if for any neighborhood U of a one an nd points b1 ; : : : ; bk 2 U r a and numbers 1 ; : : : ; k 2 R su h that Pk j f (x) f (a) i=1 i ((bi ; x) (bi ; a))j = 0: lim x!a (x; a) : Prove that the fun tion X ! R : x 7! (a; x) may be nondi erentiable at some x 6= a. Prove that this an be the ase for X = S 1 with some metri . On the other hand, if X is a subspa e of Rn equipped with the metri whi h is the restri tion of the standard metri of Rn then X ! R : x 7! (a; x) is di erentiable at ea h x 6= a. 45:2. Prove that for any metri spa e X fun tion X ! R : x 7! (a; x)r with integer r > 1 is di erentiable at a 2 X .
45 1.
45:A. Prove that if X is an open subspa e of Rn then the notion of
di erentiability introdu ed above oin ides with the lassi al di erentiability dis ussed in Se tion 44.
Let X be a metri spa e. A fun tion f : X ! R is said to be ontinuously di erentiable at a 2 X if for any neighborhood U of a there exists a neighborhood V of a and ontinuous fun tions b1 ; : : : ; bk : V ! U , 1 ; : : : ; k : V ! R (for some k) su h that for any point 2 V Pk j f (x) f ( ) i=1 i ((bi ( ); x) (bi ( ); ))j = 0: lim x! (x; )
45. DIFFERENTIAL SPACES
194
Denote by C 1 (X ) the set of fun tions X ! R ontinuously di erentiable at ea h point of X . 45:B. Prove that C 1 (X ) is a di erential stru ture of lass C 1 for any metri spa e X . : (a) (b) ( ) (d)
What is C 1 (X ) if X = f(x; y) 2 R2 j xy = 0g with metri indu ed from R2 , X = f(x; y; z ) 2 R3 j xy = 0g with metri indu ed from R3 , X = S 2 with metri indu ed from R3 X is the Cantor set with metri indu ed from R?
45 3.
Let X be a metri spa e and r a natural number. A fun tion f : X ! R is alled a fun tion of lass C r at a 2 X if for any neighborhood U of a one an nd a neighborhood V of a and ontinuous fun tions b1 ; : : : ; bk : V ! U , p : V ! R[x1 ; : : : ; xk ℄, where p takes values in polynomials of degree r su h that for any point 2 V jf (x) f ( ) p((b1 ( ); x); : : : ; (bk ( ); a))j = 0: lim x! (x; )r Denote by C r (X ) the set of fun tions X ! R of lass C r at ea h point of X . 45:C. Prove that C r (X ) is a di erential stru ture of lass C r for any metri spa e X .
Di erential Subspa es
45.H. Let X be a C r -spa e and A its subset. Consider the set of fun tions f : A ! R su h that for ea h b 2 A there exist g; h 2 C r (X ) with h(b) > 0 and f (x) = g (x) for ea h x 2 A with h(x) > 0. Prove that this is a di erential stru ture of lass C r on A.
This set of fun tion is denoted by C r (A) and alled the C r -stru ture indu ed by C r (X ). The set A equipped with C r (A) is alled a C r -subspa e of X . 45.I. (Cf. 44.O) Let U R n . Prove that the set of fun tions belonging to the C r -stru ture on U indu ed by the standard C r -stru ture of Rn
oin ides with the set of C r -fun tions U ! R . Below all the subsets of R n are onsidered as C r -spa es with the stru ture indu ed, as on subspa es, by the standard C r -stru ture of R n , unless the opposite is stated expli itely. 45.J. Prove that if A is a subset of a C r -spa e X and f 2 C r (A) then f jB : B ! R belongs to C r (B ) for any B A. 45.8. Prove that the C r -stru ture indu ed on R from the standard C r stru ture of R2 oin ides with the standard C r -stru ture of R.
45. DIFFERENTIAL SPACES
Show that the map C r (X ) ! C r (A) : f Under what onditions it is surje tive? 45.9.
195
7! f jA may be nonsurje tive.
C r -Stru tures on Subspa e of Metri Spa e 45:D. Prove that for an open subset A of a metri spa e X the C r -
stru ture indu ed from the metri C r -stru ture of X oin ides with the C r -stru ture indu ed by the restri tion to A of the metri of X . 45:E. For any subset A of a metri spa e X the C r -stru ture indu ed by
the restri tion to A of the metri of X is ontained in the C r -stru ture indu ed on A from the metri C r -stru ture of X .
45:F. Prove that for A
C r -stru ture
Rn
Rn the C r -stru ture indu ed from the stan-
dard of
oin ides with the C r -stru ture indu ed by the restri tion to A of the metri of Rn . 45:G. Constru t a subset A of a metri spa e X su h that the C r -
stru ture indu ed on A from the metri C r -stru ture of X does not oin ide with the C r -stru ture indu ed by the restri tion to A of the metri of X . 45:G:1. Embed isometri ally R1 with the standard metri to a metri spa e X su h that fun tion R ! R : x 7! jxj is di rentiable with respe t to the C 1 -stru ture indu ed from the metri C 1 -stru ture on X.
Di erentiable Maps Let X , Y be C r -spa es. A map f : X ! Y is alled a di erentiable map of lass C r or a map of lass C r or or just a C r -map if ' Æ f 2 C r (X ) for ea h ' 2 C r (Y ). A C r -map f : X ! Y de nes a homomorphism C r (Y ) ! C r (X ).
45.K General Properties of C r -Maps. Prove that: (a) (b) ( ) (d)
The omposition of C r -maps is a C r -map. The identity map of a C r -spa e is a C r -map. The in lusion of a C r -subspa e into the C r -spa e is a C r -map. A submap of a C r -map is a C r -map. 45.10. Let X be a C r -spa e. Prove that f 2 C r (X ), i f : X ! R is a C r -map (with respe t to the standard C r -stru ture of R). 45.11. Let U Rn , V Rk be open sets. Prove that f : U ! V is a C r -
map with respe t to the C r -stru tures indu ed from the standard stru tures of the ambient spa es Rn and Rk i it is a C r -map in the sense de ned in Se tion 44 (that is the ompostions of f with all the oordinate proje tions U ! R are r times ontinuously di erentiable).
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196
Di eomorphisms Let X , Y be C r -spa es. A map X ! Y is alled a di eomorphism of lass C r or C r -di eromorphism if it is an invertible C r -map, and the inverse map is also of lass C r . C r -spa es are said to be (C r -)di eomorphi if there exists a C r -di eomorphism X ! Y . 45.L General Properties of Di eomorphisms. Prove that: (a) The omposition of C r -di eomorphisms is a C r -di eomorphism. (b) The identity map of a C r -spa e is a C r -di eomorphism. ( ) The inverse map to a C r -di eomorphism is a C r -di eomorphism. 45.M. The di eomorphism relation of C r -spa es is an equivalen e relation. 45.N. Prove that C r -di eomorphisms of open subsets of R n de ned in Se tion 44 are C r -di eomorphisms in the sense dis ussed here. Prove that any di eomorphism of a semi- ubi parabola C = f(x; y) 2 R2 j x3 = y2 g onto itself preserves (0; 0) 2 C ). 45.13. Consider the angle A = f(x; y ) 2 R2 j x = 0; y 0g [ f(x; y ) 2 R2 j x 0; y = 0g, semi- ubi parabola C = f(x; y) 2 R2 j x3 = y2 g and line R. Prove that there exist C r -bije tions A ! R, R ! C and C ! R, but these C r -spa es are pairwise nondi eomorphi . 45.12.
Di erentiale Embeddings Re all that a topologi al embedding is a map f : X ! Y of a topologi al spa e X to a topologi al spa e Y su h that its submap f : X ! f (X ) is a homeomorphism. In the setup of di erential spa es this de nition has an obvious ounter-part: a map f : X ! Y of a C r -spa e X to a C r -spa e Y is alled a C r -embedding if its submap f : X ! f (X ) is a C r -di eomorphism. 45.O. The in lusion of a smooth submanifold to the smooth manifold is a di erentiable embedding. 45.P. (Cf. 45.D.) Prove that a C r -spa e X an be embedded to R n , i the algebra C r (X ) ontains n fun tions f1 , : : : , fn su h that (a) C r (X ) is the minimal C r -stru ture ontaining f1 , : : : fn , (b) for any di erent a; b 2 X there exists fi with fi (a) 6= fi (b). Whi h of the following maps are di erentiable embeddings: id : R ! R, S 1 ! R2 : ( os 2t; sin 2t) 7! (sin 2t; sin 4t), S 1 ! S 1 : z 7! z 2, R1 ! R2 : t 7! (t2 ; t3 ), R1 ! R3 : t 7! (t2 ; t3 ; t),
45.14.
(a) (b) ( ) (d) (e)
46. CONSTRUCTING DIFFERENTIAL SPACES
(f) (g) (h) (i) (j) (k) (l)
197
R ! R2 : t 7! (t2 ; t4 ), I ! R2 : t 7! (sin t; sin 2t), [0; 1) ! R2 : t 7! (sin t; sin 2t), (0; 1) ! R2 : t 7! (sin t; sin 2t), R ! S 1 S 1 : t 7! (eit ; eit ),
S 1 ! S 1 S 1 : z 7! (z 3 ; z 2), S1 ! S1 S1 ( : z 7! (z 4 ; z 2), (x; x + x5 ); if x 0 (m) R ! R2 : x 7! , (x; x x44 ); if x < 0 ( (x; 2x); if y = 0; 2 2 (n) f(x; y) 2 R j xy = 0g ! R : (x; y) 7! , (2y; y(); if x = 0 (x; 0); if y = 0; (o) f(x; y) 2 R2 j y(y x2 ) = 0g ! R2 : (x; y) 7! , (0; x); if y 6= 0 (p) f(x; y) 2 R2 j y2 = x3 g ! R3 : (x; y) 7! (x; y; y1=3 )?
Semi ubi Parabola
45.Q. The set of all C 1-fun tions on R with the rst derivative vanishing at 0 is a C 1 -stru ture on R . 45.R. Prove that for any C 1-fun tion f : R ! R with dxdf (0) = 0 there exist C 1 -fun tions : R ! R and : R ! R su h that f (x) = (x2 ) + (x3 ). 45.S. The C 1-spa e of Problem 45.Q is C 1-di eomorphi to the subspa e of R 2 de ned by equation x3 = y 2. The map x 7! (x2 ; x3 ) is the di erential embedding.
46. Constru ting Di erential Spa es Multipli ation of Di erentiable Spa es Let X and Y be C r -spa es. Denote by C r (X Y ) the minimal C r stru ture on X Y whi h ontains the ompositions of the natural proje tions X Y ! X and X Y ! Y with C r -fun tions on X and Y , respe tively. The set X Y equipped with the C r -stru ture C r (X Y ) is alled a produ t of C r -spa es X and Y . Prove that, from the point of view of C r -spa es, Rp Rq = Rp+q . 46.2. Let X , Y , A and B be C r -spa es and f : X ! Y , g : A ! B be C r -maps. Prove that (a) the Cartesian produ t f g : X A ! Y B is a C r -map, (b) if f , g are C r -di eomorphisms then f g is a C r -di eomorphisms, ( ) if f , g are C r -embeddings then f g is a C r -embedding. 46.3. Let A, X and Y be C r -spa es and f : A ! X , g : A ! Y be C r -maps. Prove that 46.1.
46. CONSTRUCTING DIFFERENTIAL SPACES
198
(a) the map h : A ! X Y : a 7! (f (a); g(a)) is a C r -map, (b) if f is a C r -embeddings then h is a C r -embedding, too.
Quotient Di erential Spa es
46.A. Let X be a C r -spa e and S a partition of X . Prove that the set of fun tions f : X=S ! R su h that f Æ pr 2 C r (X ), where pr is the
anoni al proje tion X ! X=S , is a di erential stru ture of lass C r on the quotient set X=S .
This set of fun tion is denoted by C r (X=S ) and alled the quotient of C r (X ). The quotient set X=S equipped with C r (X=S ) is alled a quotient C r -spa e of X . Let S be the partition of R2 into verti al lines (i.e., sets of the form a R). Prove that R2 =S is di eomorphi to R. 46.5. Prove that the quotient spa e of the segment [ 1; 1℄ obtained by identifying the end points 1 and 1 is not di eomorphi to the ir le S 1 (with the C r -stru ture indu ed from the standard C r -stru ture of the ambient plane R2 ). Find a C r -subspa e of R2 di eomorphi to this quotient C r -spa e. 46.6. Prove that the quotient spa e of the segment [ 1; 1℄ obtained by identifying x with x + 3=2 for x 1=2 is di eomorphi to S 1 . 46.7. Prove that the orbit spa e of involution R2 ! R2 j (x; y ) 7! ( x; y ) is di eomorphi to the one f(x; y; z ) 2 R3 j x2 + y2 = z 2; z 0g. 46.8. Prove that the orbit spa e of involution R2 ! R2 j (x; y ) 7! (x; y ) is di eomorphi to the half-plane R2+ . 46.9. Prove that the quotient spa e D 2 =S 1 (the boundary ir le S 1 of the disk D2 is ontra ted to a single point) is not di eomorphi to a subset of Eu lidean spa e of any dimension. 46.4.
There is a natural way to introdu e a C r -stru ture into a disjoint sum of C r -spa es. 46.B. Des ribe expli itely the natural onstru tion of disjoint sum of di erential spa es. As in the ase of topologi al spa es, by gluing C r -spa es one means omposition of two onstru tions: disjoint summation followed by fa torization. Prove that the result of gluing of two opies of the half-spa e Rn+ by the identity map of the boundary hyperplane is di eomorphi to f(x; y; z ) 2 Rn 1 R R j y 0 and z = 0g[f(x; y; z ) 2 Rn 1 R R j z 0 and y = 0g 46.11. Prove that the result of gluing of two opies of the half-spa e Rn + by the map of the subset f(x1 ; : : : ; xn ) j 0 x1 1g onto itself de ned by (x1 ; x2 ; : : : ; xn ) 7! (1 x1 ; x2 ; : : : ; xn ) is di eomorphi to Rn .
46.10.
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Classi al Lie Groups and Homogeneous Spa es
To be written Spa e of n-Point Subsets of Surfa e
To be written Tori Varieties
To be written
47. Smooth Manifolds C r -Manifolds A C r -spa e X is said to be lo ally modelled on a C r -spa e Y if any point of X has a neighborhood di eomorphi to an open C r -subspa e of Y . A C r -spa e is alled a smooth, or di erential, or di erentiable1 manifold of lass C r or just C r -manifold of dimension n if it is modelled on a halfspa e R n and the underlying topologi al spa e is Hausdor and se ond
ountable. As it follows immediately from the de nition, the underlying topologi al spa e of a C r -manifold of dimension n is a (topologi al) manifold of dimension n. Consider the following subsets of the plane R2 . (a) f(x; y) 2 R2 j x 0; y = 0g [ f(x; y) 2 R2 j x = 0; y 0g, (b) f(x; y) 2 R2 j x2 + y2 = 1g, ( ) f(x; y) 2 R2 j x2 + y2 = 1; x 0; y 0g [ f(x; y) 2 R2 j x = 1; y 0g [ f(x; y) 2 R2 j x 0; y = 1g, (d) f(x; y 2 R2 j y = x6 ; x 0)g [ f(x; y) 2 R2 j y = 0; x 0g, (e) f(x; y) 2 R2 j y3 = x5 g, (f) f(x; y) 2 R2 j y103 = x3 g, (g) f(x; y) 2 R2 j y103 x3 g. Equip them with the C r -stru ture indu ed from R2 . For whi h r is ea h of them a C r -manifold? 47.2. Whi h of the following C r -subspa es of an Eu lidean spa e are C r manifolds? (a) S n , (b) Dn , ( ) I n , (d) Rn+ , (e) f(x; y; z ) 2 R3 j xyz 0g, 47.1.
1A
funny term: nobody is able to di erentiate this di erentiable manifold!
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200
f(x; y; z ) 2 R3 j xyz > 0g.
A di eomorphism : U ! G of an open set U of a C r -manifold X onto an open set of R n+ or R n is alled a (lo al) oordinate system or a hart in X . The ompositions of and the oordinate proje tions G ! R are denoted by 1 , 2 , : : : n and alled oordinates in the oordinate system . The value of i at a 2 U is alled the i-th oordinate of a in .
Let : U ! G and : V ! H be harts in a C r -manifold X . Then there appear harts j : U \ V ! (U \ V ), j : U \ V ! (U \ V ) and a map ( j) Æ ( j) 1 : (U \ V ) ! (U \ V ). The latter is alled the transition map from to and denoted by t . This map al ulates the oordinates of a point in given the oordinates of this point in . 47.A. Prove that the transition map between any two harts of a C r manifold is a C r -di eomorphism. 47.B. Prove that the C r -stru ture of a C r -manifold an be re overed from a olle tion of its lo al oordinate systems if the supports of these lo al oordinate systems over the whole manifold. 47.C. Prove that the boundary of the underlying manifold of a C r manifold equipped with the indu ed C r -stru ture is a C r -manifold. 47.D. Under what onditions the produ t of two C r -manifolds ( onsidered as a produ t in the ategory of C r -spa es) is a C r -manifold? Manifolds with Corners A C r -spa e is alled an n-dimensional smooth manifold of lass C r with
orners if it is modelled on (R+ )n and the underlying topologi al spa e is Hausdor and se ond ountable. As it follows immediately from the de nition, its underlying topologi al spa e is a (topologi al) manifold of dimension n. Of ourse, any smooth manifold of lass C r is a smooth manifold of lass C r with orners. 47:A. Prove that there exists a smooth manifold of lass C r with orners whi h is not a smooth manifold of lass C r . 47:B. Prove that the produ t of any two C r -manifolds with orners is a C r -manifold with orners. In parti ular, the produ t of any two C r -manifolds (without orners) is a C r -manifold with orners.
Traditional Approa h to Smooth Manifolds The theory presented in the previous se tion is a natural generalization of the traditional theory whi h treats only smooth manifolds. The traditional theory was rst developped in full generality by H. Whitney
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(Di erentiable manifolds, Annals of Mathemati s, 37 (1936) 645{680), athough one should mention also a book by Hermann Weyl (Die Idee der Riemanns hen Fla hen, Teubner, Leipzig, Berlin, 1923) where the same s heme was applied in the ase of one-dimensional omplex manifolds. Now it is ommonly a
epted. Here we sket h it. Doing this, we rede ne several notions introdu ed above. Eventually, the newly introdu ed notions will be identi ed with their previously introdu ed versions, but for a while, to avoid onfusion, we will refer to the notions de nined above adding the words in the sense of di erential spa es. Let X be a manifold of dimension n, let U X an open set, : U ! G a homeomorphism onto an open set of either R n or R n+ . Then is alled a hart or (lo al) oordinate system in X , the set U is alled the support of . The ompositions of and the oordinate proje tions G ! R are denoted by 1 , 2 , : : : n and alled oordinates in the oordinate system . The value of i at a 2 U is alled the i-th oordinate of a in . Let : U ! G and : V ! H be harts in X . Then there appear harts j : U \ V ! (U \ V ), j : U \ V ! (U \ V ) and homeomorphism ( j) Æ ( j) 1 : (U \ V ) ! (U \ V ). The latter is alled the transition map from to and denoted by t . This map al ulates the oordinates of a point in given the oordinates of this point in . Sin e (U \ V ) and (U \ V ) are open subsets of R n or R n+ , all the notions developed in Cal ulus an be applied to t . In parti ular, t may be of lass C r . If t is a C r -di eomorphism then and are said to be C r -related. If U \ V = ?, the harts also are C r -related. 47.E. C r -related harts are C s-related for any s < r. A olle tion of mutually C r -related harts whose supports over X is
alled a C r -atlas of X . Two C r -atlases are said to be C r -equivalent if their union is a C r -atlas. 47.F. Reformulate the de nition of C r -equivalen e of atlases in terms of transition maps. 47.G. Prove that C r -equivalen e of C r -atlases is an equivalen e relation. Only transitivity in 47.G is not absolutely obvious, is it? A lass of C r -equivalent atlases of a manifold X is alled a di erential stru ture of lass C r on X , or di erentiable2 stru ture of lass C r on X , or smooth stru ture of lass C r on X , or just C r -stru ture. A pair 2 Of
ourse, nobody di erentiates this di erentiable stru ture!
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onsisting of a manifold X and a C r -stru ture on X is alled a smooth (or di erential, or di erentiable) manifold of lass C r or a C r -manifold. 47.H. (Cf. 47.B.) A di erential stru ture is determined by any atlas belonging to it. 47.I. A di erential stru ture ontains a maximal atlas. This is the unon of all the atlases of this stru ture. Sometimes the maximal C r -atlas is alled C r -stru ture. Although we do not identify them, we say that a hart belongs to a C r -stru ture and is a
oordinate system of the orresponding C r -manifold if it belongs to the maximal C r -atlas. 47.J. Let X be a C r -manifold, a 2 X , and f : X ! R a fun tion. Let : U ! G and : V ! H be harts with supports ontaining a. Then for any s r if the omposition G > 1 >> U > f jU >> R is a C sfun tion at (a) then H > 1 >> V > f jV >> R is a C s -fun tion at (a). A fun tion f : X ! R de ned on a C r -manifold X is said to be of lass C s (with s r) at a 2 X if for some (and hen e, by 47.J, for any) hart : U ! G with U 3 a the omposition G > 1 >> U > f jU >> R is a fun tion of lass C s at (a). A fun tion is said to be a C s -fun tion if it is of lass C s at ea h a 2 X .
Equivalen e of the Two Approa hes
47.K. All the C r -fun tions on a C r -manifold X omprise a C r -stru ture
on X in the sense of di erential spa es. With respe t to this C r -stru ture, all the harts of X are harts in the sense of di erential spa es. In parti ular, as a di erential spa e, X is a C r -manifold in the sense of di erential spa es.
47.L. Let X be a C r -manifold in the sense of di erential spa es. Then
its harts in the sense of di erential spa es omprise a C r -atlas turning X into a C r -manifold in the traditional sense. Cf. 47.A and 47.B. Thus we have two onversions: any C r -manifold an be onverted as indi ated in 47.K to a C r -spa e whi h is a C r -manifold in the sense of di erential spa es, and any C r -manifold in the sense of di erential spa es
an be onverted as indi ated in 47.L to a C r -manifold in the traditional sense. 47.M. These two onversions are inverse two ea h other: both of their
ompositions are identity.
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Revision of Boundary Let X be a smooth manifold of lass C r and dimension n, and let U > >> G R n+ be a hart belonging to its C r -stru ture. Then (U \ X ) = G \ R n 1 is an open set of the boundary hyperplane R n 1 of R n and j : U \ X ! G \ R n 1 is a lo al oordinate system in X . 47.N. Lo al oordinate systems in X obtained in this way from lo al oordinate systems belonging to the C r -stru ture of X de ne a C r stru ture on X . 47.O. The C r -stru ture on X de ned in 47.N oin ides with the one indu ed on X as on di erential subspa e of X , f. 47.C.
Revision of Multipli ation Let X and Y be smooth manifolds of lass C r and dimensions p and q , respe tively. Let Y = ?. For harts : U ! G and : V ! H belonging to the C r -stru tures of X and Y , respe tively, de ne map : U V ! G H : (a; b) 7! ( (a); (b)): This is a hart in X Y . 47.P. All the harts of this sort are C r -related to ea h other. The C r -stru ture de ned by an atlas, whi h onsists of harts of this type, is meant when one says on X Y as on manifold of lass C r . 47.Q. The C r -stru ture on X Y de ned by an atlas, whi h onsists of harts of the type des ribed above, oin ides with the C r -stru ture de ned as on a produ t of di erential spa es. Cf. 46.1 and 47.D.
Revision of Di erentiable Maps Let X , Y be smooth manifolds of lass C r and f : X ! Y a map. Suppose f is ontinuous at a 2 X . 47.R. Let : V ! H be a hart of Y with f (a) 2 V . Prove that there exists a hart : U ! G of X with a 2 U and f (U ) V . The map Æ (f jU;V ) Æ 1 is alled a representative of f in lo al oordinate systems and . We denote it by f . The map f is said to be of lass C s (with s r) at a if there a representative f of f : X ! Y is of lass C s with s r at (a). 47.S. Prove that this does not depend on the hoi e of oordinate sys tems: if there is a representative f of f : X ! Y whi h is of lass C s at (a) then any other representative f~~ of f is of lass C s at ~(a).
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Hen e, the lass of a map at a point is well-de ned. A map f : X ! Y is said to be of lass C s if at ea h a 2 X it is of lass C s . 47.T. Let X , Y be C r -manifolds. A map f : X ! Y is of lass C r in the sense de ned above, i it is a C r -map in the sense of di erential spa es.
Rank of Mapping
47.U. Prove that the rank of the Ja obian matrix (thematrix of the
rst order partial derivatives) at (a) of a representative f of f does not depend on the hoi e of and . This rank is denoted by rka f and alled the rank of f at a. Let f : X ! Y be a C r -map. A point b 2 Y is alled a regular value of f if rka f = dim Y at ea h point a 2 f 1 (b).
Di erential Topology Let X and Y be smooth manifolds of lass C r . Re all that a map f : X ! Y is a di eomorphism of lass C r if it is of lass C r , invertible and the inverse is also of lass C r . (Cf. above.) Information: If X , Y are C r -manifolds and there exists a di eomorphism X ! Y of lass C 1 then there exists a C r -di eomorphism X ! Y .
Smooth manifolds XS , Y are said to be di eomorphi if there exists a di eomorphism X ! Y . Information: There exists homeomorphi , but not di eomorphi smooth manifolds. The lowest dimension of su h manifolds is four. Prove that if two one-dimensional smooth manifolds are homeomorphi , they are also di eomorphi .
47.3.
Di erential topology is a bran h of mathemati s whi h studies properties of smooth manifolds preserved by di eomorphisms.
Submanifolds In Se tion 45 any subset of a C r -spa e was equipped with the indu ed C r stru ture. If we onsider only smooth manifolds then a subset, whi h an re eive a stru ture, must satisfy strong restri tions. It must be a manifold and positioned in su h a ni e way that the stru ture of C r -spa e indu ed as it was des ribed in Se tion 45 would turn it to a smooth manifold. Moreover, for some reasons usually one imposes extra onditions on the position in the ambient manifold.
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Let X be a smooth manifold of lass C r and dimension n, and A be a subset of X . One says that A is a smooth k-dimensional subset of X if at ea h b 2 A there exists a hart U > >> G of X su h that the pair ( (U ); (U \ A)) oin ides with one of the following pairs: (R n ; R k ), (R n ; R k+ ), (R n+ ; R k+ ). The submap j : U \ A ! (U \ A) is a hart of A. 47.V. Prove that all the harts obtained in this way from harts belonging to the same C r -stru ture of X are C r -related. 47.W. The C r -stru ture on a smooth subset des ribed above oin ides with the smooth stru ture indu ed on the subset from the ambient C r manifold as it was de ned in Se tion 45. The smooth subset A equipped with the C r -stru ture whi h is de ned by the harts of this sort is alled a (smooth C r -) submanifold of X . The de nition of smooth subset gives a lear idea of what a smooth subset is. It says that in a neighborhood of ea h of its points a smooth subset looks like and pla ed in the ambient manifold either as R k in R n , or R k+ in R n , or R k+ in R n+ . However, this de nition is not onvinient when one wants to he k if some spe ial set is smooth. Now we onsider its reformulations more adapted for this kind of problems. For the sake of simpli ity we restri t ourselves to the ase of proper smooth subsets, i.e., smooth subsets with A = X \ A. In the de nition of proper smooth sets one an skip the pair (R n ; R k+ ). 47.X. Prove that A is a proper smooth k-dimensional subset of a smooth manifold X , i for ea h b 2 A there exists a lo al oordinate system : U ! R n(+) of X (where R n(+) denotes either R n or R n+ ) and a di erentiable map f : R k(+) ! R n k su h that (A \ U ) is the graph of f .
47.Y. Prove that A is a proper smooth k-dimensional subset of a smooth manifold X , i for ea h b 2 A there exists a lo al oordinate system : U ! R n(+) of X and a di erentiable map f : R n(+) ! R n k su h that 0 2 R n k is a regular value of f and, if R n(+) = R n+ , a regular value of f j Rn+ and (A \ U ) = f 1 (0) Cf. Impli it Fun tion Theorem 44.L.
48. Immersions and Embeddings Immersions Let X and Y be C r -manifolds. immersion if its rank at ea h point of X is equal to the dimension of X .
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Whi h of the following di erentiable maps are immersions: (a) id : R ! R, (b) a onstant map R ! R, ( ) the proje tion R2 ! R, (d) S 1 ! R2 : ( os 2t; sin 2t) 7! (sin 2t; sin 4t), (e) S 1 ! S 1 : z 7! z 2, (f) R1 ! R2 : t 7! (t2 ; t3 ), (g) R ! R2 : t 7! (t2 ; t4 ), (h) R ! S 1 S 1 : t 7! (eit ; eit ), (i) I ! R2 : t 7! (sin t; sin 2t), (j) R2 ! R3 : (x; y) 7! ( os x; sin x os y; sin x sin y), (k) R2 ! R3 : (x; y) 7! ( os x(2 + os y); sin x(2 + os y); sin y)? 48.2. Prove that an immersion of a losed smooth manifold to a losed onne ted smooth manifold of the same dimension is a overing with a nite number of sheets. 48.3. Is the same true for ompa t manifolds with boundary? 48.4. How to generalize 48.2 to the ase of ompa t manifolds with boundary, anyway? 48.5. Does there exist an immersion S 2 ! R2 ? What about immersion S 1 S 1 ! R2 ? Find a generalization for the answers to these questions. 48.6. Does there exist an immersion of a handle (i.e., torus with a hole) to the plane? 48.1.
Di erentiable Embeddings Re all that a map f : X ! Y of a C r -spa e X to a C r -spa e Y is alled a C r -embedding if its submap f : X ! '(X ) is a C r -di eomorphism. In the traditional approa h to smooth manifolds, one should add to this an additional ondition, be ause the image f (X ) is not a smooth manifold automati ally. Thus the de nition looks as follows: A C r -map f : X ! Y is alled a di erentiable embedding if f (X ) is a smooth subsmanifold of Y and f j : X ! f (X ) is a di eomorphism. 48.A. The in lusion of a smooth submanifold to the smooth manifold is a di erentiable embedding. 48.7. Whi h of the following di erentiable maps are di erentiable embeddings: (a) id : R ! R, (b) S 1 ! R2 : ( os 2t; sin 2t) 7! (sin 2t; sin 4t), ( ) S 1 ! S 1 : z 7! z 2, (d) R1 ! R2 : t 7! (t2 ; t3 ), (e) R ! R2 : t 7! (t2 ; t4 ), (f) I ! R2 : t 7! (sin t; sin 2t), (g) [0; 1) ! R2 : t 7! (sin t; sin 2t), (h) (0; 1) ! R2 : t 7! (sin t; sin 2t), (i) R ! S 1 S 1 : t 7! (eit ; eit ), (j) S 1 ! S 1 S 1 : z 7! (z 3 ; z 2),
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(k) S 1 ! S 1 S 1 ( : z 7! (z 4 ; z 2), (x; x + x5 ); if x 0 (l) R ! R2 : x 7! ? (x; x x44 ); if x < 0
Immersions Versus Embeddings
48.B Embedding Is Immersion.
Any di erentiable embedding of a smooth manifold to a smooth manifold is an immersion. 48.C Immersion Is Embedding Lo ally. Let f : X ! Y be an immersion. Prove that ea h point a 2 X has a neighborhood U su h that f jU : U ! Y is a di erentiable embedding, unless f (a) 2 Y . 48.8. Riddle.
What if, under the onditions of 48.C, f (a) 2 Y ?
48.D Di . Embedding = Top. Embedding + Immersion. Let X and Y be C r -manifolds. A map f : X ! Y is a C r -di erentiable embedding, i f is a topologi al embedding and C r -immersion. 48.D.1. Let X and Y be C r -manifolds, a 2 X and f : X ! Y be a topologi al embedding. Then for any neighborhoods U and V of a and f (a), respe tively, there exist open subsets U0 U and V0 V su h that f (U0 ) = V0 \ f (X ). 48.D.2 Straightening Immersion. Let U be an open set of Rn and f : U ! Rk be a C r -map su h that rka f = n for some a 2 U . Then there exist a neighborhood V U of a and a neighborhood W of f (a) and ~ Rk su h that C r -di eomorphisms g : V ! V~ Rn and h : W ! W 1 W \ f (U ) = f (V ) and h Æ (f j) Æ g is the linear embedding (x1 ; : : : ; xn ) 7! (x1 ; : : : ; xn ; 0; : : : ; 0).
Embeddability to Eu lidean Spa es In early years of topology (say, in papers by Henry Poin are) by a smooth manifold one meant what is alled smooth submanifolds of an Eu lidean spa e. It was not onvinient, be ause the embedding usually is quite irrelevant, and sometimes is not easy to nd. For example, R P n with n = 2k an be embedded into R 2 n, but does not admit an embedding into R 2n 1 . The standard smooth stru ture is easier to get from the wellknown natural two-fold overing spa e, whi h is S n, than to des ribe a smooth embedding into an Eu lidean spa e. So, the transition to Whitney's abstra t de nition of smooth stru tures was well-motivated. However the transition poses the question if the set of obje ts really hanges. The answer is negative: any smooth manifold an be smoothly embedded into an Eu lidean spa e. This was proved for C r -manifolds with r 1 by Whitney in the very same paper (Di erentiable manifolds, Annals of Mathemati s, 37 (1936) 645{680), where he introdu ed di erential stru tures. The real analyti ase was done about twenty years later by Grauert and Remmert.
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Here we onsider the simplest ase of the embedding theorems. 48.E. Any ompa t C r -manifold an be C r -di erentiably embedded into
Eu lidean spa e of suÆ iently high dimension. 48.E.1. To embed a ompa t C r -manifold X to RN it is ne essary and suÆ ient to onstru t N real valued fun tions f1, : : : , fN of lass C r on X ir su h that for any a 2 X there exist i1 < i2 < < in with rk f xs (a) = n and there exist i; j with fi(a) 6= fj (a). 48.E.2. Ea h point of a C r -manifold of dimension n has a neighborhood whi h admits a C r -embedding to Rn . 48.E.3. (Cf. 44.N) Let X be an n-dimensional C r -manifold, : U ! G a lo al oordinate system su h that G ontains a ball of radius 2 entered at the origin of Rn . Let g be a C r -fun tion de ned on U . Then there exists a C r -fun tion h : X ! R su h that hj 1 (Dn ) = gj 1 (Dn ) and h(x) = 0 for x 6=2 U .
The proof of 48.E sket hed above does not work for r = a, i.e., for real analyti fun tions. 48.9.
Where does it not work for real analyti embedding?
Theorem 48.E is orre t for real analyti ase too, but requires arguments of absolutely di erent nature. As it was mentioned above, these arguments were found by Grauert and Remmert in the fties.
Information: Any n-dimensional C r -manifold an be C r -embedded into R 2n . For r 1 existense of a C r -embedding to R 2n+1 was proved by H.
Whitney in Di erentiable manifolds, Annals of Mathemati s, 37 (1936) 645{680. Eight years later he managed to de rease the dimension of the Eu lidean spa e by one. The same Whitney's results ombined with the Grauert-Remmert te hnique give embedding to R 2n for a real analyti manifold of dimension n.
49. Tangent Ve tors As smooth manifolds generalize smooth surfa es lying in Eu lidean spa e, tangent ve tors to a smooth manifold generalize ve tors in Eu lidean spa e applied to a point of a surfa e and tangent to it. In literature there are at least three ompletely di erent ways of de ning ve tors tangent to a smooth manifold. Of ourse, the results are equivalent, but this appears as a surprise. The variety of de nitions an be partially explained by advantages of di erent de nitions in di erent situations, but the main reason is a di eren e in pedagogi al prin iples and experien e.
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Di erent people think about ve tors in di erent ways. A s hool math tea her thinks that it is a dire ted segment, or a lass of parallel equally dire ted segments of the same length. A physisist would laugh at this: he knows for sure that the ele tri eld strength is a ve tor, but has next to nothing to do with a dire ted segment. A usual de nition of a ve tor for physisists is that this is a quantity whi h is hara terized by a dire tion in the spa e and magnitude (the latter is a number depending on the hoi e of unit of measurement). For a mathemati ian, ve tor is just an element of a linear spa e. At rst glan e, this is the most general point of view. Both ve tors of s hool tea her and physisist are ve tors for mathemati ian, be ause one an sum them and multiply by a number and these operations are subje t of the same axioms. But when one needs to extend a de nition of ve tor to a new situation, the axiomati point of view is not reative. It is good mostly for throwghing away wrong andidates.
Coordinate De nition A physisist would probably agree with the following de nition of a ve tor: a ve tor is a quantity whi h an be hara terised by n real numbers (its
oordinates ) if it is taken in n-dimensional spa e and a oordinate system is xed. When the oordinate system hanges, the oordinates of a ve tor
hange a
ordingly. The rst de nition of tangent ve tor that we onsider ts to this s heme. Let X be a smooth manifold of lass C r and dimension n and a be a point of X . Denote by Ca the set of lo al oordinate systems of X with supports ontaining a. A tangent ve tor of X at a is a map v : Ca ! R n su h that for any ; 2 Ca v ( ) = (22) v ( ); a
where at (a).
a
is the Ja obian matrix of the transition fun tion from to
If v ( ) = (x1 ; : : : ; xn ) and v ( ) = (y 1; : : : ; y n) then formula (22) an be rewritten as follows: n X i j xi = (23) j (a)y ; j =1 Here the upper indi es are just indi es, not exponents. The unusual position is determined by the Einstein notations for multilinear algebra whi h
49. TANGENT VECTORS
210
are explained below. The main goals of these notations is to ex lude numerous summation signs and en ode the di eren e between elements of a ve tor spa e and the onjugate spa e. In the Einstein notations formula (23) looks as follows: i xi = j (a)y j ; see Digression on Einstein Notations below. The set of all the tangent ve tors of X at a is denoted by Ta X and alled the tangent spa e of X at a. 49.A. TaX is a ve tor spa e with respe t to oordinatewise operations (v + w)( ) = v ( ) + w( ) and (av )( ) = a(v ( )). 49.B. Any oordinate system 2 Ca de nes a map Ta X ! R n : v 7! v ( ): This map is a linear isomorphism. In parti ular, a ve tor v 2 Ta X is determined by v ( ), and v ( ) an be any element of R n . The oordinates of v ( ) (with respe t to the
anoni al oordinate system in R n ) are alled the oordinates of v in (or with respe t to) the lo al oordinate system .
Digression on Einstein Notations
To be written Di erentiation of Fun tions
To be written Di erential of Map
To be written Tangent Bundle Consider the set of all the tangent ve tors of a manifold X , i.e., [a2X Ta X . It is denoted by T X and alled the tangent bundle of X . For any lo al oordinate system X U > >> G Rn put T U = [a2U Ta X and de ne a bije tion T U ! G R n by formula TxX 3 v 7! ( (x); v ( )). By this bije tion one introdu es to T U topology and smooth stru ture from G R n
To be nished
51. ORIENTATION
Tangent Ve tors in Eu lidean Spa e
To be written
Ve tors as Velo ities
To be written
50. Ve tor Bundles To be written
General Terminology of Fibrations Trivial and Lo ally Trivial Indu ed Fibrations Ve tor Bundles Constru tions with Ve tor Bundles Tautologi al Bundles Homotopy Classi ation of Ve tor Bundles Low-Dimensional
51. Orientation To be written
211
52. TRANSVERSALITY AND COBORDISMS
Linear Algebra Digression: Orientations of Ve tor Spa e Related Orientations Orientation of Ve tor Bundle Orientation and Orientability of Smooth Manifold Orientation of Boundary Orientation Covering Proje tive Spa es
52. Transversality and Cobordisms To be written Sard Theorem Transversality Embedding to R2n+1
Normal Bundle and Tubular Neighborhood Pontryagin Constru tion Degree of Map Linking Numbers Hopf Invariant Thom Constru tion Cobordisms
212
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