Introduction to Algebraic Topologyuser.math.uzh.ch/berner/files/algtop.pdf · Topology Claim: XisconnectediﬀtheonlysubsetsofXthatarebothopenandclosedare∅and X. P

Introduction to Algebraic Topology

Mitschri der Vorlesung vonDr. M. Michalogiorgaki

Tobias BernerUniversität ZürichFrühjahrssemester 2009

Contents

1 Topology 41.1 Topological spaces and continuous functions . . . . . . . . . . . . . . . . . 4

1.1.1 Topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.2 Continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Conectedness & Compactness . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.1 Connected spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.2 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Algebraic topology 162.1 Fundamental group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.1.1 Path homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 e fundamental group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Covering spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.4 Liing properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.5 e fundamental group of the circle and applications . . . . . . . . . . . . 27

2.5.1 e Fundamental eorem of Algebra . . . . . . . . . . . . . . . . 282.5.2 Deformation retracts and homotopy type . . . . . . . . . . . . . . 29

2.6 Seifert von Kampen eorem . . . . . . . . . . . . . . . . . . . . . . . . . . 352.6.1 Direct sums of abelian groups . . . . . . . . . . . . . . . . . . . . . 35

2.7 Free abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.8 Free products of groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.8.1 Free groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.8.2 e Seifert-van Kampen theorem . . . . . . . . . . . . . . . . . . . 43

2.9 CW complexes (cell complexes) . . . . . . . . . . . . . . . . . . . . . . . . 472.10 Surfaces (two-dimensional manifolds) . . . . . . . . . . . . . . . . . . . . 49

2.10.1 Fundamental group of surfaces . . . . . . . . . . . . . . . . . . . . 492.10.2 Homology of surfaces . . . . . . . . . . . . . . . . . . . . . . . . . 502.10.3 Classi cation of surfaces . . . . . . . . . . . . . . . . . . . . . . . . 51

2.11 Knot theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522.12 Classi cation of covering spaces . . . . . . . . . . . . . . . . . . . . . . . . 58

Index 66

2

CONTENTS

Literature

• Muncres: Topology 2nd edition

• Jänich: Topologie

• Massey: Algebraic Topology: An Introduction

• Stöcker, Zieschang: Algebraische Topologie

• Lickorish: An introduction to knot theory

Introduction

In calculus, you have studied Rn, n ∈ N, as well as functions f ∶ Rk → Rℓ, k, ℓ ∈ N. Youhave studied notions such as neighbourhood of a pointx ∈ Rn, as well as convergence andcontinuity of a function f at a point x ∈ Rk. For this study, you have used the Euclideanmetric.For instance if f ∶ R→ R, we say that f is continuous at x ∈ R if ∀ε > 0 ∃δ > 0 such thatif ∣x − x0∣ < δ then ∣f(x) − f(x0)∣ < ε.We used the metric d ∶ R×R→ R, with (x, y)↦ ∣x−y∣. In general the Euclidean metricd ∶ Rn ×Rn → R with (x1, . . . , xn), (y1, . . . , yn)↦

√∑n

i=1(xi − yi)2.

In topology, we study notions such as neighbourhood of a point x ∈ X , convergence,continuity for a general setX . For this study a metric is not necessary. What we will useis open sets inX .

In algebraic topology, we use abstract algebra to study topological properties.

3

1 Topology

1.1. Topological spaces and continuous functions

1.1.1. Topological spaces

Consider a setX and P(X) ∶= {U ∣ U ⊆X}.

De nition 1.1 A topology onX is a collection T ⊂ P(X) such that

1. ∅,X ∈ T .

2. If Ui ∈ T ∀i ∈ I , then⋃i∈I Ui ∈ T .

3. If Ui ∈ T , i ∈ {1, . . . , n}, then⋂i∈{1,...,n}Ui ∈ T .

(X,T ) is called a topological space (sometimes we will only writeX).U ⊆X is called open, if U ∈ T .U ⊆X is called closed, ifX/U ∈ T .

Example 1.2

1. X some set.

• Td = {U ∣ U ⊆X} discrete topology.• Tt = {∅,X} trivial topology.

2. X = {x1, x2, x3}. en the three collections

• {∅,X}• {∅,{x1},{x1, x2},X}• {∅,{x2},{x1, x2},{x2, x3},X}

are topologies onX .

3. X is a set. en the collections

• Tf = {U ⊆X ∣X/U is nite orX/U =X} nite topology.

• Tc = {U ⊆X ∣X/U is countable orX/U =X} countable topology.

are topologies onX . [Homework]

4

1.1 Topological spaces and continuous functions

De nition 1.3 Suppose that T and T ′ are topologies onX . If T ′ ⊇ T , then T ′ is calledner than T . If T ′ ⊋ T , then T ′ is called strictly ner than T . If T ′ ⊇ T or T ⊇ T ′,

then T and T ′ are comparable.

De nition 1.4 Let (X,T ) be a topological space andA ⊆X .e interior ofA is

intA ∶= ⋃U∈T ,U⊆A

U .

e closure ofA isA ∶= ⋂

X/U∈T , A⊆UU .

Basis for a topology

De nition 1.5 A basis B for a topological space (X,T ) is a collection B of open sub-setesX (i.e. B ⊆ T ), such that ∀U ∈ T ∃{Bi}i∈I ,Bi ∈ B, ∀i ∈ I , with U = ⋃i∈I Bi.

Remark 1.6 A basis is not unique.Properties: B basis for (X,T ) then B has the following properties

1. ∀x ∈X ∃B ∈ B with x ∈ B.P X ∈ T and B is a basisÔ⇒ X = ⋃i∈I Bi, Bi ∈ B, i ∈ I . So x ∈ Bj forsome j ∈ I .

2. If x ∈ B1 ∩B2,B1,B2 ∈ B then ∃B3 ∈ B such that x ∈ B3 ⊆ B1 ∩B2.P B1,B2 ∈ B ⊆ T Ô⇒ B1,B2 ∈ T Ô⇒ B1 ∩ B2 ∈ T Ô⇒ B1 ∩ B2 =⋃i∈I Bi, …

Conversely: If a collection B of subsets ofX satis es properties 1. and 2. then there is aunique topology T for which B is a basis.T is called the topology generated by B and it consists of all unions of elements of B.P We prove that T is a topology

1. X ∈ T ?Property 1Ô⇒ ifx ∈X then∃Bx ∈ Bwithx ∈ Bx ⊆X . ereforeX = ⋃x∈X Bx,i.e. X ∈ T .

∅ ∈ T ?∅ is the empty union of elements in B, so ∅ ∈ T .

2. If Ui ∈ T ∀i ∈ I , does⋃i∈I Ui ∈ T ?Ui ∈ T Ô⇒ Ui = ⋃j<inJ Bij . So⋃i∈I Ui = ⋃i∈I ⋃j∈J Bij ∈ T .

3. If Ui ∈ T for i ∈ {1, . . . , n}, is⋂i∈{1,...,n}Ui ∈ T ?We will show that if U1, U2 ∈ T then U1 ∩U2 ∈ T .U1 = ⋃λ∈ΛBλ,U2 = Uk∈KBk . Consider x ∈ U1∩U2. en x ∈ Bλx and x ∈ Bkx ,for some λx ∈ Λ and kx ∈K . us x ∈ Bλx ∩Bkx Ô⇒ [by property 2.] ∃Bx ∈ Bwith x ∈ Bx ⊆ Bλx ∩Bkx ⊆ U1 ∩U2.erefore U1 ∩U2 = ⋃x∈U1∩U2

Bx ∈ T .

5

Topology

Example 1.10

1. X = R, B = {(a, b) ∣ a, b ∈ R, a < b}. B satis es properites 1. and 2. therefore itgenerates a topology on R. is is the standard topology on R.

2. X = R, B = {(a, b) ∣ a, b ∈ Q, a < b}B satis es properties 1. and 2.Claim [Homework]: e topology T generated by B is in fact the standard topol-ogy on R.

Lemma 1.11 Let B,B′ be bases for the topologies T and T ′ on X . en the followingstatements are equivalent:

1. T ′ ⊇ T .

2. ∀x ∈X , ∀B ∈ B with x ∈ B there isB′ ∈ B′ such that x ∈ B′ ⊆ B.

P

1.⇒2.Consider x ∈ X and B ∈ B with x ∈ B. B ∈ T ⊆ T ′ Ô⇒ B ∈ T ′ Ô⇒ B =⋃λ∈ΛB

′λ,B

′λ ∈ B′.

So x ∈ B, therefore x ∈ B′λ, for some λx ∈ Λ. i.e. we have B′λx∈ B′, such that

x ∈ B′λx⊆ B.

2.⇒1.Consider U ∈ T and x ∈ U . en x ∈ U = ⋃λ∈ΛBλ, Bλ ∈ B. at is x ∈ Bλx , forsome λx ∈ Λ. 2. implies that there existsB′λx

∈ B′ such that x ∈ B′λx⊆ Bλx .

en U ⊆ ⋃x∈U B′λx⊆ U Ô⇒ U = ⋃x∈U B

′λx

Ô⇒ U ∈ T ′. erefore T ⊆ T ′.

Product topology

Let (X,Tx), (Y,TY ) be two topological spaces. e product topology is the topologywith basis the collection B = {U × V ∣ U ∈ TX , V ∈ TY }.

Metric topology

X is a set.

De nition 1.13 A metric on this set is a function d ∶X ×X → R with

1. d(x, y) ≥ 0, and d(x, y) = 0 iff x = y.

2. d(x, z) ≤ d(x, y) + d(y, z) ∀x, y, z ∈X .

3. d(x, y) = d(y, x), ∀x, y ∈X .

We call the setBd(x, ε) = {y ∈X ∣ d(y, x) < ε} the ε-ball centered at x.

{Bd(x, ε)}x∈X,ε>0 is a basis for a topology onX , the metric topology.

6

1.1 Topological spaces and continuous functions

Example 1.14 R with the standard topology (Tstand).Consider R ×R = R2.Claim: e product topology on R2 and the metric topology are the same.For this, one has to show

• Tpr ⊆ TmBy lemma 1.11 one has to show ∀x ∈ R2, ∀Bpr ∈ Bpr, x ∈ Bpr, ∃Bm ∈ Bm suchthat x ∈ Bm ⊆ Bpr.

• Tm ⊆ Tpr…

Subspace topology

De nition 1.15 (X,T ) is a topological space and Y ⊆ X . e collection TY ∶= {U ∩Y ∣ U ∈ T } is a topology on Y , called the subspace topology.

Example 1.16 Y = [0,1] ∪ {2} ⊆X = R with the standard topology.en the sets

• (a, b), with a, b ∈ [0,1],

• [0, b), with b ∈ [0,1],

• (a,1], with a ∈ [0,1],

• {2},

• [0,1]

are open sets in the subspace Y .

1.1.2. Continuous functions

De nition 1.17 Let (X,TX), (Y,TY ) be topological spaces and f ∶ X → Y . f iscalled continuous if f−1(V ) ∈ TX ∀V ∈ TY .

Claim f ∶ R→ R (with standard topology). e ε–δ de nition of continuity is equivalentto the de nition above.P

“⇐” Suppose that f ∶ R→ R is continuous with the de nition above.Consider x0 ∈ R and ε > 0. en V = (f(x0) − ε, f(x0) + ε) ∈ Tstand, sof−1(V ) ∈ Tstand by our de nition.Now x0 ∈ f−1(V ), so there exists (a, b) ∈ Bstand with x0 ∈ (a, b) ⊆ f−1(V ).Take δ = min{x0 − a, b − x0}. Clearly δ > 0. en ∣x − x0∣ < δÔ⇒ x ∈ (a, b) ⊆f−1(V )Ô⇒ f(x) ∈ VÔ⇒ f(x) ∈ (f(x) − ε, f(x) + ε)Ô⇒ ∣f(x) − f(x0)∣ < ε.

7

Topology

eorem 1.19 LetX,Y be topological spaces and f ∶ X → Y . e following are equiva-lent

1. f is continuous.

2. for every closed subset of Y the inverse image of it is a closed subset ofX .

P

1.⇒2.Let B be closed in Y , then X/f−1(B) = f−1(Y /B) which is open in X , i.e.f−1(B) is closed inX .

2.⇒1.…

Lemma 1.21 (the pasting lemma)Let X,Y be topological spaces and A,B closed subsets of X with X = A ∪ B. Letf ∶ A → Y , g ∶ B → Y be continuous with f(x) = g(x) ∀x ∈ A ∩B. en h ∶ X → Ywith

h(x) = { f(x) x ∈ Ag(x) x ∈ B

is continuous.P LetC be a closed subset of Y . en h−1(C) = f−1(C)∪g−1(C)where f−1(C)is closed inA and g−1(C) is closed inB.

(a) f−1(C) is closed inAÔ⇒ f−1(C) = A ∩G, whereG closed inX .

(b) g−1(C) is closed inBÔ⇒ g−1(C) = B ∩H , whereH closed inX .

(a)Ô⇒ f−1(C) is closed inX ,(b)Ô⇒ g−1(C) is closed inX .Ô⇒ h−1(C) = f−1(C) ∪ g−1(C) is closed inX .

De nition 1.23 LetX,Y be topological spaces. f ∶X → Y is called a homeomorphismif f is bijective, and f, f−1 are continuous.

Bijective correspondence not only between X and Y but also between the collection ofopen sets inX and the collection of open sets Y .us any property of X that is expressed in terms of its open subsets yields via f thecorresponding properties for Y . Such a property is called a topological property.

Quotient topology

Example 1.24 Torus

De nition 1.25 LetX,Y be topological spaces, p ∶X → Y a surjective map. e mapp is called a quotient map if a subset U of Y is open in Y , if and only if p−1(U) is openinX .

8

1.2 Conectedness & Compactness

De nition 1.26 LetX be a topological space,Y be some set and p ∶X → Y a surjectivemap. e quotient topology on Y induced by p is de ned as follows:A subset U ⊆ Y is open if and only if p−1(U) ⊆X is open.

e fact that this is a topology follows from p−1(∅) = ∅, p−1(Y ) = X , p−1(⋃a∈J Ua) =⋃a∈J p

−1(Ua), p−1(⋂ni=1Ui) = ⋂n

i=1 p−1(Ui).

Remark 1.27 equotient topology onY is the nest topology thatmakes p continuous.Example 1.28

1. X = {(x, y) ∈ R2 ∣ 0 ≤ x ≤ 2π, 0 ≤ y ≤ 1} = [0,2π] × [0,1] ⊆ R2. Y ={(x, y, z) ∈ R3 ∣ x2 + y2 = 1, 0 ≤ z ≤ 1}.f ∶X → Y , with (x, y)↦ (cosx, sinx, y).f is surjective. Using f we can de ne the quotient topology in Y .

2. p ∶ R→ {x1, x2, c3}

p(x) =⎧⎪⎪⎪⎨⎪⎪⎪⎩

x1 x > 0x2 x < 0x3 x = 0

equotient topology on{x1, x2, x3} induced byp is{∅,{x1},{x2},{x1, x2},{x1, x2, x3}}

1.2. Conectedness & Compactness

In calculus you have studies functions f ∶ [a, b]→ R and you have proved three theoremsfor continuous functions f ∶ [a, b]→ R.

1. Intermediate Value eorem (IVT),

2. Maximus Value eorem (MVT),

3. Uniform Continuity eorem (UCT).

ese theorems rely on the continuity of f and some topological properties of [a, b] ⊂R. In particular IVT relies on the connectedness of [a, b]. MVT and VCT rely on thecompactness of [a, b].

1.2.1. Connected spaces

De nition 1.29 Consider (X,T ) topological space. A separation of X is a pair U,Vwhere U,V ∈ T , U,V ≠ ∅, U ∩ V = ∅, andX = U ∪ V . X is called connected if thereis no separation ofX .

Remark 1.30 Connectedness is a topological property.Example 1.31

1. R, Tstand,Q = ((−∞, a) ∩Q) ∪ ((a,∞) ∩Q), a ∈ R ∖Q.

2. R, [a, b], [a, b), (a, b], (a, b) are connected.

9

Topology

Claim: X is connected iff the only subsets ofX that are both open and closed are ∅ andX .P

“⇒” Suppose, that A ⊆ X , A ≠ ∅, A ≠ X , and A ∈ TX , X ∖ A ∈ TX . en X =A ∪ (X ∖A)Ô⇒A,X ∖A is separation ofX , contradiction.

“⇐” If U,V is a separation of X . en U ∈ TX , X ∖ U = V ∈ Tx, U,X ∖ U ≠ ∅,U ∩ (X ∖U) = ∅. V ∈ TX ,X ∖U ∈ TX , U ≠ ∅, U ≠X , contradiction.ereforeX is connected.

Lemma 1.33 If C,D are a separation of X and Y is a connected subspace of X , thenY ⊆ C or Y ⊆D.P C ∩ Y , D ∩ Y are open subsets of Y . In addition (C ∩ Y ) ∩ (D ∩ Y ) = ∅. IfC ∩ Y andD ∩ Y were both nonempty, then they would form a separation of Y . But Yis connected. So C ∩D = ∅, orD ∩ Y = ∅.Ô⇒ Y ⊆ C or Y ⊆D.

eorem 1.35 Let {Ai}i∈I be connected subspaces ofX and p ∈ ∩i∈IAi. en ∪i∈IAi isconnected subspace ofX .

P Assume that Ui∈IAi is not connected, and ∪i∈IAi = C ∪D, C,D a separationof ∪i∈IAi. en p ∈ C ∪D and WLOG we can assume that p ∈ C . Ai is a connectedsubspace.Ô⇒ [Lemma]Ai ⊆ C orAi ⊆D for any i ∈ I . p ∈ ∩i∈IAi.Ô⇒ p ∈ Ai ∀i ∈ I .Ô⇒Ai ⊆ C ∀i ∈ I .Ô⇒ ∪i∈IAi ⊆ C .Ô⇒D = ∅. Contradiction.

eorem 1.37 Let A be a connected subspace of X and A ⊆ B ⊆ A. en B is alsoconnected.

P Assume that C,D is a separation of B. By Lemma A ⊆ C or A ⊆ D. WLOGA ⊆ C .A ⊆ C Ô⇒A ⊆ C .D ⊆ B ⊆ A ⊆ C Ô⇒D ⊆ C .Claim: D ∩C = ∅.Proof: e closure of C in B is C ∩B, where C is the closure of C inX . C = C ∩B =C ∩ (C ∪D) = (C ∩C) ∪ (C ∩D) = C ∪ (C ∩D).Ô⇒ C ∩D = ∅.

We have thatD ⊆ C and we showed thatD ∩C = ∅. ereforeD = ∅, contradiction.ere is no separation ofB, i.e. B is connected.

eorem 1.39 Let f ∶X → Y be a continuous function between topological spacesX andY . IfX is connected, then f(X) is connected.

10


P Consider f ∶ X → f(X), x ↦ f(x). en f is continuous. Indeed, if U openin f(X), then U = f(X) ∩ V for some V open in Y . f1(U) = f−1(f(X) ∩ V ) =f−1(f(X)) ∩ f−1(V ) =X ∩ f−1(V ) = f−1(V ) open inX .Suppose that f(X) = C∪D,C,D separation of f(X). en f−1(C), f−1(D) are opensubsets of X , disjoint and nonempty and X = f−1(C) ∪ f−1(D). Contradcition (Xconnected).

eorem 1.41 A nite (cartesian) product of connected spaces (in the product topology) isconnected.

P We start by proving that if X and Y are connected, then X × Y is connected.Consider a × b in X × Y , X × b. X × b is connected. Similarly x × Y is connected forany x ∈X . As a result, Tx = (X × b) ∪ (x × Y ) is also connected (by previous theorem,since x × b ∈ (X × b) ∩ (x × Y )).erefore, ∪x∈XTx is connected, as the union of connected subspaces, Tx with commonpoint a × b. NowX × Y = ∪x∈XTx, thereforeX × Y is connected.By induction, the general proof follows.

eorem 1.43 (IVT)Let f ∶ X → R be a continuous function and X be a connected space. If a, b ∈ X andr ∈ (f(a), f(b)), then ∃c ∈X with f(c) = r.

P A = f(X)∩ (−∞, r),B = f(X)∩ (r,∞). A,B are open in f(X),A∩B = ∅,A ≠ ∅,B ≠ ∅ (as f(a) ∈ A, f(b) ∈ B).If r ∉ f(X), then f(X) = A ∪B, i.e. A,B is a separation of f(X). However f(X) isconnected, because X is connected and f is continuous. I.e. such a separation cannotexist. I.e. r ∈ f(X), that is, ∃c ∈X such that f(c) = r.Remark 1.45 One can prove an even more generalised form of IVT, where instead of Rone considers any ordered set.

De nition 1.46 Let X be a topological space and x, y ∈ X . A path in X from x to yis a continuous map f ∶ [a, b] → X with f(a) = x and f(b) = y. X is called pathconnected, if every pair of points inX can be joined by a path in x.

Claim: IfX is path connected, thenX is connected.P Suppose X = C ∪ D, C,D a separation of X . Consider f ∶ [a, b] → X apath. f([a, b]) is connected, so f([a, b]) ⊆ C or f([a, b]) ⊆ D. is implies thatthere is no path in X joining a point in C to a point in D. Ô⇒ X not path connected,contradiction.Example 1.48 (Topologist’s sine curve S)Consider S = {(x, sin 1

x) ∣ 0 < x ≤ 1} ⊆ R2.

f ∶ (0,1] → R2, x ↦ (x, sin 1x). f is continuous, (0,1] is connected, therefore S =

f((0,1]) is connected.S = S ∪ {0 × [−1,1]}. S is connectedÔ⇒ S is connected.

Claim: S is not path connected.Proof: Suppopse f ∶ [a, c] → S continuous with f(a) = (0,0), and f(c) ∈ S. f−1(0 ×[−1,1]) is closed in [a, c] therefore it has a largest element b. en f(b) ∈ 0 × [−1,1]

11

Topology

and f((b, c]) ∈ S. Let f(t) = (x(t), y(t)), t ∈ [a, c]. en x(a) = 0, x(t) > 0 andy(t) = sin 1

x(t) for t > b. Given n ∈ N, choose un with 0 < un < x(b + 1n) such that

sin 1un= (±1)n. By the IVT there exists tn ∈ (b, b + 1

n) with x(tn) = un. en tn → b

but y(tn) does not converge. Contradicting the continuity of f .Remarks on yesterday’s example

• If X = R2 and S ⊆ X , then S = {limits of convergent sequences of points in S}.More generally, this is true ifX is rst countable, e.g. X is a metric space.

• If f ∶ [a, c] → S continuous. Of t ∈ [a, c] ∀n ∈ N, with tn → b, b ∈ [a, c], thenf(tn)→ f(b).In general, continuous implies sequentially continuous.

De nition 1.49 Given a topological space X , de ne an equivalence relation ∼ on X ,by setting x ∼ y (x, y ∈X) if there exists a connected subspace ofX with x, y ∈ Y . eequivalence classes are called the connected component ofX .De ne another equivalence relation ∼ by setting x ∼ y (x, y ∈X) if there is a path inXfrom x to y. e equivalence classes are called path components ofX .

1.2.2. Compactness

De nition 1.50 A topological space X is called compact if for any open covering{Uλ}λ∈Λ of X , i.e. Uλ open ∀λ ∈ Λ and X = ∪λ∈ΛUλ, there exist nitely manyλ1, . . . , λn ∈ Λ such thatX = ∪ni=1Uλi .

Compactness is a topological property.Example 1.51

1. R is not compact, R = ⋃r∈R(r − 12, r + 1

2), or R = ⋃n∈N(n,n + 2).

2. (a, b], a, b ∈ R is not compact. (a, b] = ⋃n∈N(a + 1n, b].

3. Subspaces ofX with nitely many points are obviously compact.

4. X = {0}∪{ 1n∣ n ∈ N} is compact. Consider {Uλ}λ∈Λ an open covering ofX and

considerλ0 ∈ Λwith 0 ∈ Uλ0 . en there existsNλ0 sucht that 1n∈ Uλ0 ∀n ≥ Nλ0 .

For each n ∈ {1, . . . ,Nλ0 − 1} choose Uλn containing it. en⋃Nλ0

i=0 Uλi =X .

Order topology

De nition 1.52 A relationC on a setX is an order relation if it has the following prop-erties:

1. If x, y ∈X , x ≠ y then xCY or yCx.

2. For no x ∈X does xCx hold.

3. If xCy, yCz then xCz, x, y, z ∈X .

12


Example 1.53 X = R, C = <.IfX is a set with an order relation < and a, b ∈X , a < b.

(a, b) = {x ∈X ∣ a < x < b},(a, b] = {x ∈X ∣ a < x ≤ b},[a, b) = {x ∈X ∣ a ≤ x < b},[a, b] = {x ∈X ∣ a ≤ x ≤ b}.

De nition 1.54 X is a set with an order relation and more than two elements. Let Bbe the collection of all sets of the following types:

• (a, b), a, b ∈X ,

• (a, b0], a ∈X , b0 the largest element (if any) inX ,

• [a0, b), b ∈X , a0 the smallest element (if any) inX ,

B is a basis for a topology, and we call the topology generated by B the order topology.

Example 1.55 R2 with the dictionary order, i.e. a × b < c × d if a < c or a = c and b < d.R ×R has no largest or smallest element. [Graph .. open sets ... ]

An ordered setX has the least upper bound property, if every subset ofX that is boundedabove has a least upper bound.

eorem 1.56 LetX an ordered set with the least upper bound property. en any closedinterval [a, b] inX is compact.

[Without proof]Corollary 1.57 In R (Tstand = Tordered), [a, b] is compact for any a, b ∈ R.

eorem 1.58 A closed subspace Y of a compact spaceX is compact.

P Consider {Uλ}λ∈Λ an open covering of Y . enUλ = Y ∩U ′λ for someU ′λ openinX , ∀λ ∈ Λ. Denote U ′ = {U ′λ}λ∈Λ. Now U ′′ = U ′ ∪ {X ∖ Y } is an open covering ofX . X is compact, so a nite subcollection ofU ′′ coversX . If this containsX∖Y , discardit. If not I leave the subcollection unchanged. What we obtain is a nite subcollection ofU ′, say {U ′λ}i∈{1,...,n} with Y ⊆ ⋃n

i=1U′λi

. is implies, that {Uλi}i∈{1,...,n} is a nitesubcovering of Y .

eorem 1.60 X compact topological spaces. Y topological space and f ∶X → Y contin-uous. en f(X) is compact.

P Let U = {Uλ}λ∈Λ be an open covering of f(X). en ∀λ ∈ Λ Uλ = U ′λ ∩ f(X)for some U ′λ open in Y . Denote U ′ = {U ′λ}λ∈Λ. {f−1(U ′λ)}λ∈Λ is an open cover-ing of X . X is compact, hence it has a nite subcovering {f−1(U ′λi

)}i∈{1,...,n}. en{Uλi}i∈{1,...,n} is a nite covering ofX .

13

Topology

eorem 1.62 e product of nitely many compact spaces is compact.

[Proof omitted]

eorem 1.63 (Extreme value theorem)X is a compact topological space andY is an ordered set in the order topology. If f ∶X → Yis continuous then ∃c, d ∈X with f(c) ≤ f(x) ≤ f(d) ∀x ∈X .

P X is compact and f is continuous, so f(X) is compact. De ne (−∞, y) ∶= {a ∈Y ∣ a < y} and note that if Y has no smallest element, then (−∞, y) = ⋃ a∈Y

a<y(a, y) and

if Y has a smallest element a0, then (−∞, y) = [a0, y).(−∞, y) is an open set in Y .If f(X) has no largest element, then {(−∞, y) ∩ f(X)}y∈f(X) is open covering off(X). Since f(x) is compact, there exist yi, i ∈ {1, . . . , n}, such that {(−∞, yi) ∩f(X)}i∈{1,...,n} covers f(X). If yj = max{yi}i∈{1,...,n}, then yj ∈ f(X), but yj ∉{(−∞, yi) ∩ f(X)}, i ∈ {1, . . . , n}, contradiction, since {(−∞, yi) ∩ f(X)}i∈{1,...,n}is an open covering of f(X).Similarly we can prove, that f(X) has a smallest element.f

Hausdorff spaces

Terminology: X topological space, x ∈X , U open set inX with x ∈ U , then U is calleda open neighbourhood of x.

De nition 1.65 A topological space X is called Hausdorff if for every pair x1, x2 ∈X with x1 ≠ x2 there exist open neighbourhoods U1, U2 of x1, x2 respectively withU1 ∩U2 = ∅.

Facts:

• Every ordered set with the order topology is a Hausdorff space.

• IfX,Y are Hausdorff spaces, thenX × Y is Hausdorff.

• A subspace of a Hausdorff space is Hausdorff.

• Every nite point set in a Hausdorff space is closed.

eorem 1.66 Every compact subspace of a Hausdorff space is closed.

P X Hausdorff space, Y ⊆X compactÔ⇒ Y is closed.We will prove, thatX ∖ Y is open.Considerx ∈X∖Y . ∀y ∈ Y choseUy, Vy open inX withx ∈ Uy , y ∈ Vy andUy∩Vy = ∅(this can be done, because X is Hausdorff). en {Y ∩ Vy}y∈Y is an open coveringof Y . Y is compact, so ∃{Y ∩ Vyi}i∈{1,...,n} a nite subcovery of Y . Note that Y ⊆Vy1 ∪ ⋅ ⋅ ⋅ ∪ Vyn =∶ V . Ux ∶= Uy1 ∩ ⋅ ⋅ ⋅ ∩ Uyn . en Ux ∩ V = ∅. Indeed, if z ∈ Ux ∩ V ,then z ∈ Uyi ∀i ∈ {1, . . . , n} and z ∈ Vyj for some j ∈ {1, . . . , n} but Vyi ∩Uyj = ∅. SoUx ∩V = ∅. We have constructed Ux open inX with x ∈ Ux andUx ∩Y ⊆ Ux ∩V = ∅,i.e. Ux ⊆X ∖ Y .SoX ∖ Y = ∪x∈X∖Y Ux andX ∖ Y is open.

14


Remark 1.68 (X,T ) topological space, x ∈ X . Some authors (e.g. Munkres) de ne: Aneighbourhood ofX is a set U ∈ T such that x ∈ U . Other authors (e.g. Janich) de ne:A neighbourhood ofX is a set U ⊆X such that ∃V ∈ T with x ∈ V ⊆ U .

eorem 1.69 Let f ∶X → Y be a continuous bijective function . IfX is compact, and Yis Hausdorff, then f is a homeomorphism.

P We have to prove that f−1 is continuous. Equivalently tat ifU is open inX , then(f−1)−1(U) is open in Y . Equivalently, that if U is closed inX , then f(X) is closed inY .Indeed, U is closed in X and X compact, so f(U) ⊂ Y is compact. Y is Hausdorff,therefore f(U) is closed in Y .

Local connectedness and local path connectedness

De nition 1.71 X is called locally (path) connected at x ∈ X if for every open neigh-bourhood U of x there is a (path) connected open neighbourhood V of x with V ⊆ U .If Y is locally (path) connected at every x ∈X , thenX is called locally (path) connected.

15

2 Algebraic topology

Determining whether two spaces are homeomorphic and studying continuous functionsbetween topological spaces are two of the central problems in topology.To show thatX,Y are homemorphic, we need to construct f ∶X → Y bijective, contin-uous, with f−1 continuous.To show thatX,Y are not homeomorphic, we can for instance show thatX has a topo-logical property (e.g. connectendes or compactness) but Y does not have this property.Any continuous map between topological spaces induces a homeomorphism betweenthere fundamental groups.x0 ∈ X ↝ π1(X,x0) fundamental group. We will show that ifX ≃ Y then their funda-mental group is isomorphic.

2.1. Fundamental group

2.1.1. Path homotopy

Notation: I ∶= [0,1].

De nition 2.1 Let f ∶ X → Y , f ′ ∶ X → Y be continuous maps. f, f ′ are calledhomeotopic if there exists a continuous map F ∶ X × I → Y with F (x,0) = f(x)∀x ∈X and F (x,1) = f ′(x) ∀x ∈X .F is called a homotopy between f and f ′.If f, f ′ are homotopic, we write f ≃ f ′.If f is homotopic to a constant map then f is called nullhomotopic.

We focus on the special caseX = [a, b].en if f ∶ [a, b] → Y is a continuous function, f is called a path. f(a) is called theinitial point, and f(b) the nal point.WLOG we can assume, that the domain of f is [0,1].

De nition 2.2 Let f ∶ I → X , f ′ ∶ I → X two paths in X . f, f ′ are called pathhomotopic if f(0) = f ′(0), f(1) = f ′(1) and there exists continuous map F ∶ I × I →X , with F (x,0) = f(x) ∀x ∈ I and F (x,1) = f ′(x) ∀x ∈ I and F (0, t) = f(0)∀t ∈ I , F (1, t) = f(1) ∀t ∈ I .If f, f ′ are path homotopic, we write f ≃p f ′.

Claim: ≃ and ≃p are equivalence relations.P We prove this for ≃p.

1. f ≃p f :F (x, t) ∶= f(x).

16

2.1 Fundamental group

2. f ≃p f ′Ô⇒ f ′ ≃p f :If F is a path homotopy from f to f ′, thenG(x, t) ∶= F (x,1− t) is a path homo-topy from f ′ to f .

3. f ≃p f ′, f ′ ≃p f ′′Ô⇒ f ≃p f ′′:If F is a path homotopy from f to f ′ and F ′ is a path homotopy from f ′ to f ′′,then

G(x, t) ∶= { F (x,2t) t ∈ [0,1/2]F ′(x,2t − 1) t ∈ [1/2,1]

is a path homotopy from f to f ′′. G(x,1/2) = F (x,1) = F ′(x,0) = f ′(x). i.e.G is well de ned. IsG continuous?G is continuous in I×[0,1/2] andG is continuous in [1/2,1]Ô⇒ [pasting lemma(lemma 1.21]G is continuous.

G(0, t) = { F (0,2t) t ∈ [0,1/2]F ′(0,2t − 1) t ∈ [1/2,1] = {

f(0) = f ′(0)f ′(0) = f ′′(0)

G(1, t) = f(1),G(x,0) = f(x) ∀x ∈ I (sinceG(x,0) = F (x,0) = f(x) ∀x ∈ I .G(x,1) = f ′′(x) ∀x ∈ I (sinceG(x,1) = F ′(x,1) = f ′′(x) ∀x ∈ I .

If f is a path, we denote its homotopy class by [f].

Example 2.4

1. Let f ∶ I → R2, g ∶ I → R2, paths with f(0) = g(0) and f(1) = g(1). enF ∶ I × I → R2 with (x, t)↦ (1 − t)f(x) + tg(x) is a path homotopy between fand g.F is called a linear homotopy.

2. f ∶ I → R2, s↦ (cos(πs), sin(πs)), g ∶ I → R2, t↦ (cos(πt),3 sin(πt)).f(0) = (1,0) = g(0), and f(1) = (−1,0) = g(1).f, g are path homotopic. Indeed, consider the linear homotopy F described in 1.

3. f ∶ I → R2 ∖ {(0,0)} with s ↦ (cos(πs), sin(πs)), and g ∶ I → R2 ∖ {(0,0)}with t↦ (cos(πt), sin(πt).Are f, g path homotopic? Yes. Prove using the linear homotopy.

4. f ∶ I → R2 ∖ {(0,0)}, with s ↦ (cos(πs), sin(πs)), and g ∶ I → R2 ∖ {(0,0)}with t↦ (cos(πt),−3 sin(πt)).e linear homotopy is not a path homotopy between f and g.Inf fact, there exists no path homotopy between f and g.

We de ne the following operation:

De nition 2.5 Let f ∶ I → X path in X with f(0) = x0, f(1) = x1, x0, x1 ∈ X .g ∶ I →X path inX with g(0) = x1, g(1) = x2, x2 ∈X .We de ne the product f ⋅ g of f, g as the path h with

h(s) = { f(2s) s ∈ [0,1/2]g(2s − 1) s ∈ [1/2,1].

17

Algebraic topology

Remark 2.6 h is well de ned, since f(1) = g(0) = h(1/2).h is continuous by the pasting lemma (lemma 1.21).e product f ⋅ g induces a well de ned product on path homotopy equivalce classesde ned by [f] ⋅ [g] = [f ⋅ g].Indeed, if f, f ′ paths inX with [f] = [f ′], and g, g′ paths inX with [g] = [g′]. Considerpath homotopies F,G between f and f ′, g and g′ respectively. en

H(s, t) ∶= { F (2s, t) t ∈ [0,1/2]G(2s − 1, t) s ∈ [1/2,1]

is a path homotopy between f ⋅ g and f ′ ⋅ g′ (homework).erefore [f ⋅ g] = [f ′ ⋅ g′], i.e. [f] ⋅ [g] = [f ′] ⋅ [g′] and the induced product is well-de ned.

Reparametrisation

De ne a reparametrisation of a path f to be a composition f ○ φ, where φ ∶ I → Icontinuous map, such that φ(0) = 0, and φ(1) = 1.Reparametrising a path preserves its homotopy class (i.e. [f ○φ] = [f]). Indeed, considerthe path homotopy f ○ φt whereφt(s) ∶= (1 − t)φ(s) + ts.φ0(s) = φ(s)↝ f ○ φ0 = f ○ φ, φ1(s) = s↝ f ○ φ1 = f .Note that (1 − t)φ(s) + ts lies between φ(s) and s, hence (1 − t)φ(s) + ts lies in I , sof ○ φt is de ned.

2.2. e fundamental group

We restrict our attention to paths f ∶ I → X with f(0) = f(1). Call x0 ∶= f(0) = f(1).Such paths are called loops inX at the basepoint x0. e set {[f] ∣ f ∶ I → X, f(0) =f(1) = x0} is denoted by π1(X,x0).If [f], [g] ∈ π1(X,x0), then f, g are loops inX at the basepoint x0, so f ⋅ g is de ned.e induced multiplication [f] ⋅ [g] = [f ⋅ g] is well-de ned.Proposition 2.7 (π1(X,x0), ⋅) is a group.P

1. associativity: [f] ⋅ ([g] ⋅ [h]) = ([f] ⋅ [g]) ⋅ [h], [f], [g], [h] ∈ π1(X,x0).⇐⇒ [f] ⋅ [g ⋅ h] = [f ⋅ g] ⋅ [h]⇐⇒ [f ⋅ (g ⋅ h)] = [(f ⋅ g) ⋅ h].Claim: f ⋅(g ⋅h) is a reparametrisation of (f ⋅g) ⋅h by the piecewise linear functionφ: where

φ(t) =⎧⎪⎪⎪⎨⎪⎪⎪⎩

12t t ∈ [0,1/2]t − 1

4t ∈ [1/2,3/4]

2t − 1 t ∈ [3/4,1]Proof:

(f ⋅(g⋅h))(t) = { f(2t) t ∈ [0,1/2](g ⋅ h)(2t − 1) t ∈ [1/2,1] =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

f(2t) t ∈ [0,1/2]g(4t − 2) t ∈ [1/2,3/4]h(4t − 3) t ∈ [3/4,1]

Analogously

((f ⋅ g) ⋅ h)(s) = { (f ⋅ g)(2s) s ∈ [0,1/2]h(2s − 1) s ∈ [1/2,1] =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

f(4s) s ∈ [0,1/4]g(4s − 1) s ∈ [1/4,1/2]h(2s − 1) s ∈ [1/2,1]

18

2.2 e fundamental group

So if one easily checks, that ((f ⋅ g) ⋅ h)(φ(t)) = (f ⋅ (g ⋅ h))(t).

2. Identity: Consider c ∶ I → X with c(s) = x0 ∀s ∈ I . en f ⋅ c is a reparametri-sation of f via φ:

φ(t) = { 2t t ∈ [0,1/2]1 t ∈ [1/2,1]

I.e. f ⋅ c = f ○ φ.[f] = [f ○ φ] = [f ⋅ c].[f] = [f] ⋅ [c].

3. inverse: Consider [f] ∈ π1(X,x0) and f(s) = f(1 − s).Claim: f ⋅ f ≃ c ≃ f ⋅ f .Proof: Consider the homotopy

H(s, t) =⎧⎪⎪⎪⎨⎪⎪⎪⎩

f(2s) s ∈ [0, (1 − t)/2]f(1 − t) [(1 − t)/2, (1 + t)/2]f(2 − 2s) [(1 + t)/2,1]

H(0, t) = f(0) = x0,H(1, t) = f(0) = x0.

H(s,0) = { f(2s) s ∈ [0,1/2]f(2 − 2s) s ∈ [1/2,1] = f ⋅ f

H(s,1) = x0.

H is a homotopy between f ⋅ f and c i.e. [f ⋅ f] = [c],i.e. [f] ⋅ [f] = [c].Analogously [f] ⋅ [f] = [c].

Ô⇒ [f] is the inverse of [f].

De nition 2.9 π1(X,x0) is called the fundamental group ofX at x0.

Example 2.10 ConsiderX ⊆ Rn convex. x0 ∈X .π1(X,x0) = {[f] ∣ f ∶ I →X path, f(0) = f(1) = x0}.If f0, f1 are two paths inX at the basepoint x0, then ft(s) = (1 − t)f0(s) + tf1(s) is alinear homotopy between f0 and f1.Let x0, x1 ∈X , π1(X,x0), π1(X,x1).Proposition 2.11 Ifx0, x1 are in the samepath component ofX , thenπ1(X,x0),π1(X,x1)are isomorphic.P x0, x1 ∈ X . Consider h ∶ I → X a path with h(0) = x0, h(1) = x1 andh ∶ I →X , h(s) = h(1 − s).De ne h ∶M1(X,x1)→ π1(X,x0) where [f]↦ [h ⋅ f ⋅ f].

• h is well-de ned: If [f] = [g] ∈ π1(X,x1) and ft is a homotopy between f and gthen h ⋅ ft ⋅h is a homotopy between h ⋅ f ⋅h and h ⋅ g ⋅ g i.e. [h ⋅ f ⋅h] = [h ⋅ g ⋅h].

• h is a homomorphism: h([f] ⋅ [g])h([f ⋅ g]) = [h ⋅ f ⋅ g ⋅h] = [h ⋅ f ⋅h ⋅h ⋅ g ⋅h] =[hfh][hgh] = h([f])h([g]).

19

Algebraic topology

• h is bijective: h = (h)−1.(hh)[f] = h(h([f])) = h([hfh]) = ⋅ ⋅ ⋅ = [f].

De nition 2.13 X is simply connected if it is path connected and π1(X) is trivial.

Proposition 2.14 X is simply connected iff there is a unique homotopy class of pathsconnecting any two points inX .P Path connectedness is the existence of paths connecting every pair of points inX . So we only need to check uniqueness.

“⇒” Suppose π1(X) = 0. Let x0, x1 ∈ X . If f, g are two paths from x0 to x1, thenf ≃ fgg ≃ g. Ô⇒ [f] = [g].

“⇐” If there is a unique homotopy class of paths connecting any two points, then thereis a unique homotopy class of paths connecting a point to itself. Ô⇒ π1(X) = 0.

Induced homomorphisms

X,Y topological spaces, x0 ∈ X , y0 ∈ Y and h ∶ X → Y a continuous map withh(x0) = y0. We will denote this by h ∶ (X,x0)→ (Y, y0).If f ∶ I →X is aloop inX , based at x0, then h ○ f is a loop in Y based at y0.

De nition 2.16 Let h ∶ (X,x0) → (Y, y0) continuous. De ne h∗ ∶ π1(X,x0) →π1(Y, y0) where [f] ↦ [h ○ f]. h∗ is called the homomorphism induced by h relativeto x0.

Remark 2.17 h∗ is well-de ned.If f, g loops inX based at x0 with [f] = [g]. ConsiderH a homotopy from f to g. enh ○H is a homotopy from h ⋅ f to h ⋅ g.I.e. [h ⋅ f] = [h ⋅ g]⇐⇒ h∗([f]) = h∗([g]).Remark 2.18 h∗ is a group homomorphism.Just calculate and compare the following two expressions:

h∗([f] ⋅ [g]) = h∗([f ⋅ g]) = [h ○ (f ⋅ g)]h∗([f]) ⋅ h∗([g]) = [h ○ f] ⋅ h[⋅g] = [(h ○ f) ⋅ (h ○ g)].

Remark 2.19 h∗ depends on the basepoint x0. So strictly speaking we should have writ-ten (hx0)∗.

Properties of induces homomorphisms

1. Ifh ∶ (X,x0)→ (Y, y0), g ∶ (Y, y0)→ (Z, z0) continuous, then (g○h)∗ = g∗○h∗.

2. If 1X ∶ (X,x0)→ (X,x0), then (1x)∗ = 1π1(X,x0).

Proposition 2.20 IfH ∶ (X,x0)→ (Y, y0) a homoeomorphism,t henh∗ ∶ π1(X,x0)→π1(Y, y0) is a grou isomorphism.

20

2.3 Covering spaces

P Consider h−1(Y, y0)→ (X,x0) and (h−1)∗ ∶ π1(Y, y0)→ π1(X,x0)(h−1)∗ ○ h∗ = (h−1 ○ h)∗ = 1∗ = 1.h∗ ○ (h−1)∗ = (h ○ h1)∗ = 1∗ = 1.π1 gives a covariant functor from the category with objects topological spases with base-point and morphism basepoint preserving continuous maps to the category with objectsgroups and morphisms group homomorphisms.

A category D consists of

1. a collection of objects Ob(D),

2. sets of morphismsMor(X,Y ) for eachX,Y ∈ Ob(D)with a distinguished iden-tity morphism 1X in Mor(X,X),

3. a composition of morphisms ○ ∶Mor(X,Y )×Mor(Y,Z(→Mor(X,Z) for eachtripel X,Y,Z ∈ Ob(D) with the properties f ○ 1 = 1 ○ f = f and (f ○ g) ○ h =f ○ (g ○ h).

Example 2.22

1. Ob(D) = {G ∣ G group}.Mor(G,H) = {f ∶ G→H ∣ f group homomorphism}.

2. Ob(D) = {(X,x0) ∣X topological space, x0 basepoint}.Mor((X,x0), (Y, y0)) = {f ∶ (X,x0)→ (Y, y0) ∣ f continuous}.

A coveriant functor F from a categoriy C to a category D assigns to each X ∈ Ob(C)an F (X) ∈ Ob(D) to each f ∈ Mor(X,Y ) a F (f) such that F (1X) = 1F (X) andF (f ○ g) = F (f) ○ F (g).Example 2.23 π1 is a covariant functor from category C2 in example 2 to the category C1in example 1.(X,x0) ∈ Ob(C2)

π1↝ π1(X,x0) ∈ Ob(C1).f ∈Mor((X,x0), (Y, y0))

π1↝ f∗ ∈Mor(π1(X,x0), (Y, y0))(1(X,x0))∗ = 1π1(X,x0).(f ○ g)∗ = f∗ ○ g∗.

2.3. Covering spaces

• useful for computing π1,

• algebraic features of π1 can be translated into geometric features of the spaces.

De nition 2.24 Consider p ∶ X → X a continuous and surjective map. If there existsan open cover {Ua}a∈A ofX such that ∀a ∈ A p−1(Ua) = ∪b∈BaV

ba , where V b

a ∩V b′

a =∅ for any b, b′ ∈ Ba, b ≠ b′. p ∶ V b

a → Ua is a homeomorphism ∀b ∈ Ba and V ba open

in X ∀b ∈ Ba, then p is called a covering map. X is called a covering space ofX .

Example 2.25

21

Algebraic topology

1. S1 = {(x, y) ∈ R2 ∣ x2 + y2 = 1}. p ∶ R→ S1 where s↦ (cos 2πs, sin 2πs).p is a covering map, p is continuous (calculus), and p is surjective.U1 ∶= {(x, y) ∈ S1 ∣ x > −

√2/2}

U2 ∶= {(x, y) ∈ S1 ∣ x <√2/2}.

..− 1

2

.12

{U1, U2} is an open cover of S1p−1(U1) = ∪n∈ZVn, where Vn ∶= (n − 3/8, n + 3/8).Vn is open in R ∀n ∈ Z.Vn ∩ Vn = ∅, for any n,m ∈ Z, n ≠m.p ∶ Vn → U1 is a homeomorphism ∀n ∈ Z.We can make the analogous construction for U2. So p is a covering map.Remark 2.26 You can prove that p is a covering by using any open cover of S1 bytwo open subsets of S1 (≠ S1).

2. S1 = {z ∈ C ∣ ∣z∣ = 1}, n ∈ N, n ≥ 1.pn ∶ S1 → S1, z ↦ zn. pn is a covering map (problem sheet 5).f ∶X → Y is called an embedding if f ∶X → f(X) is a homeomorphism.

Consider the solid torus S1 ×D2 where D2 = {(x, y) ∈ R2 ∣ x2 + y2 ≤ 1}, and itsboundary ∂(S1 ×D2) = S1 × S1.Consider f ∶ S1 → ∂(S1 ×D2) an embedding such that f(S1) wraps around therst S1 three times.

Lastly consider the projection π ∶ S1×D2 → S1×{(0,0)} and restrict it to f(S1).

3. f ∶ R+ → S1, where s↦ (cos 2πs, sin 2πs) is not a covering map.Consider an open cover{Ua}a∈A ofS1. en∃U ∈ {Ua}a∈A withx ∈ U . p−1(U) =V0 ∪ (∪b∈BaV

ba ).

p ∶ V0 → U is not a homeomorphism.

eorem 2.27 If p ∶ X → X is a covering map, X0 ⊆ X , and X0 = p−1(X0), thenp0 ∶ X0 →X0 obtained by restricting p to X0 is a covering map.

P p0 is continuous (restricting the domain or the range of a continuous functiongives a continuous function). p0 is surjective.Consider an open cover {Ua}a∈A ofX with the properties in the de nition of coveringmap.

22

2.3 Covering spaces

en ∀a ∈ A p−1(Ua) = ∪b∈BaVba .

If x0 ∈X0, then there exists a ∈ A with x0 ∈ Ua. Now Ua ∩X0 is an open set inX0.p−10 (Ua∩X0) = p−10 (Ua)∩p−10 (X0) = p−1(Ua)∩X0 = (∪b∈BaV

ba )∩X0 = ∪b∈Ba(V b

a ∩X0).We have

• V ba ∩ X0 open in X0,

• (V ba ∩ X0) ∩ (V b′

a ∩ X0) = ∅ for any b, b′ ∈ Ba, b ≠ b′.

• p0 ∶ V ba ∩ X0 → Ua ∩X0 is a homeomorphism.

{Ua ∩X0}a∈A is an open cover ofX0 with the desired propertiesÔ⇒ p0 ∶ X0 → X0 isa covering map.

eorem 2.29 If p ∶ X → X , p′ ∶ X ′ → X ′ are covering maps, then p × p′ ∶ X × X ′ →X ×X ′ is a covering map.

Example 2.30

1. Consider p ∶ R→ S1, where s↦ (cos 2πs, sin 2πs).e map p × p ∶ R ×R→ S1 × S1 is a covering map.

2. X0 = (S1 × p(0)) ∪ (p(0) × S1).X0 = (p × p)−1(X0) = (R ×Z) ∪ (Z ×R).(p × p)0 ∶ X0 →X0 is a covering map (by theorem).Two circles S1 with a one common point is called wedge of two circles (notationS1 ∨ S1).

3. Other covering spaces of S1 ∨ S1.

..

a

.

b

. x

Consider X as in the following picture

..x1 . x2.

a

.

b

.

a

en p ∶ X → S1 ∨ S1, where x1 ↦ x and x2 ↦ x. pmaps each edge of X to theedge of X with the same label by a map that is a homeomorphism and preservesthe orientation.

23

Algebraic topology

4. Consider p×1R+ ∶ R×R+ → S1 ×R+ (where p ∶ R→ S1: s↦ (cos 2πs, sin 2πs))and f ∶ S1 ×R+ → R2 ∖ {(0,0)}, where (x, t)↦ t ⋅ x.f is a homeomorphism.f ○ (p × 1R+) ∶ R ×R+ → R2 ∖ {(0,0)} is a covering map.

P p × p′ is continuous (problem sheet 3). p × p′ is surjective (if (x,x′) ∈ X ×X ′,then x ∈ X and x′ ∈ X ′ Ô⇒ [p, p′ is surjective] ∃x ∈ X , x′ ∈ X ′, such that p(x) = xand p′(x′) = x′. Ô⇒ ∃(x, x′) ∈ X × X ′ such that (p × p′)(x, x′) = (x,x′)).Consider {Ua}a∈A, {U ′c}c∈C open covers of X and X ′ as in the de nition of coveringmaps.If (x,x′) ∈ X × X ′ then x ∈ X and x′ ∈ X ′ i.e. ∃U ∈ {Ua}a∈A with x ∈ U and∃U ′ ∈ {Uc}c∈C with x′ ∈ U ′.p−1(U) = ∪b∈BVb, Vb open Vb ∩ Vb′ = ∅ for b ≠ b′ and p ∶ V ∶b→ U homeomorphism∀b ∈ B. (p′)−1(U ′) = ∪d∈DV ′d , V ′d open, V ′d ∩ V ′d′ = ∅ for d ≠ d′ and p′ ∶ V ′d → U ′

homeomorphism ∀d ∈D.(x,x′) ∈ U ×U ′, U ×U ′ is open inX ×X ′,(p × p′)−1(U ×U ′) = ∪ b∈B

d∈D(Vb × V ′d), where Vb × V ′d open inX ×X ′, Vb × V ′d disjoint,

p × p′ ∶ Vb × V ′d → U ×U ′ is a homeomorphism.I.e. {Ua×U ′c} a∈A

c∈Cis the open cover ofX ×X ′. at shows that p×p′ is a covering map.

2.4. Liing properties

We will discuss two properties of covering spaces.

De nition 2.32 Let p ∶ X →X be a map. If f ∶ Y →X is continuous a liing of f is amap f ∶ Y → X such that p ○ f = f .

Example 2.33 p ∶ R → S1, s ↦ (cos 2πs, sin 2πs), f ∶ I → S1 the path f(s) =(cosπs, sinπs).en f ∶ I → R, f(s) = s

2is a liing of f .

Proposition 2.34 (path liing property)Let p ∶ X →X be a coveringmapwith p(x0) = x0. If f ∶ I →X is a pathwith f(0) = x0,then ∃! liing of f to a path f ∶ I → X with f(0) = x0.P

1. Consider the open cover {Ua}a∈A ofX (as in the de ntion of covering map). Lett ∈ (0,1), and ft) ∈ X Ô⇒ f(t) ∈ Ua for some a ∈ A. f ∶ I → X is continuousÔ⇒ ∃(at, bt) ⊆ (0,1) with t ∈ (at, bt) and f([at, bt]) ⊆ Ua.

Analogously for t = 0 ∃[0, b0) ⊆ [0,1] with f([0, b0]) ⊆ Ua, and for t = 1∃(a0,1] ⊆ [0,1] with f([a0,1]) ⊆ Ua.I is compact, and {(at, bt)}t∈(0,1) together with [0, b0) and (b1,1] is an opencover of I . We can choose a nite subcover, say [0, b0), (a0,1], (a1, b1), . . . , (am, bm)}.Consider {ai, bi}i∈{0,...,m} and orther its elements. (Possibly) rename the ele-ments as follows: 0 < t1 ≤ ⋅ ⋅ ⋅ ≤ t2m+2 < 1. this gives a subdivision 0 = s0 < s1 <⋅ ⋅ ⋅ < sn = 1 of I with the property, that ∀i ∈ {0, . . . , n− 1} f([si, si+1]) ⊆ Ua forsome a ∈ A.

24

2.4 Liing properties

2. De ne f(0) = X0. Suppose that f is de ned in [0, si]. De ne f in [si, si+1] asfollows: f([si, si+1]) ⊆ U for some U in {Ua}a∈A.Let p−1(U) = ∪b∈BVb. Now (p ○ f)(si) = f(si) ∈ U .Ô⇒ f(si) ∈ p−1(U) = ∪b∈BVbÔ⇒ f(Si) ∈ V0 for some V0 ∈ {Vb}b∈B .De ne f(s) = (p∣V0

)−1(f(s)), s ∈ [si, si+1].

f is continuous in [si, si+1], since p∣V0is a homeomorphism. uswe have de ned

f ∶ [0,1]→ X continuous with f(0) = x0.

3. Suppose f is another liing of f with f(0) = x0. en f(0) = f(0) = x0.

Suppose that f(s) = f(s) in [0, si]. p ○ f([si, si+1]) = f([si, si+1]) ⊆ Ua forsome a ∈ A.Ô⇒ f([si, si+1]) ⊆ p−1(Ua) = ∪b∈BVb.f([si, si+1]) is connected and f(si) = f(si) ∈ V0.Ô⇒ f([si, si+1]) ⊆ V0.

For s ∈ S, p ○ f(s) = f(s)Ô⇒ f(s) ∈ p−1(f(s)).

Also f ∈ V0. So f(s) = (p∣V0)−1(f(s)) = f(s).

I.e. f(s) = f(s).

Proposition 2.36 Let p ∶ X →X be a covering map with p(x0) = x0. Let F ∶ I ×I →Xbe continuous with F (0,0) = x0 en ∃! liing ofF to a continuous map F ∶ I ×I → Xwith F (0,0) = x0. If F is a path homotopy then F is a path homotopy.P Consider {Ua}a∈A an open cover ofX with the properties as in the de nition ofcovering map.

1. De ne F ((0,0)) = x0.Extend F to I × {0} and {0} × I using the proposition from last time. Extend toI × I , as follows: Choose subdivision s0 < s1 < ⋅ ⋅ ⋅ < sm, t0 < t1 < ⋅ ⋅ ⋅ < tn ofI with the property that for each such rectangle Ii × Jj = [si−1, si] × [tj−1, tj],F (Ii×Jj) ⊆ Ua for some a ∈ A. De ne F rst in I1×Ij , then I2×J1,. . . , Im×J1,then I1 × J2, I2 × J2 and so on.Consider Ii0 × Jj0 and suppose that F is de ned on the unionA of the rectanglesIi × Jj with j < j0 or j = j0 and i < i0.Consider C = A ∩ (Ii0 × Jj0). Choose U ∈ {Ua}a∈A with F (Ii0 × Jj0) ⊆ U .p−1(U)) ∪b∈B Vb, F is de ned in C , C is connected. So, F (C) is connected, i.e.∃V0 such that F (C) ⊆ V0.If p0 = p∣x0

∶ V0 → U , then p0 ○ F (x) = p ○ F (x) = F (x) Ô⇒ F (x) =(p−10 )(F (x)).De ne F (x) = p−10 (F (x)) ∀x ∈ Ii0 × Jj0 .F is continuous, according to the pasting lemma.

2. Check that at every step, there is a unique way to de ne F .

25

Algebraic topology

3. If F is a path homotopy, then F ({0} × I) = x0. Also, F ({0} × I) ⊆ p−1({x0}).p−1({x0}) has the discrete topology as a subspace of X . Anlose F ({0} × I) isconnected. us F ({0} × I) = {x0}.Similarly F ({1} × I) is a set with one point.

us, F is a path homotopy.

Proposition 2.38 (Homotopy liing property)p ∶ X → X covering map with p(x0) = x0. Consider f, g the paths inX from x0 to x1and f , g their liings to paths in X with f(0) = g(0) = x0. If f ≃ g, then f ≃ g.P Consider F the path homotopy between f and g. en F ((0,0)) = x0. LetF ∶ I × I → X the liing of f to X with F ((0,0)) = x0. en F ({0} × I) = {x0} andF ({1} × I) = {x1}.F∣I×{0} is a path i X starting at x0 that is a liing ofF∣I×{0}. I.e. FI×{0} = f (uniquenessof path liings).F∣I×{1} = g.So f(1) = g(1) = x1 and F ∶ I × I → X is a path homotopy from f to g.

De nition 2.40 Let p ∶ X →X be a covering map with p(x0) = x0. Given a loop f inX based x0, let f be the liing of f in X with f(0) = x0. De ne

ϕ ∶ π1(X,x0)→ p−1({x0}),[f]↦ f(1).

ϕ is called the liing correspondence derived from the map p

Claim: ϕ is well-de ned.P If [f] = [f ′], then f, f ′ are path homotopic. We can li this homotopy to a pathhomotopy between f and f ′Ô⇒ f(1) = f ′(1).

eorem 2.42

1. If p ∶ X → X be covering map with p(x0) = x0. If X is path connected, then ϕ issurjective.

2. If X is simply connected, then ϕ is bijective.

P

1. Let x1 ∈ p−1({x0}). en since X is path connected, there is a path f ∶ I → Xwith f(0) = x0 and f(1) = x1. en f = p ○ f is a loop in X based at x0, sincef(0) = p(x0) = x0, f(1) = p(x1) = x0 and ϕ([f]) = f(1) = x1.

2. We only need to check that ϕ is injective. Indeed consider ä[f], [g] ∈ π1(X,x0)with ϕ([f]) = ϕ([g]). en f(1) = g(1), where f , g are the liings of f, g withf(0) = g(0) = x0. Since X is simply connected, we have that [f] = [g], i.e.there is a path homotopy F between f and g. en F = p ○ F is a path homotopybetween f and g, i..e [f] = [g].

26

2.5 e fundamental group of the circle and applications

2.5. e fundamental group of the circle and applications

Remark 2.44 S1 is path connected, so we will write π1(S1).

eorem 2.45(π1(S1), ⋅) ≃ (Z,+).

P Let p ∶ R → S1, p(t) = (cos 2πt, sin 2πt). Denote p(0) = (1,0) = x0. enp−1({x0}) = Z. Consider ϕ ∶ π1(S1)→ Z, where [f]↦ f(1).By exercise 1 of sheet 3 R is simply connected. By the second point of the theorem ϕ isbijective.It remains to show that ϕ is a group homomorphism,i.e. ϕ([f] ⋅ [g]) = ϕ([f]) + ϕ([g]).⇐⇒ ϕ([f ⋅ g]) = ϕ([f]) + ϕ([g])⇐⇒ f ⋅ g(1) = f(1) + g(1),where f is the li of f with f(0) = 0 ∈ R and g the li of g with g(0) = 0 ∈ R and f ⋅ gthe li of f ⋅ g where f ⋅ g(0) = 0 ∈ R.Let g′ ∶ I → R be the path g′(s) = g(s) + f(1). en (p ○ g′)(s) = p(g(s) + f(1)) =(p ○ g)(s) = g(s).Ô⇒ g′ is a liing of g with g′(0) = f(1).Furthermore, p((f ⋅ g′)(s)) = (f ⋅ g)(s)Ô⇒ f ⋅ g′ is a liing of f ⋅ g with f ⋅ g′(0) = 0.(f ⋅ g′)(+) = f(1) + g(1)Ô⇒ f ⋅ g(1) = f(1) + g(1).

Retractions

De nition 2.47 A retraction of aX onto A ⊂ X is a continuous map r ∶ X → A withr(a) = a ∀a ∈ A.If such a map existsA is called a retract ofX .

Claim: IfA is a retract ofX and i ∶ A↪X is the inclusion map, then i∗ is injective.P r ○ i ∶ A→ A. r ○ i = 1AÔ⇒ (r ○ i)∗ = (1A)∗.Ô⇒ r∗ ○ i∗ = 1π1(A,x0)Ô⇒ i∗ is injective.Proposition 2.49 ere is no retraction of D2 onto S1.P If S1 was a retract of D2 and i ∶ S1 → D2 the inclusion. en i∗ ∶ π1(S1) →π1(D2) would be injective. But π1(S1) ≃ Z and π1(D)2 is trivial as a convex subset ofR2.Lemma 2.51 Let h ∶ S1 →X be a continuous map. e following are equivalent:

1. h is nullhomotopic.

2. ere exists a continuous map h ∶ D2 →X with h∣S1 = h.

3. h∗ is trivial.

P

27

Algebraic topology

1. “1.⇒2.”: Consider the homotopy H ∶ S1 × I → X between h and the constantmap c. Let q ∶ S1 × I → D2 with (x, t)↦ t ⋅ x.q is continuous, q is open and surjective.

2.

3.

Vorlesung von Di 17. März fehlt nochRemark 2.53 Ha the corollary: If S2 = A1 ∪ A2 ∪ A3, A1 closed ∀i ∈ {1,2,3}, then∃x ∈ S2, i ∈ {1,2,3} such that {x,−x} ⊆ Ai.

Inscribe a sphere in the tetrahedron. Project each face of the tetrahedral radially onto thesphere. en one obtains four closed setsAi, i ∈ {1, . . . ,4}with S2 = A1∪A2∪A3∪A4,but none of theAi contains a pair of antipodal points.

2.5.1. e Fundamental eorem of Algebra

(Any non-constant polynomial in C[x] has a root in C).

• proof in algebra,

• proof in complex analysis (corollary of Liouville’s theorem),

• proof in topology.

eorem 2.54 (Fundamental eorem of Algebra)If anxn + ⋅ ⋅ ⋅ + a1 + a0 ∈ C[x], n > 0, then ∃x0 ∈ C with anxn0 + ⋅ ⋅ ⋅ + a1x0 + a0 = 0.

P Let anxn + ⋅ ⋅ ⋅ + a1x + a0 ∈ C[x] and n > 0.

1. We can assume an = 1 (an(xn + an−1an

xn−1 + ⋅ ⋅ ⋅ + a0

an)).

We can assume ∑n−1i=0 ∣ai∣ < 1. Indeed choose c ∈ R>0 and set x = cy is gives

cnyn + ⋅ ⋅ ⋅ + a1cy + a0 = cn(yn + an−1cyn−1 + ⋅ ⋅ ⋅ + a0

cn). Choose c large enough

such that ∣an−1c∣ + ⋅ ⋅ ⋅ + ∣ a0

cn∣ < 1. Now if y0 is a root of yn + an−1

cyn−1 + ⋅ ⋅ ⋅ + a0

cn,

then cy0 is a root of xn + ⋅ ⋅ ⋅ + a0.

2. Consider f ∶ S1 → S1, z ↦ zn and p ∶ I → S1, s ↦ (cos 2πs, sin 2πs) = e2πis.en f∗ ∶ π1(S1, (1,0)) → π1(S1, (1,0)). f∗([p]) = [f ○ p], where f ○ p(s) =(cos 2πns, sin 2πns).is shows that f∗ is injective. Also if ι ∶ S1 → R2 ∖ {(0,0)} (inclusion), then ι∗is injective, since S1 is a retract of R2 ∖ {(0,0)}.I.e. ι∗ ○ f∗ = (ι ○ f)∗ is injective, which implies (Lemma ??) that ι ○ f is notnullhomotopic (a).

3. Suppose, that our polynomial has no root inD2. Consider h ∶ D2 → R2∖{(0,0)},z ↦ zn + an−1zn−1 + ⋅ ⋅ ⋅ + a1z + a0. By lemma ?? ↝ h∣S1 is nullhomotopic (b).Furthermore: F ∶ S1 × I → R2 ∖ {(0,0)} where (z, t)↦ zn + t(an−1zn−1 + ⋅ ⋅ ⋅ +a1z + a0). is is a homotopy between h∣S1 and ι ○ f (c).

28


∣F (z, t)∣ = ∣zn + t(an−1zn−1 + ⋅ ⋅ ⋅ + a0)∣ ≥ ∣zn∣ − ∣t∣∣an−1zn−1 + ⋅ ⋅ ⋅ + a0∣ ≥1 − ∣t∣∑n−1

i=0 ∣ai∣ > 0.

(a),(b),(c) contradiction, i.e. our polynomial has a root in D2.

Exercise: Any root of such a polynomial zn + ⋅ ⋅ ⋅ + a0 with∑ ∣ai∣ < 1 is in D2.

2.5.2. Deformation retracts and homotopy type

Covering spaces↝ fundamental group computation (e.g. π1(S1)).

De nition 2.56 Let A ⊆X ,A is called a deformation retract ofX , if there is a contin-uous mapH ∶ X × I < raX withH(x,0) = x ∀x,H(x,1) ∈ A ∀x ∈ X ,H(a, t) = a∀a ∈ A, ∀i ∈ I . e homotopyH is called a deformation retraction ofX ontoA.

Remark 2.57

1. r ∶X → A, r(x) =H(x,1) is a retraction ofX ontoA.

2. H is a homotopy between 1X and ι ○ r, where ι ∶ A↪X inclusion map.

Example 2.58

1. {(x, y, z) ∈ R3 ∣ (x, y) ≠ (0,0), z = 0} =∶ A is a deformation retract of{(x, y, z) ∈ R3 ∣ (x, y) ≠ (0,0)} =∶X .H ∶X × I →X ,H(x, y, z, t) ∶= (x, y, (1 − t)z).

• H is continuous,• H(x, y, z,0) = (x, y, z) ∀(x, y, z) ∈X ,• H(x, y, z,1) = (x, y,0) ∈ A ∀(x, y, z) ∈X ,• if (x, y, z) ∈ A, then z = 0 and thusH(x, y, z, t) = (x, y,0) ∈ A.

2. Let p, q ∈ R2, p ≠ q. en S1 ∨ S1 is a deformation retract of R2 ∖ {p, q}.

..p. q

3. p, q ∈ R2, p ≠ q, R2 ∖ {p, q} deform retracts onto the theta space:

.

(Homeomorph to S1 ∪ (0 × [−1,1])).

4. Möbius band

.

29

Algebraic topology

eorem 2.59 Let A be a deformation retract ofX , x0 ∈ A, and ι ∶ (A,x0) → (X,x0)the inclusion map. en ι∗ ∶ (A,x0)→ π1(X,x0) is an isomorphism.

P Let r ∶ (X,x0) → (A,x0), r(x) = H(x,1)en r ○ ι ∶ (A,x0) → (A,x0) isthe id map 1A and (r ○ ι)∗ = (1A)∗. Ô⇒ r∗ ○ ι∗ = 1π1(A,x0).Furthermore ι∗ ○ r∗ = (ι ○ r)∗. ι ○ r ∶ (X,x0) → (X,x0) and there is a homotopy Hbetween 1X and i ○ r. H(x0, t) = x0 ∀t ∈ I .If f ∶ I →X loop , f(0) = f(1) = x0, thenH ○(f ×1I)) ∶ I×I →X is a path homotopybetween 1X ○ f and (ι ○ r) ○ f . (path homotopy: H ○ (f × 1I)(0, t) = H(x0, t) = x0∀t ∈ I ,H ○ (f × 1I)(1, t) =H(x0, t) = x0 ∀t ∈ I).So [1X ○ f] = [(ι ○ r) ○ f]Ô⇒ (1∗(f) = (ι ○ r)∗(f).Ô⇒ 1π1(X,x0) = (ι ○ r)∗Ô⇒ ι∗r∗ = 1π1(X,x0).Corollary 2.61 If ι ∶ Sn → Rn+1∖{0} the inclusionmap,x0 ∈ Sn, then ι∗ ∶ π1(Sn, x0)→π1(Rn+1 ∖ {0}) is isomorphism.P De neH ∶ (Rn+1 ∖ {0}) × I → Rn+1 ∖ {0},H(x, t) ∶= (1 − t)x + t x

∥x∥ .

• H is continuous,

• H(x,0) = x ∀x ∈ Rn+1 ∖ {0} =∶X ,

• H(x,1) = x∥x∥ ∈ S

n =∶ A.

• If x ∈ Sn, thenH(x, t) = (1 − t)x + tx = x ∀t ∈ I .

H is a deformation retraction of Rn+1 ∖ {0} onto Sn. Now use theorem.Example 2.63 We can look at example 2.58 again:

1. π1(X)m≃ π1(A)

Cor≃ π1(S1) ≃ Z.

2.,3. “ gure eight“ space is deformaton retraction ofR2∖{p, q} mÔ⇒π1(A1) ≃ π1(R2∖{p, q}),“theta” space is deformation retraction of R2 ∖ {p, q} mÔ⇒ π1(A2) ≃ π1(R2 ∖{p, q}).Ô⇒ π1(A1) ≃ π1(A2).Comment: “ gure eight” and “theta” space have isomorphic fundamental group,but none is a deformation retract of the other (exercise).

De nition 2.64 Let f ∶X → Y be a continuous map. If there exists a continuous mapg ∶ Y → X such that g ○ f ∶ X → X is homotopic to 1X ∶ X → X and f ○ g ∶ Y → Yis homotopic to 1Y ∶ Y → Y , then f is called a homotopy equivalence.g is called homotopy inverse of f . X ,Y are called homotopy equivalent.IfX,Y are homotopy equivalent, we say thatX,Y have the same homotopy type.

Check: Homotopy equivalence is an equivalence relation.Remark 2.65 Homotopy equivalence is more general than deformation retraction.

• Suppose thatH ∶X × I →X is a deformation retraction ofX ontoA. Let ι ∶ A→X the inclusion map and r ∶ X → A, r(x) = H(x,1). en r ○ ι ∶ A → A is theid map 1A, andH is a homotopy between ι ○ r and 1X .

30


Lemma 2.66 Let h, k ∶ X → Y homotopic continuous maps with h(x0) = y0, k(x0) =y1. If H ∶ X × Y → Y is the homotopy between h and k and if a(t) = H(x0, t), thenk∗ = a ○ h∗.

π1(X,x0)h∗ //

k∗ &&MMMMM

MMMMM

π1(Y, y0)

a

��π1(Y1, y1)

P …der Beweis ist noch sehr fehlerha, noch fehlerhaer als gewohnt :P …Let f ∶ I →X a loop based at x0.We want to show that k∗([f]) = a ○ h∗([f])⇐⇒ [k ○ f] = a([h ○ f])⇐⇒ [k ○ f] = [a][h ○ f][a]⇐⇒ [a][k ○ f] = [h ○ f][a]⇐⇒ [a(k ○ f)] = [(h ○ f)a].

h(x) =H(x,0)∀x ∈X , soh(f(s)) =H(f(s),0) = (H○f0)(s)where f0 ∶ I →X×I ,f0(s) = (f(s),0). I.e. h ○ f =H ○ f0.k(x) =H(x,1)∀x ∈X , so k(f(s)) =H(f(s),1) = (H○f1)(s)where f1 ∶ I →X×I ,f1(s) = (f(s),1). I.e. k ○ f = ○f1.Consider c ∶ I →X × I , c(t) = (x0, t). enH(c(t)) = a(t).F ∶ I × I →X × I , F (s, t) ∶= (f(s), t).G0γ1 is homotopic to γ0B1, callG the path homotopy between them.

enF○G ∶ I×I →X×I , continuous, (F○G)(s,0) = F (G(s,0)) = { F (β0(s),0), s ∈ [0, 12 ]F (1, γ1(s)) s ∈ [ 1

2,1] =

{ (f ○ β0(s),0) s ∈ [0, 12]

(f(1), γ1(s)) s ∈ [ 12,1] = {

f0(s) s ∈ [0, 12]

(x0, s) s ∈ [ 12,1] .

(F ○G)(s,1) = ⋅ ⋅ ⋅ = (c ⋅ f1)(s), (F ○G)(0, t) = F ((0,0)) = (x0,0), (F ○G)(1, t) =F ((1,1)) = (x0,1)Ô⇒ (F ○G is a path homotopy from c ⋅ f1 to c ⋅ f1.Lastly,H ○ (F ○G) is a path homotopy fromH(f0 ⋅ c) =H(f0) ⋅H(c) = (h ○ f) ⋅ a toH(c ⋅ f1) =H(c) ⋅H(f1) = a ⋅ (k ○ f)Corollary 2.68 Let h, k ∶ (X,x0) → (Y, y0) homotopic continuous maps. If H ∶ X ×I → Y is the homotopy between h and k andH(x0, t) = y0 ∀t ∈ [0,1], then k∗ = h∗.

eorem 2.69 Let f ∶ X → Y homotopy equivalence with f(x0) = y0. en f∗ ∶π1(X,x0)→ π1(Y, y0) is an isomorphism.

P Let g ∶ Y →X be a homotopy inverse of f , and let g(y0) = x1 and f(x1) = y1.

(X,x0)f // (Y, y0)

g // (X,x1)f // (Y, y1)

π1(X,x0)(fx0

)∗ // π1(Y, y0)g∗ // π1(X,x1)

(fx1)∗ // π1(Y, y1)

g ○fx0 is homotopy to 1X Ô⇒[Lemma] there is a path a ∶ I →X , such that (g ○fx0)∗ =a ○ (1X)∗ = a.

31

Algebraic topology

a is a group isomorphism, so g∗ ○ f∗ is a group isomorphism.Ô⇒ g∗ is surjective (1).Similarly (fx1 ○ g)∗ is a group isomorphism.⇐⇒ (fx1)∗ ○ g∗ is a group isomorphism.Ô⇒ g∗ is injective (2).

(1),(2)↝ g∗ is a group isomorphism,↝ (fx0)∗ = (g∗)−1 ○ a is a group isomorphism.

eorem 2.71 X,Y have the same homotopy type iffX,Y are homeomorphic to to defor-mation retracts of a space Z .

[Proof: hard, we omit it]

eorem 2.72 π1(S1 ∨ S1) is not abelian.

P X ∶= S1 ∨ S1.

.. x0.B . A

Consider the following covering space

.. A0

.

B0

.

B2

.

B−1

.

B−2

.

A1

.

A2

.

A−1

.

A−2

p ∶ X → X , p maps each Ai homeomorphically onto A ∀i ∈ Z ∖ {0}. and Bi homeo-morphically to B ∀i ∈ Z ∖ {0}. p maps all tangency points to X0. p wraps A0 aroundA,B0 aroundB.p is a coveringmap (check). Consider f ∶ I → X , f(s) = (s,0). g ∶ I → X , g(s) = (0, s)and f = p ○ f , g = p ○ g.Also consier f ⋅ g and g ⋅ f and li these two paths to paths f ⋅ g, g ⋅ f in X both startingat (0,0). f ⋅ g is a path in X that goes from (0,0) to (1,0) along x-axis and then oncearound B1. g ⋅ f is a path in X that goes from (0,0) to (0,1) along the y-axis and thenonce aroundA1.f ⋅ g(1) ≠ g ⋅ f(1). So f ⋅ g and g ⋅ f are not path-homotopic. [f ⋅ g] ≠ [g ⋅ f] i.e.[f] ⋅ [g] ≠ [g] ⋅ [f]Ô⇒ π1(S1 ∨ S1) is not abelian.

32


eorem 2.74 π1(X × Y,x0 × y0) ≃ π1(X,x0) × π1(Y, y0).

P Consider the projections p ∶X×Y →X , x×y ↦ x and q ∶X×Y → Y ,x×y ↦ y,and p∗ ∶ π1(X × Y,x0 × y0)→ π1(X1, x0 and q∗ ∶ π1(X × Y,x0 × y0)→ π1(Y, y0).De ne Φ ∶ π1(X × Y,x0 × y0)→ π1(X,x0) × π1(Y, y0) , [f]↦ (p∗([f]), q∗([f])).Claim: Φ is a group homomorphism.Proof: Φ([f][g]) = Φ([fg]) = (p∗([fg]), q∗([fg])) = ([p ○ (fg)], [g ○ (fg)]) =([(p ○ f][p ○ g], [g ○ f][q ○ g)]).Φ([f])Φ([g]) = ([p ○ f], [q ○ f])([p ○ g][q ○ g]) = ([p ○ f][p ○ g], [g ○ f][q ○ g]).

Claim: Φ is surjective.Proof: Consider g ∶ I → X Loop based at x0, and h ∶ I → Y loop based at y0. De-ne f ∶ I → X × Y , f(s) = (g(s), h(s)). en f is a loop based at (x0, y0) and

Φ([f]) = (p∗([f]), q∗([f])) = ([g], [h]).

Claim: Φ is injective.Proof: Suppose that f ∶ I → X × Y is a loop based at (x0, y0) with Φ([f]) is theidentity element in π1(X,x0) × π1(Y, y0). en p ○ f is path homotopic to cx0 via apath homotopy G and q ○ f is path homotopic to cy0 via a path homotopy H . De neF ∶ I × I → X × Y by F (s, t) ∶= (G(s, t),H(s, t)). F is a path homotopy from f toc(x0,y0). Ô⇒ [f] trivial.Corollary 2.76 Consider the torus T ≃ S1 × S1. Ô⇒ π1(T) ≃ Z ×Z.

De nition 2.77

1. Let {Xi}i∈I an indexed collection of sets. e disjoint union of these sets is⊔i∈IXi = ∪i∈IXi × {i}.

2. Suppose that Xi is a topological space ∀i ∈ I . e disjoint union topology on⊔i∈IXi is the nest topology, such that all of the following maps are continuous:φi ∶Xi → ⊔i∈IXi, x↦ x × {i}.Explicitly, U is open in ⊔i∈IXi iff φ−1(U) is open ∀i ∈ I .

3. Let X,Y be topological spaces, A ⊆ X , and f ∶ A → Y continuous. e ad-junction spaceX ∪f Y (adjunction ofX to Y along f (alongA)) is the quotientspace

X ⊔ YÒ∼,

where a ∼ f(a) ∀a ∈ A. f is called the attaching map.

Remark 2.78 IfXi ∩Xj = ∅ for any i, j ∈ I , i ≠ j, then ⊔i∈IXi ≃ ∪i∈IXi.Example 2.79 X1,X2 ⊂ {R,T,S2},X1 ⊔X2.

De nition 2.80 e wedge sum ofX and Y is

X ∨ Y ∶=X ⊔ YÒx0 ∼ y0,

for x0 ∈X , y0 ∈ Y .

33

Algebraic topology

IfXi are topological spaces, xi ∈Xi, i ∈ I , then

∨i∈IXi = ⊔i∈IXiÒ{xi ∼ xj , i, j ∈ I}.

De nition 2.81 LetX,Y n-manifolds. Consider U,V , open discsD2 inX,Y respec-tively, and f ∶ ∂U → ∂V a homeomorphism. e connected sum ofX and Y is

X#Y ∶= (X ∖U) ⊔ (Y ∖ V )Ò∼,

where x ∼ y if x ∈ ∂U , y ∈ ∂V and f(x) = y.

Remark 2.82 Although this construction involves the choice of the discsU,V the result-ing space is unique up to homeomorphism.

Claim: Y ∶= S1 ∨ S1 is a retract ofX ∶= T#T.P (hier bräuchte es ne Graphik…)Consider f ∶X → Y , where f maps the dotted circle to y, and f restricted to S1∨S1 ⊆Xis a homeomorphism h, and f 1 ∶ 1 everywhere else. Further consider r, where r retractsY onto S1 ∨ S1 by mapping each cross-sectional circle to the point where it intersetsS1 ∨ S1. Use h−1 to map S1 ∨ S1 in Y to S1 ∨ S1 inX .↝ retraction of T#T onto S1 ∨ S1.Corollary 2.84 π1(T#T) is not abelian.P If ι ∶ S1∨S1 → T#T the inclusionmap, then ι∗ is injective (follows from claim).

ι∗ ∶ π1(S1 ∨ S1)→ π1(T#T).

So π1(T#T) is not abelian.Corollary 2.86 T, T#T are not homotopy equivalent.P obvious.Lemma 2.88 If P = (0, . . . ,0,1) ∈ Sn ⊂ Rn+1, then Sn ∖ {P} is homeomorphic to Rn.P De ne f ∶ Sn ∖{P}→ Rn (stereographic projection) , where (x1, . . . , xn+1)↦

11−xn+1

(x1, . . . , xn).f is obviously continuous and furthermore g ∶ Rn → Sn ∖ {P} where (y1, . . . , yn) ↦( 21+∥y∥2 y1, . . . ,

21+∥y∥2 yn,1−

21+∥y∥2 ) is continuous and f○g = 1Rn and g○f = idSn∖{P}.

Proposition 2.90 π1(Sn) = 0 ∀n ∈ N, n ≥ 2.P Consider f ∶ I → Sn a loop based at x0 ∈ Sn. Let x ∈ Sn ∖ {x0} and B a smallopen ball in Sn ∖ {x0} with x ∈X .en f−1(B) ⊆ (0,1) and f1−(B) is open., So f−1(B) = ∪i∈I(ai, bi).f−1({x}) is a closed subset of the compact space I , i.e. f−1({x}) is compact.f−1({x}) ⊆ f−1(B) = ∪i∈I(ai, bi),Ô⇒ ∃(a1, b1), . . . , (an, bn) such that f−1({x}) ⊆(a1, b1) ∪ ⋅ ⋅ ⋅ ∪ (an, bn).Consider fi = f∣[ai,bi]. en f((ai, bi)) ⊆ B and f([ai, bi]) = f((ai, bi)) ⊆ f((ai, bi)) ⊆B.f(ai), f(bi) ∈ ∂B.Since n ≥ 2, choose gi ∶ I → ∂B path, such that gi(0) = f(ai) and gi(1) = f(bi).Furthermore,B is convex, so π1(B) = 0. Ô⇒ fi ≃ gi.

34

2.6 Seifert von Kampen eorem

Repeating this process for all intervals [ai, bi], i ∈ {1, . . . , n}, we obtain a loop g ∶ I →Sn, homotopic to f , and with the property that g(I) ⊆ Sn ∖ {x}, [f] = [g].e lemma implies that [g] = [cx0], so [f] = [cx0] and π1(Sn, x0) = 0.Corollary 2.92 R2 and Rn are not homeomorphic for any n ≠ 2.P Suppose that f ∶ R2 → Rn a homeomorphism (n ≠ 2). en

• n = 1: R2 ∖ {0} ≃ R ∖ {f(0)}, but R2 ∖ {0} is path connected, but R ∖ {f(0} isnot.

• Recall π1(R2 ∖ {P}) ≃ π1(S1) ≃ Z, π1(Rn ∖ {f(p)}) ≃ π1(Sn−1) = 0 for n ≥ 3.f∣R2∖{P} ∶ R2 ∖ {P}→ Rn ∖ {f(P )} homeomorphism, contradiction.

Recall:

space π1 usingX ⊆ Rn convex, x0 ∈X 0 de nitionS1 Z covering spacesR2 ∖ {(0,0)} Z deformation retractsR3 ∖ {z-axis} Z deformation retractsS1 ∨ S1 π1(S1 ∨ S1) ≃ π1(R2 ∖ {p, q}) deformation retractsθ-space π1(θ) ≃ π1(R2 ∖ {p, q}) deformation retractsMöbius-bandM Z deformation retractsS1 ∨ S1 not abelian covering spacesT#T not abelian retractionSn, n ∈ N, n ≥ 2 0T Z ×Z

2.6. Seifert von Kampen eorem

2.6.1. Direct sums of abelian groups

De nition 2.94 Let G be an abelian group and {Ga}a∈A a family of subgroups of G.We say that the groupsGa generatesG if every element x ofG can be written as a nitesum of elements of the groupsGa.Since G is abelian, we can always write this sum in the form X = Xa1 + ⋅ ⋅ ⋅ + Xan ,Xa ∈ Ga, ai ≠ aj , if i ≠ j.In this case we oen write x = ∑a∈AXa.If the groupsGa generateG, thenG is called the sum ofGa and we writeG = ∑a∈AGa

orG = G1 + ⋅ ⋅ ⋅ +Gn.If Ga generates G and ∀x ∈ G ∃!A-tuple (xa)a∈A with xa = 0, for all but nitelymany a in A, and G = ∑a∈AXa then G is called the direct sum of Ga, and we writeG = ⊕a∈AGa, orG = G1 ⊕ ⋅ ⋅ ⋅ ⊕Gn.

Example 2.95

1. R2,G1 = {(x,0) ∣ x ∈ R},G2 = {(0, y) ∣ y ∈ R}, R2 = G1 ⊕G2.

2. R∞: the set of all sequences of real numbers that are eventually zero. R∞ is a groupunder coordinate addition. Gn = {(0, . . . ,0, xn,0, . . . ∣ xn ∈ R} are subgroups ofR∞ ∀n ∈ N and R∞ = ⊕n∈NGn.

35

Algebraic topology

Lemma 2.96 (Extension condition)LetG be an abelian group and {Ga}a∈A be a family of subgroups. IfG = ⊕a∈AGa, thenG stais es the following condition:

(∗) IfH is any abelian group and ha ∶ Ga → H is a family of homomorphisms, thenthere exists a homomorphism h ∶ G→H , such that h∣Ga

= ha ∀a ∈ A.

Furthermore, this is unique. Conversely if Ga generate G and (∗) holds, then G =⊕a∈AGa.P Suppose thatG = ⊕a∈AGa. ConsiderH abelian group., andha ∶ Ga →H familyof homomorphisms. De ne h ∶ G→H in the following way: If x ∈ G, then x = ∑n

i=1 xai

and h(x) = ∑ni=1 hai(Xai). h is well-de ned, because there is a unique way of writing x

as∑ni=1 xai . h is a homomorphism and h∣Ga

= ha.Suppose that h′ ∶ G → H with h′Ga

= ha, ∀a ∈ A. en h′(x) = h′(∑ni=1 xai) =

∑ni=1 h

′(xai) = ∑ni=1 hai(xai) = h(x).

Converse: Suppose that x = ∑a∈A xa = ∑a∈A ya for some x ∈ G. Choose b ∈ A andlet H = Gb. De ne ha ∶ Ga → H by ha = 1Gb

if a = b, and ha trivial if a ≠ b. Leth ∶ G → H be the extension of the homomorphisms ha. en h(x) = ∑a∈A h(xa) =∑a∈A ha(xa) =Xb, h(x) = ∑a∈A h(ya) = ∑a∈A ha(ya) = Yb.Ô⇒Xb = Yb, b ∈ A,Ô⇒Xa = Ya ∀a ∈ A.

Corollary 2.98 Let G = G1 ⊕ G2 and G1 = ⊕a∈AHa, and G2 = ⊕b∈BHb, where theindex setsA,B are disjoint. enG = ⊕γ∈A∪BHγ .[Proof: Problem Set 7, Exercise 4.a)]

Corollary 2.99 IfG = G1 ⊕G2, thenGÒG2≃ G1.

[Proof: Problem Set 7, Exercise 4.b)]

Given a family of abelian groups {Ga}a∈A, nd a groupG that contains subgroupsG′a ≃Ga ∀a ∈ A andG = ⊕a∈AG

′a.

De nition 2.100 Let {Ga}a∈A be a family of abelian groups, G be an abelian group,and ia ∶ Ga → G a family of monomorphisms such that G = ⊕a∈Aιa(Ga). en G iscalled the external direct sum of the groupsGa, relative to monomorphisms ia.

Notation: ≤: Subgroup.

eorem 2.101 If {Ga}a∈A is family of abelian groups, then there exists an abelian groupG, and a family of monomorphisms ia ∶ Ga → G, such thatG = ⊕a∈Aia(Ga).

P Consider the product ∏a∈AGa (with coordinate wise addition), which is anabelian group. LetG = {(xa)a∈A ∣ xa = 0 for all but nitely many indices inA} ≤∏Ga.Given β ∈ A, de ne iβ ∶ Gβ → G, by letting iβ(x) the tuple with x as its β-th coordinateand 0 in every other coordinate. iβ is a monomorphism (follows from the de nition),and if x ∈ G, then x has nitely many non zero coordinates, i.e. x = (x1, . . . , xn,0, . . . )and x = i1(x1) + ⋅ ⋅ ⋅ + in(xn). Furthermore, this expression is unique.

36

2.7 Free abelian groups

Lemma 2.103 (extension condition)Let {Ga}a∈A be a family of abelian groups, G be an abelian group, ia ∶ Ga → G familyof homomorphisms. If ia is a monomorphism ∀a ∈ A, and G = ⊕a∈Aia(Ga), then thefollowing condition holds:

(∗) Given any abelian group H and any family of homomorphisms, hA ∶ Ga → H ,then there exists a homomorphism h ∶ G→H , with h ○ ia = ha ∀a ∈ A.

Futhermore h is unique.Conversely if ia(Ga) generate G and (∗) holds, then ia is a monomorphism ∀a ∈ A,andG = ⊕a∈Aia(Ga).[Proof: Problem Set 7, Exercise 4.c)]

eorem 2.104 (Uniqueness of direct sums)Let {Ga}a∈A be a family of abelian groups, G,G′ be abelian groups, ia ∶ Ga → G,i′a ∶ Ga → G′ families of monomorphisms, such that G = ⊕a∈Aia(Ga) and G′ =⊕a∈Ai

′a(Ga). en ∃!φ ∶ G→ G′ isomorphism, such that φ ○ ia = i′a ∀a ∈ A.

P

• Lemma 2.103 (H = G′)Ô⇒ ∃!φ ∶ G→ G′, such that φ ○ ia = i′a ∀a ∈ A.

• Lemma 2.103 (H = G)Ô⇒ ∃!ψ ∶ G′ → G, such that ψ ○ i′a = ia ∀a ∈ A.

Now (Lemma 2.103) ψ ○ φ ∶ G→ G satis es (ψ ○ φ) ○ ia = ia ∀a ∈ A.Ô⇒ ψ ○ φ = 1G, analogously φ ○ ψ = 1G′ .

2.7. Free abelian groups

De nition 2.106 Let G be an abelian group and {xa}a∈A a family of elements in G.Let Ga = ⟨xa⟩, i.e. the subgroup generated by xa. If Ga generate G, we also say thatxa generateG.IfGa = ⟨xa⟩ is in nite cyclic∀a ∈ A, andG = ⊕a∈A⟨xa⟩, thenG is called a free abeliangroup with basis {xa}a∈A.

Lemma 2.107 (extension condition)LetG be an abelian group and {xa}a∈A a family of elements ofG that generatesG. enG is a free abelian group with basis {xa}a∈A iff for any abelian groupH and any family{ya}a∈A of elements inH , there is a homomorphism h ∶ G → H , such that h(xa) = ya∀a ∈ A. In such a case h is unique.P

“⇒” Given H , {ya}a∈A de ne homomorphisms ha ∶ Ga → H , with ha(xa) = ya.(Ga is cyclic, so the condition h(xa) = ya determines uniquely a homomorphismha ∶ Ga → H). By lemma 2.96 ∃h ∶ G → H homomorphism, with h∣Ga

= ha∀a ∈ A. Furthermore h is unique.

“⇐” Suppose that for some β ∈ A, ⟨xβ⟩ is nite. en forH = Z and ya = 1 ∀a ∈ A,there is no homomorphism h ∶ G→H with h(xa) = 1 ∀a ∈ A.By lemma 2.96, we get that G = ⊕a∈AGa, i.e. G is a free abelian group with basis{xa}a∈A.

37

Algebraic topology

Remark 2.109 Let G be free abelian with basis {x1, . . . , xn}. G = ⟨x1⟩ ⊕ ⋅ ⋅ ⋅ ⊕ ⟨xn⟩ ≃Z⊕ ⋅ ⋅ ⋅ ⊕Z.Proposition 2.110 IfG is free abelian with basis {x1, . . . , xn}, then n is uniquely deter-mined byG.P G ≃ Z⊕ ⋅ ⋅ ⋅ ⊕Z ≥ 2Z⊕ ⋅ ⋅ ⋅ ⊕ 2Z,

Z⊕ ⋅ ⋅ ⋅ ⊕ZÒ2Z⊕ ⋅ ⋅ ⋅ ⊕ 2Z ≃ Z2 ⊕ ⋅ ⋅ ⋅ ⊕Z2.

∣Z2 ⊕ ⋅ ⋅ ⋅ ⊕Z2∣ = 2n.

De nition 2.112 If G is a free abelian group with basis {x1, . . . , xn}, then n is calledthe rank ofG.

2.8. Free products of groups

De nition 2.113 Let G be a group and {Ga}a∈A be a family of subgroups of G. Wesay that Ga generate G, if ∀x ∈ G ∃(x1, . . . , xn) with xi element of some Ga ∀i ∈{1, . . . , n}, and x = x1 . . . xn.Such a sequence is called a word of length n in the groupsGa that represents x.

If xi, xi+1 ∈ Ga for some a ∈ A, we group them together to obtain the word

(x1, x2, . . . , xi−1, xixi+1, xi+2, . . . , xn)

of length n − 1. If xi = 1, we delete xi from our sequence.Repeating these reduction operations, we obtain (y1, . . . , ym) representing x, such thatno group Ga contains yi and yi+1 and, yi ≠ 1 ∀i ∈ {1, . . . ,m}. We call (y1, . . . , ym) areduced word.Convention: ∅ is a reduced word representing 1 ∈ G.

De nition 2.114 Let G be a group and {Ga}a∈A a family of subgroup that generatesG and Ga ∩Gb = {1} ∀a, b ∈ A, a ≠ b. If ∀x ∈ G there is only one reduced word inthe group {Ga}a∈A, that represents that element, then G is called the free product ofthe groups {Ga}a∈A.G = ∗a∈AGa, orG = G1 ∗ ⋅ ⋅ ⋅ ∗Gn.

Example 2.115 G1 = {1, x1}, G2 = {1, x2}. x ∈ G = G1 ∗G2. Elements of G are forexample x1, x2, x1x2, x2x1, x1x2x1, x2x1x2, . . .Example 2.116 G = {φ ∶ {0,1,2} → {0,1,2} ∣ φ bijection}. φ1 ∶ {0,1,2} → {0,1,2},φ1(2) = 2, φ1(1) = 0, φ1(0) = 1φ2 ∶ {0,1,2}→ {0,1,2}, φ2(0) = 0, φ2(1) = 2, φ2(2) = 1.

∣⟨φ1⟩∣ = 2, ∣⟨φ2⟩∣ = 2,⟨φ1⟩, ⟨φ2⟩ generateG, butB ≠ ⟨φ1⟩ ∗ ⟨φ2⟩.Indeed the reduced words (φ1, φ2, φ1) and (φ2, φ1, φ2) represent the same element inG.

38

2.8 Free products of groups

Claim: Suppose that Ga generate G. Ga ∩ Gb = {1} ∀a, b ∈ A with a ≠ b. If therepresentation of 1 by the empty word is unique thenG = ∗a∈AGa.P Let x ∈ G and suppose that (x1, . . . , xn), (y1, . . . , ym) are two reduced wordsrepresenting x, where xi ∈ Gai , yi ∈ Gbj , ai ∈ A, i ∈ {1, . . . , n}, bj ∈ A, j ∈ {1, . . . ,m}.x1 . . . xn = x = y1 . . . ymÔ⇒ x1 . . . xny

−1m . . . y−11 .

Ô⇒ an = bm (i.e. xn, ym ∈ Gan = Gbm).We obtain (x1, . . . , xn−1, xny−1m , y−1n−1, . . . , y

−11 ).

Now we must have xny−1m = 1Ô⇒ xn = ym.Continue this process, to conclue that n =m and xi = yi, ∀i ∈ {1, . . . , n}.Lemma 2.118 (extension condition / universal mapping property) Let G be a group,{Ga}a∈A be a family of subgroups. If G = ∗a∈AGa, then it satis es the following ex-tension condition: Given any groupH and any family of homomorphisms ha ∶ Ga →Hthre exists a homomorphism h ∶ G→H , such that h∣Ga

= hA. In addition h is a unique.P De ne h ∶ G→H as follows:

• h(1) = 1,

• If x ∈ G, x ≠ 1, let (x1, . . . , xn) be the (unique) reduced word representing x. Seth(x) = ha1(x1) . . . han(xn), where ai is the index for which xi ∈ Gai .

en

• h is well-de ned,

• h∣Ga= ha ∀a ∈ A,

• h is a group homomorphism (problem sheet 8).

In fact, uniqueness ofh follows from the fact thathmust statisfy h(x) = h(x1, . . . , xn) =h(x1) . . . h(xn) = ha1(x1) . . . han(xn).

External free product

De nition 2.120 Let {Ga}a∈A be a family of groups. Suppose that G s a group and{ia ∶ Ga → G}a∈A a family of monomorphisms with G = ∗a∈Aia(Ga), then G iscalled the external free product of the groupsGarelative to the onomorphisms ia.

Does such a groupG exist?

eorem 2.121 Let {Ga}a∈A be a family of groups. ere exists a group G and a familyof monomorphisms ia ∶ Ga → G, such thatG = ∗a∈Aia(Ga).

Lemma 2.122 (extension condition){Ga}a∈A family of groups, G group. ia ∶ Ga → G family of homomorphisms. If ia is amonomorphism ∀a ∈ A and G = ∗a∈Aia(Ga), then the following extension conditionholds: Given any groupH and a family of homomorphisms ha ∶ Ga → H , ∃ ∶ G → Hhomomorphism such that h ○ ia = ha ∀a ∈ A. In addition h is unique.[Proof ommited]

39

Algebraic topology

eorem 2.123 (uniqueness of free product)Let {Ga}a∈A be a family of gropus, G,G′ be groups, {ia ∶ Ga → G}a∈A, {i′a ∶ Ga →G′}a∈A families of monomorphism such that {ia(Ga)}a∈A, {i′a(Ga)} generate G,G′repsectively.IfG,G′ satisfy the extension condition, then ∃! isomorphismφ ∶ G→ G′ such thatφ○ia =i′a ∀a ∈ A.

[Proof: analogous to the proof of uniqueness of direct sums (problem sheet 8)]

Lemma 2.124 Let {Ga}a∈A be a family of groups, G be a gropu and ia ∶ Ga → G bea family of homomorphisms. If ia(Ga) generate G, and the extension condition holds,then ia is a monoomorphism ∀a ∈ A, andG = ∗a∈AGa.P Consider b ∈ A. SetH = Gb, ha ∶ Ga → H the identity homomorphism 1Gb

ifa = b, and ha ∶ Ga →H the trivial homomorphism, if a ≠ b.Let h ∶ G→H be the homomorphism given by the extension condition.en h ○ ib = hbÔ⇒ h ○ ib = 1Gb

. Ô⇒ ib is 1:1.I.e. ia is a monomorphism ∀a ∈ A.Existence theorem Ô⇒ there exists a group G′ and {i′a ∶ Ga → G′}a∈A a family ofmonomorphisms such thatG′ = ∗a∈Ai′a(Ga). G,G′ both satisfy the extension conditionand are generated by {ia(Ga)}a∈A, respectively {i′a(Ga)}a∈A.Uniqueness theoremÔ⇒ there is an isomorphism φ ∶ G → G′ with φ ○ ia = i′a. G′ =∗a∈Ai′a(Ga) .Ô⇒G = ∗a∈Aia(Ga).Corollary 2.126 If G = G1 ∗ G2, , G1 = ∗a∈AHa, G2 = ∗b∈BHb, A ∩ B = ∅, thenG = ∗γ∈A∪BHγ .N normal subgroup ofG: N ⊴ G.

eorem 2.127 Let G = G1 ∗ G2, Ni ⊴ Gi, i ∈ {1,2}. If N is the smallest normalsubgroup ofG that containsN1, andN ∶ , then

GÒN ≃ (G1ÒN1

) ∗ (G2ÒN2) .

P Consider the inclusion homomorphism i ∶ G1 → G1 ∗G2, i′ ∶ G2 → G1 ∗G2,and the projection homomorphism p ∶ G1 ∗G2 → G1 ∗G2/N , g ↦ gN .Let n1 ∈ N1. en (p ○ i)(n1) = n1N = N . is implies thatN1 ≤ ker(p ○ i) and p ○ iinduces (p ○ i)′ ∶ G1ÒN1

→ G1 ∗G2ÒN .Analogously, (p ○ i′)′;G2ÒN2

→ G1 ∗G2ÒN .We will apply lemma 2.124 for the homomorphisms i1, i2.

• Check that the extension condition holds.Let h1 ∶ G1/N1 → H1, h2 ∶ G2/N2 → H arbitrary homomorphisms and p1 ∶G1 → G1/N1, p2;G2 → G2/N2 the projection homomorphism. en h1 ○ p1 ∶G→H , h2 ○p2 ∶ G2 →H homomorphisms. e extension condition forG1∗G2

implies that ∃h ∶ G1∗G2 →H , with hGi = hi○pi, i ∈ {1,2}. Ifni ∈ Ni, i ∈ {1,2},then h(ni) = 1H . I.e. Ni ≤ kerh ∀i ∈ {1,2}.Ô⇒N ≤ kerh.is implies thath ∶ G1∗G2 →H induces a homomorphismh′ ∶ G1∗G2/N →H

40


with h′ ○ i1 = h1, and h′ ○ i2 = h2.(h′ ○ i1(g1N1) = h′ ○ (p ○ i)′(g1N1) = h′((p ○ i)(g1)) = h′(p(i(g1))) =h′(p(g1)) = h′(g1N) = h(g1) = (h1 ○ p1)(g1) = h1(g1N1)).

• Check that i1(G1/N1), i2(G2/N2) generateG1 ∗G2/N .i1(G1/N1) = (p ○ i)′(G1/N1) = (p ○ i)(G1) = p(G1) = G1/N , and similarlyi2(G2/N2) = G2/N2.(Lemma2.124)Ô⇒ i1, i2 aremonomorphisms and (G1 ∗G2)ÒN = i1(G1/N1)∗i2(G2/N2).

Corollary 2.129 IfN is the smallest normal subgroup ofG1 ∗G2 that containsG1, thenG1 ∗G2ÒN ≃ G2.P N1 = G1,N2 = {1G2}.G group, {Ga}a∈A family of groups, {ia ∶ Ga → G}a∈A family of homomorphisms.en the following statements are equivalent:

1. (ia is a monomorphism ∀a ∈ A and)G = ∗a∈Aia(Ga) .

2. Given any group H and any family of homomorphisms ha ∶ Ga → H , ∃! homo-morphism h ∶ G→H with h ○ ia = ha, ∀a ∈ A.

2.8.1. Free groups

De nition 2.131 LetG be a group and {xa}a∈A a family ofGwith ⟨xa⟩ in nite cylclic∀a ∈ A. IfG = ∗a∈A⟨xa⟩, thenG is called a free group and {xa}a∈A is called a systemof free generators.In this case, if x ∈ G ∖ {1}, then x can be written uniquely as x = (xai)n1 ⋅ (xak

)nk

where ai ≠ ai ≠ ai+1, ∀i ∈ {1, . . . , k − 1} and n1 ∈ Z ∖ {0} ∀i ∈ {1, . . . , k}.

Free groups are characterized by the following:Lemma 2.132 (exension condition)Let G be a group and {xa}a∈A be a family of elmeents in G. If G is a free group with{xa}a∈A system of free generators, thenG satisi es

(∗) Given any groupH and any family {ya}a∈A of elements ofH there exists a homo-morphism h ∶ G→H with h(xa) = ya ∀a ∈ A.

In addition h is unique. Conversely, if {xa}a∈A generates G, and (∗) holds, then G is afree group with system of free generators {xa}a∈A.P Lemma 2.118.For the converse, consider b ∈ A. en there exists a homomorphism hb ∶ G → Z withhb(xb) = 1, and hb(xa) = 0 ∀a ∈ A, a ≠ b. (H = Z, {ya}a∈A = {0,1}). is implies,that ⟨xb⟩ is in nite cyclic. erefore ⟨xa⟩ in nite cyclic ∀a ∈ A.en lemma 2.124.

eorem 2.134 Let G = G1 ∗ G2, G1,G2 are free groups with {xa}a∈A, {xa}a∈B re-sepective free sytems of generators withA ∩B = ∅, thenG is a fre group with {xa}a∈A∪Bfree system of generators.

41

Algebraic topology

De nition 2.135 Let {xa}a∈A be an arbitrary indexed family and Ga be the set of allsymbols xna , n ∈ Z. De ne xna ⋅ xma = xn+ma . is makes our setGa a group.e external free product of the groupsGa is called the free group on the elements xa.

Notation: We will identify the elements xna ofGa with their images ia(xna) inG.

Relation between free groups and free abelian groups.

eorem 2.136 If G is a free group and {xa}a∈A is a system of free generators, thenGÒ[G,G] is the free abelian group with basis the {[xa]}a∈A (where [xa] is the cosetxa[G,G]).

P We wil show, that 1. {[xa]}a∈A generate GÒ[G,G] and 2. if H is an abeliangroup and {ya}a∈A is a family of elements in H , then there is a homomorphism h′ ∶GÒ[G,G] → H , with h′([xa]) = ya ∀a ∈ A. Lemma 2.107 will them directly implythe proof.

1. {xa}a∈A generateG↝ {[xa]}a∈A generateGÒ[G,G].

2. Consider an abelian groupH and {ya}a∈A a family of elements inH . G freeÔ⇒∃h ∶ G→H homomorphism with h(xa) = ya ∀a ∈ A.

H abelian Ô⇒ [G,G] ≤ kerh. Ô⇒ there exists an induced morphism h′ ∶GÒ[G,G]→H with h′([xa]) = ya ∀a ∈ A.

Corollary 2.138 If G is a free group with n free generators, then an system of free gen-erators ofG has n elements.P GÒ [G,G] free abelian group with basis {[xi]}i=1,...,n.

De nition 2.140 IfG is a free group, then

rank(G) ∶= rank(GÒ [G,G]).

Properties:G free abelian group G free groupH ≤ G→H , is free abelian H ≤ G↝H is free.rank(G) = n, and H ≤ G Ô⇒rank(H) ≤ n.

Consider the free group G and two ele-ments {x1, x2}, [G,G] ≤ G is freely gen-erated by [xn1 , xn2 ], m,n ∈ Z ∖ {0}. Infact a free group with rank greater than 1has subgroups of all countable orders.

Groups up to isomorphism:G free abelian, then the isomorphism type is determined by the cardinality of the basis.G free, then the isomorphism type is determined by the cardinality of systems of gener-ators.G nitely generated abelian group, then G ≃ H ⊕ T , where H is free-ablian and T istorsion subgroup, so the isomorphismtype is de ned by the rank of H and elementarydivisors.

42


Group representation Let G be a group and {xa}a∈A be a family that generates G.Consider the free group F on {xa}a∈A. De ne a epimorhpishm h ∶ F → G, xa ↦ xa.G ≃ FÒkerh.

N ∶= kerh. N is called the relation subgroup. If n ∈ N , then n is called a relation on F .N ⊴ F and we can specifyN by specifying a family {rb}b∈B of elements in F such thatN is the smalest normal subgroup containing {rb}b∈B . (Exercise 4, problem sheet 8).Such a family {rb}b∈B is called a complete set of relations.

De nition 2.141 If G is a group, a presentation of G is family of generators {xa}a∈Aof generators for G together with a complete set of relations {rb}b∈B for G. If both{xa}a∈A and {rb}b∈B are nite, thenG is called nitely presented. If {xa}a∈A is nite,thenG is called nitely generated.

A presentation determinesG uniquelly up to isomorphism. Given two different presen-tations, it is hard to decide whether they determine isomorphic groups (Unsolvability ofthe isomorphism problem for groups).Example 2.142

1. G cyclic group of order n. Let x ∈ G be generator. ⟨x ∣ xn⟩.

2. D2n dihedral group. ⟨r, s ∣ rn, s2, (rs)2⟩.

3. G ∶= ⟨a, b ∣ a3b−2⟩,G′ ∶= ⟨x, y ∣ xyxy−1x−1y−1⟩. IsG ≃ G′?F = ⟨a, b⟩,F ′ = ⟨x, y⟩. De ne h ∶ F → G′ group homomorphismwith h(a) = xyand h(b) = xyx.

h(a3b−2 = (h(a))3(h(b))−2 = xyxyxyx−1y−1x−1y−1x−1 = xyxxyxx−1y−1x−1x−1y−1x−1 =1G.If N is the smallest normal subgroup containing a3b−2 then (exercise 4, problemsheet 8)N = ⟨ga3b−2g−1, g ∈ F ⟩.enN ≤ kerh.h′ ∶ F Ò N → G′, i.e. h′ ∶ G→ G′.De ne g ∶ F ′ → G group homomorphism, with g(x) ∶= a−1b, and g(y) ∶= b−1a2.Do the same as above, then, g induces a homomorphism g′ ∶ G′ → G.Check that h′ ○ g = 1G′ ∶ G′ → G′ and g′ ○ h′ = 1G ∶ G→ G.

2.8.2. e Seifert-van Kampen theorem

LetX be a topological space, x0 ∈X , andX = ∪α∈AAα, whereAα open path-connectedsubset of X with x0 ∈ Aα ∀α ∈ A. en we can consider the homomorphism jα ∶π1(Aα) → π1(X) induced by the inclusion Aα ↪ X (for brevity we write π1(Aα)instead of π1(Aα, x0) and we write π1(X) instead of π1(X,x0)). ese homomomor-phism extend to a homomorphism Φ ∶ ∗α∈Aπ1(Uα)→ π1(X) (extension condition forfree product).e Seifert-van Kampen theorem tells us, thatΦ is “very oen” surjective, but in generalnot injective.Consider iαβ ∶ π1(Aα∩Aβ)→ π1(Aα) the homomorphism induced byAα∩Aβ ↪ Aα.en jα ○ iαβ = jβ ○ iβα. Let ω ∈ π1(Aα ∩ Aβ). en Φ(iαβ(ω) ⋅ (iβα(ω))−1) =

43

Algebraic topology

Φ(iαβ(ω))(Φ(iβα(ω)))−1 = (jα ○ iαβ)(w) ⋅ (jβ ○ iβα(w))−1 = 1π1(X). I.e. iαβ(ω) ⋅i−1βα ∈ kerΦ.

eorem 2.143 (Seifert-van Kampen)If X = ∪α∈AAα. Aα open and path connected with x0 ∈ Aα, ∀α ∈ A, and Aα ∩ Aβ ispath connected for any pair {α,β} inA, thenΦ is surjective. If furthermoreAα∩Aβ ∩Aγ

is path-connected, for any triple {α,β, γ} in A, then kerΦ = N , where N is the small-est normal subgroup containing all the elements of the form iαβ(ω)i−1βα, so Φ induces an

isomorphism ∗απ1(Aα)ÒN ≃ π1(X).

P

• surjectivity of Φ ∶ ∗απ1(Aα)→ π1(X).Let f be a loop in π1(X) based at x0 f is continuous and Aα is open ∀α ∈ A, so∀s ∈ I there is an open interval VS ⊂ I with s ∈ Vs and f(V s) ⊆ Aα for someα ∈ A. We have I = ∪s∈IVs and I is compact, so it is covered by nitelymany of theVs. en endpoints of these intervalls de ne a partition 0 = s0 < s1 < s2 < ⋅ ⋅ ⋅ <sm = 1 of I such that for each i there is some subsetAα with f([si−1, si] ⊆ Aα.

Denote the Aα containing f([si−1, si]) byAi and the path f∣[si−1,si] = fi. enf = f1 ⋅ ⋅ ⋅ ⋅ ⋅ fm, fi path inAi.

Now Ai ∩Ai+1 is path-connected, so we can choose a path gi in Ai ∩Ai+1 fromx0 to f(si).Consider the loop (f1 ⋅ g1) ⋅ (g1 ⋅ f2 ⋅ g2)⋯(gm−1 ⋅ fm), which is homotopic to f .

is is a composition of loops based at x0 each lying in a single Ai. Hence [f] =[f1 ⋅ g1] ⋅ [g1 ⋅ f2 ⋅ g2]⋯[gm−1 ⋅ fm] ∈ imΦ.

Notation: A factorization of [f] ∈ π1(X) is a fromal product [f1]⋯[fk], whereeach fi is a loop in someAα based at x0, and f is homotopic to f1⋯fk.A factorization of [f] is aword in∗α∈Aπ1(Aα), possibly unreduced, withΦ([f1]⋯[fk]) =[f].Two factorizatons of [f] are equivalent, if they are related by a sequence of the fol-lowing two kinds ofmoves or their inverses: 1. Combine adjacent terms [fi]⋅[fi+1]into [fi ⋅ fi+1] if [fi], [fi+1] lie in the same π1(Aα), and 2. regard the term[fi] ∈ π1(Aα) as lying in π1(Aβ), if fi is a loop inAα ∩Aβ .

LetQ ∶= ∗α∈Aπ1(Aα)ÒN . e rstmmove, doesn’t change the element of∗α∈Aπ1(Aα)de ned by the factorization. e second move doesn’t change the image of this el-ment inQ.

• kerΦ.Recall that N ≤ kerΦ. If I show that any two factorizations of f are equivalent,this will imply that Φ′ ∶ Q→ π1(X) is injective, hence kerΦ = N .

Let [f1]⋯[fk] and [f ′1]⋯[f ′ℓ] be two factorizations of [f] and let F ∶ I ×I →X ahomotopy from f1⋯fk to f ′1⋯f ′ℓ. en there exists partions 0 = s0 < ⋅ ⋅ ⋅ < sm = 1and 0 = t0 < ⋅ ⋅ ⋅ < tn = 1, such that each rectangle Rij = [si−1, si] × [tj−1, tj] ismapped by f into a singleAα, which we label byAij . ese partitions are obtainedby

44


– covering I × I with nitely many rectangles [a, b]× [c, d] each mapped intoa singleAα,

– partioning I × I by the union of all horizontal and vertical lines containingedgese of the rectangle above.

We can also assume that the s-partition subdivides the partitions given by the prod-ucts f1⋯fk, f ′1⋯f ′ℓ. We may also perturb the vertical sides of the rectangles Rij ,so that each point in I × I lies in at most three Rij ’s. Relable the new rectanglesRmn ordering them as in the picture

..R1

.R2

.R3

.R4

.

R9

.

R10

.

R11

.

R8

.

R5

.

R6

.

R7

.

R12

Consider γr , r ∈ {0,1, . . . ,mn}, the path in I × I from the le edge {0} × I tote right edge {1}× I that separatesR1, . . . ,Rr fromRr+1, . . . ,Rmn. We will callthe corners of the rectangles vertices. For each vertex v with F (v) ≠ x0 let gv bea path from x0 to F (v) in the intersection of the two or three Aij ’s correspond-ing to the rectangles containing v. Insert into F∣γr

the appropriate paths gv ⋅ gvat successive vertices (same idea as in the rst part of the proof). is gives us afactorization of [F∣γr

] by regarding the loop corresponding to a horizontal or ver-tical segment between two adjacent vertices as lying in the Aij for either of therectangles containing the segment.Diufferent choices of Aij give equivalent factorizations. Also the factorizationsassociated to successive paths γr, γr+1 are equivalent, because pushing γr to γr+1acrossRr+1 changes F∣γr

to F∣γr+1 by the homotopy within Aij corresponding toRr+1.We can arrange that the factorization associtated to γ0 is equivalent to [f1]⋯[fk]and by choosing gv for each vertex v along I × {0} ⊆ I × I , to lie not just in thethe twoAij ’s corresponding to the rectangles containing v, but also to to lie in theAα for the fi containing v in its domain.Similarly, the factorization associated to γmn is equivalent to [f ′1]⋯[f ′ℓ]. is givesus that [f1]⋯[fk], [f ′1]⋯[f ′ℓ] are equivalent.

Corollary 2.145 LetX = U∪V ,U,V open,U,V,U∩V path connected. Let x0 ∈ U∩V .enΦ induces an isomorphism π1(X) ≃ π1(U) ∗ π1(V )ÒN , whereN is the normalsubgroup generated by all elements i1(ω) ⋅ i2(ω)−1, where i1 ∶ π1(U ∩ V ) → π1(U),i2 ∶ π1(U ∩ V )→ π1(V )morphisms induced by the inclusion and ω ∈ π1(U ∩ V ).[Proof: Immediate from Seifert-van Kampen theorem].Corollary 2.146 Consider the assumption of corollary 2.145. If in addition U ∩ V issimply connected, then Φ induces an isomorphism π1(X) ≃ π1(U) ∗ π1(V ).[Proof: Immediate from corollary 2.145].Example 2.147

1. LetX be a theta space. Recall π1(X) is not abelian. (S1 ∨ S1 and theta space aredeformation retracts of R2 ∖ {p, q} therfore π(S1 ∨ S1) ≃ π1(X)).

45

Algebraic topology

..b .a . c

U ∶=X∖{c},V ∶=X∖{a}. U,V are open and path connected. U∩V =X∖{a, c}is path connected, in addition U ∩ V is contractible.Ô⇒ π1(X) ≃ π1(U) ∗ π1(V )ÒN . By corollary 2.146 π1(X) ≃ π1(U)∗π1(V ).U,V have the homotopy type of S1Ô⇒ π1(U) ≃ Z, π1(V ) ≃ Z.Ô⇒ π1(X) ≃ Z ∗Z.

2. LetX be the wedge of two circles, i.e. X = S1 ∨ S1. U ∶=X ∖ {a}, V ∶=X ∖ {c}.π1(X) ≃ Z ∗Z.

Remark 2.148

1. Necessity of path connectedness ofAα ∩Aβ .Consider

..S1.

V

And U analogously by re ection on the vertical through the center. en U,Vopen, path connected, but U ∩V is not path connected the morphism induced byΦ is not surjective, π1(U) ∗ π1(V )→ π1(S1).

2. Necessity of path-connectedness ofAα ∩Aβ ∩Aγ .Consider the theta space

..b .a . c

Aa ∶=X ∖ {a},Ab ∶=X ∖ {b},Ac ∶=X ∖ {c}, thenAa ∩Ab ∩Ac =X ∖ {a, b, c}not path connected, π1(X) ≃ Z ∗Z∗.

Example 2.149 (e shrinking wedge of circles)ConsiderCn the circle with center (1/n,0) and radius 1/n inR2. LetX = ∪n∈NCn withthe subspace topology.

.

46

2.9 CW complexes (cell complexes)

Consider the retractions rn ∶ X → Cn collapsing all Ci’s except Cn to (0,0). Each rninduces a surjection (rn)∗ ∶ π1(X) → π1(Cn) (basepoint (0,0)). e product of thesesurjections (i.e. ρ[f]) = (r1[f], r2[f], . . . , )) gives a homomorphism ρ ∶ π1(X) →∏∞Z (=direct product of countably in nite many copies of Z).Claim: ρ is surjective (Ô⇒ π1(X) is uncountable).Proof: For any sequence of integers (Kn)n∈N we can construct a loop f ∶ I → X basedat (0,0) and goingKn times around the circle Cn in time [1 − 1/n,1 − 1/(n + 1)].f is indeed continuous:

• ∀t ∈ [0,1), f is continuous at t✓.

• f is contiuous at 1 ∈ I ⇐⇒ (∀ neighbourhoods V of f(1) = (0,0) ∃ neighbour-hood U of 1 such that f(U) ⊆ V ).V contains all but nitely many of the circles Cn, and this proves continuity of fat 1.

2.9. CW complexes (cell complexes)

We have seen, that the torus T can be constructed from

..

R

.a

.

a

.b

.b

e interior of the rectangle can be though of as an open disc or a 2-cel attached to theunion of two circles a, b.e union of the two circles can be thought of as obtained from their point of intersectionP by attachin to it two open arcs, 1-cells.In general construct a spaceX by the following procedure:

1. Start with a discrete setX0, whose points are regarded as 0-cells.

2. Inductively construct the n-skeletonXn from the (n−1)-skeletonXn−1 via con-tinuous maps φα ∶ Sn−1 →Xn−1.

Xn =Xn−1∐

α

enαÒ∼,

where ∼ the equivalence relation generated by x ∼ φα(x), x ∈ ∂enα.

3. Either stop this inductive process aer nitely many steps, setting X = Xn, orcontinue inde nitely, setting X = ∪n∈NXn. In latter case, X is given the weaktopology, i.e. A ⊂X is open, iffA ∩Xn is open inXn ∀n.

A space constructed in this way is called a CW-complex, where C stands for closure-niteness: the closure of each cell intersects only nitely many other cells, W stands for

weak-topology.Example 2.150 S2.

47

Algebraic topology

Suppose we attach a collection e2α of 2-cells to a path connected spaceX

Y ∶=X∐α

e2αÒ∼,

where x ∼ φα(x) for x ∈ ∂e2α.If s0 is a base-point of S1, then φα determines a loop inX based at φα(s0).Proposition 2.151 If i ∶ X → Y is the inclusion map, then i∗ ∶ π1(X,x0) → π1(Y,x0)is surjective and kerΦ∗ = N , i.e. π1(Y,x0) ≅ π1(X,x0)ÒN .X path-connected space, Y space constructed from X by attaching a family of 2-cellse2α via maps φα ∶ S1 → X . If s0 is a basepoint of S1, then φa determines a loop in Xbased at φα(s0). Let x0 ∈ X a basepoint and let γα ∶ I → X a path with γ(0) = x0 andγ(1) = φα(s0).en γα ∶ φαγα is a loop inX based at x0, γαφαγα is nullhomotopic in Y ∀α.LetN be the normal subgroup of π1(X,x0) generated by all elements γαφαγα.If i∗ ∶ π1(X,x0) < raπ1(Y,x0) is the homomorphism induced by i ∶ X ↪ Y , thenN ≤ ker i∗.Proposition 2.152 If i ∶ X → Y is the inclusion map, then i∗ ∶ π1(X,x0) → π1(Y,x0)is surjective, and ker i∗ = N .P Consider the spaceZ obtained from Y by attaching rectangular stripsSα = I×Iwith I×{0} is attached along γα, I×{1} is not attached to anything, {0}×I are identi edfor all α, {1} × I is attached along an arc in e2α.Z deformation retracts onto Y .A ∶= Z ∖ {yα}α, where yα is a point in e2α not in the arc along which Sα is attached.B ∶= Z ∖X .We apply the Seifert-van Kampen theorem for Z = A ∪B. A = Z ∖ {yα}α deformationretracts ontoX . B is contractible. A,B statisfy te conditions of the Seifert-van Kampentheorem.Ô⇒ π1(Y ) ≃ π1(Z) ≃ π1(A)ÒN , where N is the normal subgroup generated by theimage of π1(A ∩B)→ π1(A).Consider the coverAα = A ∩B ∖ ∪β≠αe2β .Aα deformation retracts onto a circle in e2α ∖ {yα}, so π1(Aα) ≃ Z and π1(Aα) is gen-erated by a loop homotopic to γαφαγα. So π1(A ∩B) is generated by loops homotopicto γαφαγα for all α.Example 2.154 T has a cell-structure with one 0-cell, two 1-cells and one 2-cell. LetX =X1 be the 1-skeleton of T ,X1 = S1 ∨S1 and attach toX1 one 2-cell e21 via φ1 ∶ S1 →Xwith φ1(S1) = a ⋅ b ⋅ a ⋅ b, to produce Y = T .

..

R

.a

.

a

.b

.b

It follows from the proof, that π1(T ) ≃ π1(X1)ÒN , where N is the normal subgroup

of π1(X1) generated by φ1.Nowπ1(X1) ≃ Z∗Z, with generators [a], [b], soπ1(T ) = ⟨[a], [b] ∣ [a][b][a]−1[b]−1⟩.α ∶= [a], β ∶= [b]. Let F = ⟨α,β⟩ be the free group on two generators α,β andN be thesmallest normal subgropu containing [α,β]. en N ≤ [F,F ]. Furthermore FÒN isabelian, then [F,F ] ≤ N .Ô⇒N = [F,F ], and ⟨α,β ∣ [α,β]⟩ = FÒN =

FÒ[F,F ] =Z⊕Z.

48

2.10 Surfaces (two-dimensional manifolds)

2.10. Surfaces (two-dimensional manifolds)

2.10.1. Fundamental group of surfaces

Tn ∶= T# . . .#T

n

n-fold Torus.

Example 2.155 T#T has CW-cell structure one 0-cell, four 1-cells, one 2-cell.Example 2.156 T#T#T

Example 2.157 n-fold torus: 4n-gonwith labelling (a1b1a1b1)(a2b2a2b2) . . . (anbnanbn)

Proposition 2.158

π1(Tn) ≃ ⟨α1, β1, . . . , αn, βn ∣ [α1, β1][α2, β2] . . . [αn, βn]⟩.

P Completly analogous to the case n = 1, i.e. T. Tn has a cell structure with one0-cell, 2n 1-cells and one 2-cell. e attaching map φ1 ∶ S1 → (Tn)1 of the 2-cell e21determines a loop in Tn with [φ1] = [a1][b1][a1]−1[b1]−1 . . . [an][bn][an]−1[bn]−1.

e real projective planeRP2 Notation: P ∶= RP2.

RP2 ∶= R3 ∖ {0}Ò∼,

where x ∼ y ∶⇐⇒ ∃λ ≠ 0 with x = λy.We can also look at it as

S2Ò∼where p, q ∈ S2, p ∼ q ∶⇐⇒ p = −q.

RP2 can also be descried as the quotient of D2 with antipodal points of ∂D2 identi ed.As U = {(x, y, z) ∈ S2 ∣ z ≥ 0} ≃ D2.Proposition 2.160 π1(RP2) ≃ ⟨α ∣ α2⟩ ≃ Z2.P Method 1: Using the cell structure of RP2 given by the picture

..

a

.

a

Method 2: p ∶ S2 → RP2 is a covering map, since π1(S2) = 0, ϕ ∶ π1(RP2, x)→ p−1(x)x ∈ RP2 is bijective.Consider Pn ∶= P# . . .#P.P2 = P#P. has cell structure

.. a.a

.

b

.b.

c

Proposition 2.162 π1(Pn) ≃ ⟨α1, . . . , αn ∣ α21 . . . α

2n⟩.

P Using cell-structure of Pn.Remark 2.164

49

Algebraic topology

1. For every group G, there exists a CW-complexXG with π1(XG) ≃ G: Consider⟨gα∣rβ⟩ a representation of G and take VαS1α. en attach 2-cells e2β along theloops speci ed by hte relations rβ .

2. RP2 = P

• P cannot be embedded in R3 (i.e. there does not exist a map f ∶ P → R3,such that f ∶ P→ f(P ) is a homeomorphism).

• P can be immersed in R3 (i.e. there exists a differentiable map f ∶ P → R3

with Dpf ∶ TpP→ Tf(p)R3 is injective ∀p ∈ P .

Question: Can we deduce, that Tn, Tm are not homotopy equivalent, if n ≠m?Answer: Yes, forT andT2. In general not yet. (It is not a tirivial matter to compare twogroup presentations).We turn to study π1Ò[π1, π1].

2.10.2. Homology of surfaces

Let X be a path connected space, x0, x1 ∈ X , a ∶ I → X path from a(0) = x0 toa(1) = x1. en we can construct an isomorphism

a ∶ π1(X,x0)→ π1(X,x1),

where [f]↦ [a][f][a] and

aab ∶ π1(X,x0)Ò[π1(X,x0), π1(X,x0)]→π1(X,x1)Ò[π1(X,x1), π1(X,x1)].

If b ∶ I →X is a path inX with b(0) = x0 and b(1) = x1 and g = ab, then

gab ∶ π1(X,x0)Ò[π1(X,x0), π1(X,x0)]→π1(X,x1)Ò[π1(X,x0), π1(X,x0)],

where

[f][π1(X,x0), π1(X,x0)]↦ [g][f][g][π1(X,x0), π1(X,x0)]= [f][π1(X,x0), π1(X,x0)].

Sogab = 1π1(X,x0)Ò[π1(X,x0),π1(X,x0)]

.

at is, the isomorphism aab is independent of the choice of path a.

De nition 2.165 IfX is a path connected space and x0 ∈X , let

H1(X) ∶= π1(X,x0)Ò[π1(X,x0), π1(X,x0)].

H1(X) is called the rst homology group ofX .

[We omit the basepoint, since there is in fact a unique isomorphism between the corre-sponding groups of different basepoints].One can de ne homology groupsHn(x) ∀n ∈ N.

50

2.10 Surfaces (two-dimensional manifolds)

Lemma 2.166 Let F be a group andN ⊴ F . en

F Ò NÒ[F Ò N,F Ò N] ≃F Ò [F,F ]Ò{n[F,F ] ∣ n ∈ N}.

[Proof omitted].

Let F be a free group with free generators α1, . . . , αn. Let x ∈ F andN be the smallestnormal subgroup containing x. Finally, let G = F Ò N . F Ò [F,F ] is free abeliangropu with basis α1[F,F ], . . . , αn[F,F ]. N Ò [F,F ] is the subgroup generated byx[F,F ].e lemma implies thatGÒ [G,G] is isomorphic to the quotient of a free abelian groupwith basis {α1[F,F ], . . . , αn[F,F ]} by the subgroup ⟨x[F,F ]⟩.

eorem 2.167 H0(Tn) is a free abelian group of rank 2n ≃ Z⊕ ⋅ ⋅ ⋅ ⊕Z2n

.

P π1(Tn) = ⟨α1, β1, . . . , αn, βn ∣ [α1, β1]⋯[αn, βn]⟩.H1(Tn) = π1(Tn) Ò [π1(Tn), π1(Tn) is the quotient of a free abelian group of rank2n by the group generated by

[α1, β1]⋯[αn, βn][π1(Tn), π1(Tn)] = 1π1(Tn)Ò[π1(Tn),π1(Tn)].

I.e. H1(Tn) ≃ Z⊕ ⋅ ⋅ ⋅ ⊕Z.

eorem 2.169 H1(Pn) has a torsion subgroup T (Pn) of order 2 and H1(Pn) ÒT (P(n) is a free abelian group of order n − 1.

P π1(Pn) = ⟨α1, . . . , αn ∣ α21 . . . α

2n⟩. H1(Pn) = π1(Pn) Ò [π1([Pn), π1(Pn)]

is the quotient of a free abelian group of rank n (basis {α1, . . . , αn}) by the subgroupgenerated by α2

1 . . . α2n[π1(Pn), π1(Pn)]. Since we ompute in an abelian group we can

use additive notation and write 2α1 + ⋅ ⋅ ⋅ + 2αn + [. . . ]. Change the basis {α1, . . . , αn}to the basis {α1, . . . , αn−1, . . . , α1 + ⋅ ⋅ ⋅ + αn}. is shows that H1(Pn) is isomorphicto the quotient of the free abelian group with basis {α1, . . . , αn−1, α1 + ⋅ ⋅ ⋅ + αn} by thesubgroup ⟨2(α1 + ⋅ ⋅ ⋅ + αn)⟩. SoH1(Pn) = Z⊕ ⋅ ⋅ ⋅ ⊕Z

n−1

⊕Z2.

eorem 2.171 e surfaces S2, T, T2, …, P, P2, …are topologically distinct (i.e. anytwo of these have different homotopy type).

[Proof: Follows immediatly from the previous theorems].

2.10.3. Classi cation of surfaces

eorem 2.172 (Classi cation of closed connected surfaces)Let S be a closed surface (i.e. compact without boundary). en S is either homeomorphicto S2, or to Tn for some n ∈ N, or to Pm for somem ∈ N.

Ideas of the proof:

51

Algebraic topology

Triangulation of a closed surface

De nition 2.173 A triangulation of a closed surface S is a nite family of closed sub-setes {T1, . . . , Tn} that cover S together with homeomorphisms φi ∶ T ′i → Ti whereeach T ′i is a triangle in R2. e subsets Ti are called triangles. e subsetes of Ti thatare images of vertices and edges are called vertices, respectively edges. We require thatif Ti, Tj are distinct, i, j ∈ {1, . . . , n}, then Ti ∩ Tj = ∅, or Ti ∩ Tj is vertex or Ti ∩ Tjis a common edge.

Example 2.174 Torus:

..1.

1

. 2.

2

. 3.

3

.

7

.

7

.

4

eorem 2.175 (T. Radó, 1925)Any closed surface admits a triangulation.

[Proof: Using the Jordan curve theorem].

Cutting and pasting polygonal regions inR2

eorem 2.176 If S is a closed surface, then S is homeomorphic to a space obtained froma polygonal region in R2 by gluing its edges together in pairs.

eorem 2.177 IfX is the quotient space obtained from a polygonal region inR2 by gluingits edges together, thenX is homeomorphic either to S2 or to Tn, for some n ∈ N or to Pm

for somem ∈ N.

2.11. Knot theory

First attempt to de ne the notion of a knot A knot is a simple closed curve in R3 (i.e.∃γ ∶ S1 → R3 continuous and injective, further, as S1 is compact and R3 Hausdorff, wehave f is closed)⇐⇒ γ ∶ S1 → γ(S1) is a homeomorphism⇐⇒ γ is an embedding ofS1 in R3.Problem: “wild” knots are allowed in this de nition.

52

2.11 Knot theory

De nition 2.178 A link L ofm components is a subset of R3 consisting ofm disjoint,piecewise linear, simple closed curves. A link of one component is a knot K .

Each of the curves is a union of nitely many line segments attached end to end.Example 2.179

1. e unknot U :

.

(U is the only knot bounding a disc embedded in R3).

2. e trivial knot ofm components:

.

3. Hopf Link

.

4. Trefoil knot

5. Borromean rings

De nition 2.180 Too links L1, L2 in R3 (respectively in S3 = R3 ∪ {∞}) are calledequivalent if there is an orientation preserving piecewise linear homeomorphism h ∶R3 → R3 (respectively h ∶ S3 → S3) with h(L1) = L2.

Question: Given two links L1, L2 how can we decide if they are equivalent.

• If L1, L2 are equivalent, then try to deform one into the other.

• If L1 and L2 are not equivalent, use invariants to distinguish them.

If L1, L2 are equivalent, ten R3 ∖L1 and R3 ∖L2 are homeomorphic. e homeomor-phism type of R3 ∖L is such an invariant.How strong an invariant is it?

eorem 2.181 (Gordon-Luecue, 1989)e knotsK1,K2 are equivalent⇐⇒ S3 ∖K1 and S3 ∖K2 are homeomorphic.

[Proof: Very difficult; omited].Note: e theorem says, that the homeomorphism type ofS3∖K is a complete invariant.

53

Algebraic topology

Remark 2.182 e theorem does not hold for links (with more than one component).

Determining the hoeomorphism type of S3∖L is not so simple. We turn to studyπ1(S3∖L).at is π1(S3 ∖ L) ≠ π1(S3 ∖ L2) Ô⇒ S3 ∖ L1 ≠ S3 ∖ L2 Ô⇒ L1 and L2 are notequivalent.

De nition 2.183 If L is a link in S3, then π1(S3 ∖ L) is called the group of the link L.It is sometimes denoted by π1(L).

Remark 2.184 e inclusion R3 ↪ S3 induces an inclusion i ∶ R3 ∖ L ↪ S3 ∖ L withwith i∗ ∶ π1(R3 ∖ L) → π1(S3 ∖ L) an isomorphism. So it is therefore equivalent toconsider L ⊂ R3 or L ⊂ S3.

Torus knots Consider T ⊂ R3 (rotat C1 = {(x, y, z) ∈ R3 ∣ y = 0, (x − 1)2 + z2 = 19}

about the z-axis). We want to study knots on T.T can be thought of as the quotient space of {(x, y) ∈ R2 ∣ 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} withopposite edges identi ed or as the quotient space obtained from R2 by identifying twopoints (x, y), (x′, y′) if x − x′ ∈ Z and y − y′ ∈ Z.Let q ∶ R2 → T be the identi cation map described and L = {(x, y) ∈ R2 ∣ ∣y =mnx, m,n ∈∈ Z,1 <m < n, (m,n) = 1}. Take T(m,n) ∶= q(L) ⊆ T.

is is a simple closed curve on T that wrapsm times around a line through the hole inthe torus and n-times around a circle inside T. T(m,n) is called an (m,n)-torus knot.T(m,n) can be given by the following parametrisation:

x = (2 + cos(nφ/m)) cosφ,y = (2 + cos(nφ/m)) sinφ,z = sin(nφ/m),

with φ ∈ [0,2mn]. is lies on the torus given by (r − 2)2 + z2 = 1 in cylindrical coor-dinates.

Remark 2.185

1. T(1,n) is equivalent to U the unknot.

2. T(m,n), T(n,m) are equivalent.

3. (m,n) ≠ 1: T(m,n) torus link.

We want to compute π1(S3 ∖K), whereK is the unknot U , orK is a torus knot.

Decomposition of S3 A ∶= {(x1, x2, x3, x4) ∈ S3 ∣ x21 + x22 ≤ x23 + x24} and B ∶={(x1, x2, x3, x4) ∈ S4 ∣ x21 + x22 ≥ x23 + x24}.enA,B are closed subsets of S3 and S3 = A∪B. FurtherA∩B = {(x1, x2, x3, x4) ∈S3 ∣ x21 + x22 = x23 + x24}.SoA∩B is the cartesian products of the circle {(x1, x2) ∈ R2 ∣ x21 + x22 = 1/2} with thecircle {(x3, x4) ∈ R2 ∣ x23 + x24 = 1/2}, that isA ∩B is a torus.

54

2.11 Knot theory

Now letD2 = {(x1, x2) ∈ R2 ∣ x21+x22 ≤ 1/2} and S1 = {(x3, x4) ∈ R3 ∣ x23+x24 = 1/2},and de ne f ∶ D2 × S1 → A by

(x1, x2, x3, x4)↦ (x1, x2,√2x3

√1 − (x21 + x22),

√2x4

√1 − (x21 + x22)).

f is obviously continuous and injective. One can easily verify, that it is also surjective.Further (as D2 × S1 compact,AHausdorff) f homeomorphism.Analogously, we can prove thatB is homeomorphic to D2 × S1.

S3 = A ∪B.

We can now compute:

1. π1(S3 ∖U):Take U to be the core circle of the A torus, i.e. U = {(x1, x2, x3, x4) ∈ A ∣ x1 =x2 = 0}. e boundary A ∩B of A is a deformation retract of A ∖ U . So B is adeformation retract of S3 ∖U = (A ∪B) ∖U .Ô⇒ π1(S3 ∖U) ≃ π1(B) ≃ Z.

2. π1(S3 ∖K) whereK is an (m,n) torus knot T(m,n):ConsiderK as a subset ofA ∩B in S3. S3 ∖K = (A ∖K) ∪ (B ∖K). (A ∖K),(B ∖ K), (A ∖ K) ∩ (B ∖ K) path connected. However (A ∖ K), (B ∖ K),(A ∖K) ∩ (B ∖K) are not open subsets so we need to modify them in order toapply the Seifert-van Kamepen theorem.

Choose ε > 0, such that there exists a tubular neighbourhood N ofK in S3 withradius ε. S3 ∖ N is a deformation retract of S3 ∖ K . en consider 1

2ε open

neighbourhoodsA′,B′ ofA andB respectively.

A′,B′ are homeomorphic to the product of S1 with an open disk. AndA′ ∩B′ ishomemorphic to to (A ∩B) × (− 1

2ε,− 1

2ε). Lastly A,B are deformation retracts

ofA′ andB′ respectively.

We can apply the Seifert-van Kampen theorem for S3 ∖N = (A′ ∖N)∪ (B′ ∖N)to compute π1(S3 ∖N) ≃ π1(S3 ∖K).

• A′ ∖N deformation retracts onto the core circle of the torus A i..e π1(A′ ∖N) ≃ Z,B′ ∖N deformation retracts onto the wire cricle ofB, i.e. π1(B′ ∖N) ≃ Z.

• (A′∖N)∩(B′∖N) = (A′∩B′)∖N has the homotopy type of (A∩B)∖K .So π1((A′ ∖N) ∩ (B′ ∖N)) ≃ π1((A ∩B) ∖K) ≃ Z.

• π1(A′ ∩B′ ∖N) = ⟨γ⟩ → π1(A′ ∖N) = ⟨α⟩, π2(A′ ∩B′ ∖N) = ⟨γ⟩ →π1(B′ ∖N) = ⟨β⟩.i1(γ) = am, i2(γ) = βn.

• Applying the Seifert-van Kamen theorem gives

Proposition 2.186 π1(S3 ∖T(m,n)) = ⟨α,β ∣ αmβ−n⟩.[Proof: Follows from the above analysis].Proposition 2.187 If the knotsT(m,n),T(m′,n′) are equivalent, thenm =m′ andn = n′.Furthermore T(m,n) and U are not equivalen.t

55

Algebraic topology

P (Schreier, 1923)Consider the element αm = bn =∶ z ∈ π1(S3 ∖ T(m,n) =∶ G, and the subgroup N of Ggenerated by z. z ∈ Z(G), i.e. zg = gz∀g ∈ G. SoN ≤ Z(G).en N ⊴ G and we can consider the quotient G Ò N . G Ò N has the followingpresentation: ⟨αN,βN ∣ (αN)m, (βN)n⟩. us G Ò N ≃ Zm ∗ Zn, which in turnimplies that Z(GÒ N) = {1GÒN}.Sine Z(G)Ò N ≤ Z(GÒ N), it follows thatN = Z(G), soGÒ Z(G) ≃ Zm ∗Zn.Now if T(m,n),T(m′,n′) are equivalent, π(T(m,n)) ≃ π1(T(m′,n′)).GÒ Z(G) ≃ G′ Ò Z(G′)Ô⇒ Zm ∗Zn ≃ Zm′ ∗Zn′ .Ô⇒m =m′ and n = n′.Lastly π1(U) ≃ Z, that is π1(U) Ò Z(π1(U)) ≃ Z Ò Z ≠ Zm ∗ Zn for anym,n ∈ Zwith (m,n) = 1

Wirtinger presentation Procedure for writing a presentation of the group of a knotKstarting from a diagram forK ⊂ R3.

..........

Label the arcs by a1, . . . , an, so that each ai is ocnnected to ai−1 and ai + 1 (mod n).Assume that the arcs are oriented compatibly with their labellling.Draw an arrow xi passing under each ai in a right-le direction. Each xi represents aloop in R3 ∖K as follows. Suppose that the black board is the xy-plane P . Consider asbasepoint ∗ the point (0,0,1). e loop consists of a segment from ∗ to the tail of xi,then the arrow, then a segment from the head of xi to ∗.At each crossing, there is a ceratain relation:

. .

In total there are n relations.

eorem 2.189 π1(R3 ∖K) = ⟨x1, . . . , xm ∣ r1, . . . , rn⟩.

Example 2.190 Trefoil

π1(trefoil) = ⟨x1, x2, x3 ∣ x3x1 = x2x3, x2x3 = x1x2, x1x2 = x3x1⟩ = ⟨x1, x2 ∣x1x2x1 = x2x1x2⟩.

56

2.11 Knot theory

Remark 2.191 In example 3 of group presentations we proved that ⟨x1, x2 ∣ x1x2x1 =x2x1x2⟩ = ⟨α,β ∣ α3 = β2⟩.P (of theorem 2.189)We will write f instead of [f].We will apply the Seifert-van Kampen theorem for R3 ∖K . Assume that K lies on thexy-plane except where it goes down by distance ε > 0, at each crossing. Set the basepointat (0,0,1).

..

z = −2ε/3. z = 0

Let A = {(x, y, z) ∈ R3 ∣ z > −2ε/3} −K . en A ≃homeo open 3-dim Ball ∖{n un-knotted arcs with endpoints on the boundery of the ball}.

So π1(A) = ⟨x1, . . . , xn⟩.For each crossin

..ai

.ai+1

.

ak

or.

let Yℓ be an open rectangular box around it at level −2ε < z < −ε/3, and set Bℓ =(Yℓ ∖K) ∪ { open neighbourhood of an arc from a point on ∂Yℓ to ∗ }.enBℓ ≃homeo open 3-dimball∖{ one unknotted arcwith endpoints on the boundary}.So π1(Bℓ) = ⟨yℓ⟩.Finally C = open neighbourhood of (R3 ∖A ∖ ∪nℓ=1Bℓ) ∪ ( open neighbourhood of anarc to ∗ ).

R3 ∖K = A ∪B1 ∪ ⋅ ⋅ ⋅ ∪Bℓ ∪C .

π1(A ∪B1) =?.A ∩B1 =?.i1 ∶ π1(A ∩B1)→ π1(A), c1 ↦ xkxix

−1k , b1 ↦ xi+1,

i2 ∶ π1(A ∩B1)→ π1(B1), c1 ↦ y1, c1 ↦ y1.So we have i1(c1)(i2(c1))−1 = xkx1x−1k y−11 , i1(b1)(i2(b1))1− = xi+1y−11 .Soπ1(A∪B1) = ⟨x1, . . . , xn, y1 ∣ xi+1y−11 , xkxix

−1k y−11 ⟩ = ⟨x1, . . . , xn ∣ xkxix−1k x−1i+1⟩.

Repeat this process for eachBi to get π1(A∪B1 ∪ ⋅ ⋅ ⋅ ∪Bn) = ⟨x1, . . . , xn ∣ r1, . . . , rn⟩.Since C and C ∩ (A ∪ B1 ∪ ⋅ ⋅ ⋅ ∪ Bn) are simply connected, applying the Seifert-vanKampen theorem we get that π1(K) = ⟨x1, . . . , xn ∣ r1, . . . , rn⟩.

Study π1(S3 ∖K) (≃ π1(R3 ∖K)) to show that any one of the ri can be ommited.

LetA′ = A ∪ {∞}. C ′ = C ∪Bn ∪ {∞}. en π1(A′) ≃ π1(A) and π1(A′ ∪B1 ∪ ⋅ ⋅ ⋅ ∪Bn−1) = ⟨x1, . . . , xn ∣ r1, . . . , rn−1⟩.S3 ∖K = A′ ∪B1 ∪ ⋅ ⋅ ⋅ ∪Bn−1 ∪C ′.Nowwe can compute that π1(C ′) = ⟨yn⟩ and also that π1(C ′∩(A′∪B1∪⋅ ⋅ ⋅∪Bn−1)) =⟨bn⟩.

57

Algebraic topology

π1(C ′ ∩ (A′ ∪B1 ∪ ⋅ ⋅ ⋅ ∪Bn−1))→ π1(C ′), bn ↦ yn.π1(C ′ ∩ (A′ ∪B1 ∪ ⋅ ⋅ ⋅ ∪Bn−1))→ π1(A′ ∪B1 ∪ ⋅ ⋅ ⋅ ∪Bn−1), bn ↦ xi+1.

Applying the Seifert-vanKampen theoremweobtain that ⟨x1, . . . , xn ∣ r1, . . . , rn−1⟩)π1(S3∖K) = π1(R3 ∖K).Remark 2.193 e above reasoning applies also to compute π1(S3∖L)whereL is a linkand π1(S3 ∖L) = ⟨x1, . . . , xn ∣ r1, . . . , rm⟩.Example 2.194

1.

2.

eorem 2.195 e group of a knot is not a complete knot invariant.

2.12. Classi cation of covering spaces

p ∶ X →X covering map with p(x0) = x0, p∗ ∶ π1(X, x0)→ π1(X,x0).We will study H ∶= p∗(π1(X, x0)) ≤ π1(X,x0) to deduce several statements aboutcovering maps.Lemma 2.196 Let p ∶ X →X be a covering map with p(x0) = x0. en

1. p∗ ∶ π1(X, x0)→ π1(X,x0) is a monomorphism.

2. e liing correspondance ϕ ∶ π1(X,x0) → p−1(x0) where [f] ↦ f(1) inducesan injective map Φ ∶ π1(X,x0) Ò H → p−1(x0), where by π1(X,x0) Ò H wedenote all the right cosets of H in π1(X,x0). If X is path connected, then Φ isbijective.

3. If f ∶ I → X is a loop based at x0, then [f] ∈ H iff f lis to a loop in X based atx0.

P

1. Immediate, it follows from homotopy liing property.

2. If f ∶ I → X , g ∶ I → X are loops based at x0, then let f ∶ I → X , g ∶ I → Xbe their lis with f(0) = g(0) = x0. If [f] ∈ H[g], then [f] = [g], whereh ∈H ∶= p∗(π1(X, x0)) i.e. h = p ○ h for some loop h ∶ I → X based at x0. Nowh ⋅ g is de ned and it is a li of h ⋅ g. Since [f] = [h ⋅ g], f(1) = h ⋅ g(1).Ô⇒ f(1) = g(1).Ô⇒ ϕ([f]) = ϕ([g]). at shows Φ is well-de ned.Check that Φ is injective: If ϕ([f] = ϕ([g]), then f(1) = g(1) and we can de nef ⋅g loop based at x0. erefore [(f ⋅ g) ⋅g] = [f]. If F is a path homotopy between(f ⋅ g) and f , then F ∶= p ○ F is a path homotopy between p((f ⋅ g) ⋅ g) and p(f).at is [(p(f ⋅ g)) ⋅ g] = [f].Ô⇒ [f] ∈H[g].Ô⇒ Φ is injective.If X is path connected, then ϕ is surjective. So Φ is also surjective.

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2.12 Classi cation of covering spaces

3. Φ is injective, hence ϕ([f]) = ϕ([g])⇐⇒ [f] ∈ H[g]. For g the constant loopbased at x0 this gives [f] ∈H ⇐⇒ ϕ([f]) = x0⇐⇒ f(1) = x0.

Lemma 2.198 LetX be path connected and locally path connected. If Xa is path com-ponent of X , then p∣Xa

∶ Xa →X is a covering map.

P X is locally homeomorphic to X , so X is also locally path connected. Xa is apath component of X , so Xa is open. Covering maps are open maps, so p(Xa) is open.Claim: p(Xa) is closed.Proof: Let x ∈ p(Xa) and U be a path connected open neighbourhood of x such thatp−1(U) = ∪αVα, where Vα pairwise disjoint open, p∣Vα

∶ Vα → U homemorphism ∀α.SinceU contains a point of p(Xa), Vα ∩ Xa ≠ ∅ for some α. Now Vα is pat hconnected,hence Vα ⊆ Xα. en p(Vα) = U ⊆ p(Xa), i.e. x ∈ p(Xa). So p(Xa) = p(Xa).p(Xa) ⊆X is both open and closedÔ⇒ p(Xa) =XÔ⇒ p∣Xa

is surjective.Now let x ∈X and chose an open neighbourhoodU of x as before. If Vα ∩ Xa ≠ ∅, thenVα ⊆ Xa. erefore (p∣Xa

)−1(U) is the union of those Vα’s that intersect Xα. Each ofthese is ope with p∣Vα

∶ Vα → U is a homeomorphism.p ∶X →X covering map. We restrict to spacesX thate are locally path connected. enX is also locally path connected. We can also assume that X is path connected. Lastlywe assume, that X is also path connected. So, we can determine all covering spaces ofa locally path connected space X by determining all path connected covering spaces ofpath components ofX .From onw on, whenever we write p ∶ X → X is a covering map, we imply thatX, X arelocally path connected, and path connected.Lemma 2.200 (liing lemma) Let p ∶ X → X covering map with p(x0) = x0, andf ∶ Y → X continuos. With Y path connected and locally path connected. f(y0) = x0.en f can be lied to a continous map f ∶ Y → X with f(y0) = x0 iff f∗(π1(Y, y0)) ⊆p∗(π1(X, x0)). Also if such a li exists, it is unique.P

“⇒” If f exists, then f∗(π1(Y, y0)) = (p ○ f)∗(π1(Y, y0)) = p∗(f∗(π1(Y, y0))) ⊆p∗(π1(X, x0)).

“⇐” Given y1 ∈ Y , choose a pathα ∶ I → Y withα(0) = y0, α(1) = y1. Li f ○α ∶ I →X to f ○ α ∶ I → X with f ○ α(0) = x0. De ne f ∶ Y → X by f(y1) = f ○ α(1).

– f is well-de ned:Let β ∶ I → Y be a path with β(0) = y0 and β(1) = y1. Li f ○ β ∶ I → X

to a path f ○ β ∶ I → X with f ○ β(0) = f ○ α(1). en f ○ α ⋅ f ○ β is ali of the loop f ⋅(αβ). By assumption f∗(π1(Y, y0)) ⊆ p∗(π1(X, x0)). So[f ○ (αβ)] ∈ p∗(π1(X, x0)). Lemma 2.196 implies that its li f ○ α ⋅ f ○ βis a loop based at x0. So f ○ α(1) = f ○ β(1).

– f is constinuous:Let y1 ∈ Y and N an open neighbourhood of f(y1). We have to show thatthere is an open neighbourhoodW of y1 with f(W ) ⊆ N . Start by choos-ing a path connected open neighbourhood U of f(y1) such that p−1(U) =∪αVα, Vα disjoint open sets. p∣Vα

∶ Vα → U homeomorhpism ∀α. f is

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Algebraic topology

continuous at y1 and Y is locally path connected, so there is a path con-nected open neighbourhood W of y1 with f(W ) ⊆ U . We will show thatf(W ) ⊆ V0, where V0 is the subset containing f(y0). Let y ∈W , and choosea path b ∶ I →W with b(0) = y1 and b(1) = y. since f is well-de ned f(y)can be obtained by taking the path αβ from y0 to y, liing f ○(aβ) to a pathf ○ (aβ) ∶ I → X with f ○ (aβ)(0) = X0 and setting f(y) = f ○ (αβ)(1).

Now f ○ α is a li of f ○ β with f ○ α(0) = x0. Since f ○ β(I) ⊆ U , the path(p∣V0

)−1 ○ f ○ β is a li of f ○ β with ((p∣V0)−1 ○ f ○ β)(0) = f(y1). en

(f ○ α)((p∣V0)−1○f○β) is a li of f○(αβ)with (f○α)((p∣V0

)−1○f○β)(0) =x0. But (f ○ α)((p∣V0

)−1 ○ f ○ β)(1) = ((p∣V0)−1 ○ f ○ β)(1) ∈ V0, hence

f(W ) ⊆ V0.

Let y1 ∈ Y and α ∶ U → Y , with α(0) = y0, and α(1) = y1. Consider the path f ○ α ∶I → X and li it to a path f ○ α ∶ I → X with f ○ α(v) = x0. en y1 = f(α(1) andf(α(1)) = f ○ α(1) since f ○ α is a li of f ○ α with f(α(0)) = f(y0) = x0. So we seethat if such an f exists, it is unique.Example 2.202 Space that is path connected, but not locally path connected.

..0.

1.

12

.13

.14

And consider a point (0, y), where y > 0.Example 2.203 Space that is locally path connected, but not path connected.

.

Equivalent covering spaces

De nition 2.204 Let p ∶ X → X , p′ ∶ X ′ → X be covering maps, p, p′ are calledequivalent if there exists a homeomorphism h ∶ X → X ′ with p = p′ ○ h. h is calledequivalence between the covering spaces.

eorem 2.205 Let p ∶ X → X , p′ ∶ X ′ → X be covering maps with p(x0) = p′(x′0) =x0.ere is an equivalence h ∶ X → X ′ with h(x0) = x′0 ⇐⇒ p∗(π1(X, x0) =p′∗(π1(X ′, x′0)). If such an h exists, then it is unique.

P

“⇒” h homeomorphismÔ⇒ h∗(π1(X, x0)) = π1(X ′, x′0).Ô⇒ p′∗(h∗(π1(X, x0))) = p′∗(π1(X ′, x′0)).Ô⇒ p∗(π1(X, x′0)) = p′∗(π1(X, x′0)).

60


⇐” – p′ covering map, p continous, X path connected and locally path connectedp∗(π1(X, x0)) = p′∗(π1(X ′, x′0)).Ô⇒ ∃!h ∶ X → X ′ with p′ ○ h = p and h(x0) = x′0.

– p covering map, p′ continuos, X ′ path connected and locally path conectedp′∗(π1(X ′, x′0)) = p∗(π1(X, x0)).Ô⇒ ∃!k ∶ X ′ → X continous with p ○ k = p′ and k(x′0) = x0.

– k ○ h ∶ X → X is a li of p since p ○ k ○ h = p′ ○ h = p iwth k ○ h(x0) = x0.1X ∶ X → X is another such liing, so by uniqueness of lis k ○ h = 1X .

– Analogously, we can prove that h ○ k = 1X′ .

What about equivalences h ∶ X → X ′ that do not necessarily satisfy h(x0) = x′0?Lemma 2.207 Let p ∶ X → X be a covering map and x0, x1 ∈ p−1(x0) and Hi ∶=p∗(π1(X, xi)), i ∈ {0,1}.

1. If γ ∶ I → X is a path with γ(0) = x0 and γ(1) = x1, and α ∶= p ○ γ, then[α]H1[α]−1 =H0.

2. Conversly given x′0 and a subgroupH ≤ π1(X,x0) conjugate toH0, there existsa basepoint x1 ∈ p−1(x0) withH1 =H .

P

1. “⊆” If [h] ∈ H1, then [h] = p∗([h]) for some h ∶ I → X loop based at x1. Letk = (γ ⋅ h) ⋅ γ. en p∗([k]) = p∗([γ ⋅ h ⋅ γ]) = [α ⋅ h ⋅ α] = [α][h][α]−1.at is [α][h][α]−1 = p∗([k]) ∈ p∗(π1(X, x0)) =H0.

“⊇” Apply the above for γ and α = p ○ γ. We that that [α]H0[α]−1 ⊆ H1. Ô⇒H0 ⊆ [a]H1[a]−1.

2. Let x0 ∈ X andH ≤ π1(X,x0) conjugate toH0. enH0 = [α]H[α]−1 for someα ∶ I → X based at x0. et let γ be the li of α to X with γ(0) = x0. Letγ(1) = x1. en by 1. Ô⇒H0 = [α]H1[α]−1.

eorem 2.209 Let p ∶ X → X , p′ ∶ X ′ → X be covering maps with p(x0) = p′(x′0) =x0. p, p′ are quivalent⇐⇒H0 ∶= p∗(π1(X, x0)) andH ′0 ∶= p′∗(π1(X ′, x′0)) are con-jugate.

P

“⇒” If h ∶ X → X ′ is an equivalence, let x′1 = h(x0) and H ′1 = p∗(π1(X ′, x′1)). Bythe previous theroem H0 = H ′1. And by the lemma H ′0,H ′1 are conjugate. Ô⇒H0,H

′0 are conjugate.

“⇐” If H0,H′0 are conjugtae, then according to the lemma there exists x1′ ∈ X ′ with

H ′1 = H0. en by previous theorem there is an equivalence h ∶ X → X ′ withh(x0) = x10.

Example 2.211 X = S1, x0 ∈ S1, π1(S1, x0) ≃ Z.Subgroups of Z: nZ, n ∈ N.Covering spaces of S1: p ∶ R→ S1 where t↦ (cos 2πt, sin 2πt).p∗(π1(R)) is trivial. at is this covering space of S1 corresponds to the trivial subgroup

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Algebraic topology

of π1(S1).An other covering map we saw was pn ∶ S1 → S1, z ↦ zn. en (pn)∗(π1(S1)) = nZ.Question: Is there an a covering map p ∶ X → S1, that is not equivalent to any of theabove.No, as p∗(π1(X)) ≤ π1(S1).

e universal covering space

De nition 2.212 Let p ∶ X →X be a covering map. If X is simply connected, then Xis called a universal covering space ofX .

Remark 2.213 p∗(π1(X)) trivialÔ⇒ any two universal covering spaces ofX are equiv-alent. We can therefore speak of the universal covering space ofX .Lemma 2.214 Let p, q, r be continuous maps with p = r ○ q. If p, r are covering maps,then q is also a covering map.P Let x0 ∈X , z0 ∈ p(x0), y0 = q(x0).Claim 1: q is surjective.Let y ∈ Y . Choose a path α ∶ I → Y with α(0) = y0 and α(1) = y. en α = r ○ α is apath in Z iwth α(0) = r(α(0)) = r(y0) = r(q(x0)) = p(x0) = z0. Let α be the li ofα to a path in X with α(0) = x0. en q ○ α is a li of α to Y with q ○ α(0) = y0. Byuniqueness of path liings q ○ α = α. So q ○ α(1) = α(1) = yÔ⇒ y = q(α(1)). Ô⇒ qis surjective.Claim 2: Given y ∈ Y , there exists n open neighbourhood V of y with q−1(V ) = ∪γWγ ,whereWγ pairwise disjoint open sets with q∣Wγ

∶Wγ → V homeomorphism ∀γ. So, lety ∈ Y . Set z ∶= r(y). Since p, r are covering maps we can nd an open path-connectedneighbourhood of z such that p−1(U) = ∪αUα, Uα disjoint open with p∣Uα

∶ Uα → Uhomeomorphism ∀α. r−1(U) = ∪βVβ , Vβ open, r∣Vβ

∶ Vβ → U homeomorphism ∀β.Call V the member of the family Vβ that contains the point y. q(Uα) ⊆ r−1(U) ∀αand Uα is connected, hence q(Uα) ⊆ Vβ for some β. Also q−1(V ) ⊆ ∪αUα. Indeed letx ∈ q−1(V ) then q(x) ∈ V Ô⇒ r(q(x)) ∈ r(V ) = U Ô⇒ p(x) ∈ U Ô⇒ x ∈ p−1(U)Ô⇒ x ∈ ∪αUα.erefore q−1(V ) is the union of those Uα’s with q(Uα) ⊆ V .In fact, q∣Uα

∶ Uα → V is a homeomorphism for each such α.Since p∣Uα

, r∣V are homeomorphisms and q∣Uα= (r∣V )−1 ⋅ p∣Uα

.Call this family of Uα’sWγ .

eorem 2.216 Let p ∶ X →X be a coveringmap, and X simply connected. If r ∶ Y →Xis a covering map. en there exists a covering map q ∶ X → Y with r ○ q = p.

P Let x0 ∈ X . Choose x0 ∈ X and g0 ∈ Y with p(x0) = x0, r(y0) = x0.p∗(π1(X, x0)) ⊆ r∗(π1(Y, y0)). So we can apply the liing lemma for the coveringmap r ∶ Y →X . at is ∃!q ∶ X → Y continuous with q(x0) = y0 and r ○ q = p. p, r arecovering mapsÔ⇒ q is covering map.Remark 2.218 is theorem justi es the use of the term the universal covering space.Question: Does every spaceX have a universal covering space?Answer: No.Lemma 2.219 Let p ∶ X → X be a covering map with p(x0)x0 and X is simply con-nected. en x0 has an open neighbourhood U such that the map i∗ induced by theinclusion i ∶ (U,x0)↪ (X,x0) is trivial.

62


P Let U be an open neighbourhood of x0 with p−1(U) = ∪αUα, Uα pairwise dis-joint open sets, p∣Uα

∶ Uα → U homeomorphism∀α. LetU0 be theUα, that contains x0.Consider f ∶ I → U a loop based at x0. And let f ∶= (p∣U0

)−1(f) be the li of f in X .f is a loop based at x0. π1(X, x0) trivialÔ⇒ ∃ a path homotopy F between f and theconstant loop at x0. en F = p ○ F is a path homotopy inX between p ○ f = f and theconstant path at p(x0) = x0. at is, i∗ is trivial.Example 2.221 X = ∪n∈NCn (Hawaiian earrings). Let rn ∶ X → Cn mapping every Ci

with i ≠ n to (0,0) with rnCn= 1Cn .

Let U be an open neighbourhood of x0 ∈ X . Choose n large enough such Cn ⊆ U .j ∶ Cn ↪X , k ∶ Cn ↪ U , i ∶ U ↪X .rn ○ j = 1Cn Ô⇒ (rn)∗ ○ j∗ = 1π(Cn,x0)Ô⇒ j∗ is injective.Now j∗ = i∗ ○ k∗, hence i∗ ○ k∗ ∶ π1(Cn, x0) → π1(X,x0) is injective. I.e. i∗ is nottrivial.LemmaÔ⇒e shrinking of circles has no universal covering space.

Existence of covering spaces

De nition 2.222 X is called semilocally simply connected if∀x ∈X ∃ open neighbour-hood U of x with i∗ ∶ π1(U,x) → π1(X,x), induced by the inclusion i ∶ (U,x) ↪(X,x), is trivial.

Remark 2.223

1. If U is an open neighbourhood of x with i∗ ∶ π1(U,x) → π1(X,x) trivial andV is an open neighbourhood of x with V ⊆ U , then obviously j∗ ∶ π1(V,x) →π1(X,x) is trivial.

2. IfX is locally simply connected, i.e. ∀x ∈ X and ∀ open neighbourhoods U of xthere exists a simply connected open neighbourhood V of x with V ⊆ U , thenXis semilocally simply connected.

Semilocal simply connectedness is in fact a necessary and sufficient condition for thecorresopondance

“(Covering maps ofX)Ò ∼ 1∶1→ conjugacy classes of subgroups of π1(X,x0)”

to be surjective.

eorem 2.224 Let X be path connected, locally path and semilocally simply connected.Let x0 ∈X andH ≤ π1(X,x0). en ∃p ∶ X →X covering map with p∗(π1(X.x0)) =H .

Corollary 2.225 X has a universal covering space iffX is parth connected, locally pathconnected and semilocally simply connected.P (of the theorem)We do this in seven steps:Step 1 Construction of X :De ne P ∶= {γ ∶ I → Y path with γ(0) = x0}. For α,β ∈ P de ne a ∼ β if α(1) = β(1)and αβ ∈ H . Now de ne X ∶= P Ò ∼ = {γ# ∣ γ ∶ I → X path with γ(0) = x0}, andde ne p ∶ X →X by p(γ#) = γ(1).Note:

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Algebraic topology

• p is surjective, sinceX is path connected.

We will de ne a topology on X so that p ∶ X →X is a covering map.Remark 2.227 If [α] = [β], then α# = β#. Furhter, if α# = β# and δ ∶ I → X withδ(0) = α(1), then (αδ)# = (βδ)#Step 2 Let α ∈ P and U a path connected neighbourhood of α(1). De ne B(U,α) ∶={(αδ)# ∣ δ ∶ I → U path with δ(0) = α(1)}. en α# ∈ B(U,α).Claim: e setsB(U,α) form a basis for a topology on X .Proof: We rst prove that if β# ∈ B(U,α), thenα# ∈ B(U,β) andB(U,α) = B(U,β).For this consider β# ∈ B(U,α). en β# = (αδ)#, for some δ ∶ I → U with δ(0) =α(1) and (βδ)# = (αδδ)# = α#. So α# = (βδ)# ∈ B(U,β). Not let (βγ)# ∈B(U,β). en (βγ)# = ((αδ)γ)# = (α(δγ))# ∈ B(U,α). at is B(U,β) ⊆B(U,α), and by analogous argumentB(U,α) = B(U,β).Now we can prove that the setsB(U,α) form a basis for a topology on X . Indeed:

• Letα# ∈ X . en choose a path connected open neighbourhoodU ofα(1). enα# ∈ B(U,α).

• Letα# ∈ B(U,α1)∩B(U,α2). Choose a path connected open neighbourhood Vof α(1)with V ⊆ U1∩U2. enB(V,α) ⊆ B(U1, α)∩B(U2, α) = B(U1, α1)∩B(U2, α2).

erefore the setsB(U,α) form a basis for a topology on X .Step 3 p is open and continuous:p is open: We will prove that p(B(U,α)) = U , which implies p is open.

“⊇” Let x ∈ U and choose a path δ ∶ I → U with δ(0) = x, δ(1) = x. en (αδ)# ∈B(U,α) and p((αδ)#) = (αδ)(1) = δ(1) = x. I.e. x ∈ p(B(U,α)), i.e. U ⊆p(B(U,α)).

“⊆” p(B(U,α)) ⊆ U (follows form the de nition of p andB(U,α).

p is continuos: Let α# ∈ X and W an open neighbourhood of p(α#). Choose a pathconnected open neighbourhood U of p(α#) = α(1) with U ⊆ W . en B(U,α) is anopen neighbourhood of α# in X with p(B(U,α)) = U ⊆W .Step 4 ∀x ∈ X ∃ an open neighbourhood U of x with p−1(U) = ∪αVα where Vα pair-wise disjoint open sets, and p∣Vα

∶ Vα → U homemomorphism.Letx ∈X . Choose a path connected openneighbourhoodU ofX such that i∗ ∶ π1(U,x)→π1(X,x) is trivial.Claim 1: p−1(U) = ∪α∶I→X,α(0)=x0,α(1)=xB(U,α).Proof:

“⊇” p(B(U,α)) = U Ô⇒B(U,α) ⊆ p−1(U)Ô⇒ ∪α∶I→X,...B(U,α) ⊆ p−1(U).

“⊆” Let β# ∈ p−1(U). en p(β#) ∈ U , i.e. β(1) ∈ U . Choose a path δ ∈ U fromx to β(1) and let α = βδ. en [β] = [αδ]Ô⇒ β# = (αδ)# ∈ B(U,α). I.e.p−1(U) ⊆ ∪α∶I→X,...B(U,α).

Claim 2: Distinct setsB(U,α) are disjoint.Proof: Suppose β# ∈ B(U,α) ∩B(U,α2). enB(U,α1) = B(U,β) = B(U,α2).Claim 3: p∣B(U,α) ∶ B(U,α)→ U is bijective.Proof: We have shown that p(B(U,α)) = U , i.e. p is surjective. To check injectiv-ity suppose p((αδ1)#) = p((αδ2)#). δi ∶ I → U , δi(0) = α(1), i ∈ {1,2}. en

64


(αδ1)(1) = (αδ2)(1)Ô⇒ δ1(1) = δ2(1). So we can de ne δ1δ2 ∶ I → U (loop basedat δ1(0) = δ2(0) = δ2(1) = x). Since i∗ ∶ π1(U,x) → π1(X,x) is trivial, there is a pathhomotopy inX between δ1δ2 and the constant loop at x. [αδ1] = [αδ2]. Now p∣B(U,α)is bijective, continuous and open. Ô⇒ p∣B(U,α) homeomorphism.Step 5 Liing a path inX to a path in X .Let e0 be the equivalence class in X of the constant path at x0. en p(e0) = x0. Letα ∶ I → X path with α(0) = x0. We want to compute its li to a path α ∶ I → Xwith α(0) = e0 and show that α(1) = α#. Given c ∈ [0,1], let αc ∶ I → X thepath de ned by αc(t) ∶= α(ct), t ∈ I . De ne α ∶ I → X by α(c) = (αc)#. enp(α(c)) = (p(αc)#) = αc(1) = α(c). Ô⇒ p ○ α = α and α(0) = (α0)# = e0,α(1) = (α1)# = α#.Claim: α is continuous.Proof: omitted.Step 6 p ∶ X →X is convering map.p is surjective (step 1), p stais es the covering condition (step 4). X is path and locallypath connected by assumption, X is path connectec (step 5). (X is locally path connectedasX is).Step 7H = p∗(π1(X, e0)):Let α ∶ I → X be a loop based at x0 and α ∶ I → X its li with α(0) = e0. en[α] ∈ p∗(π1(X, e0))⇐⇒ α(1) = α(0) = e0 ⇐⇒ α# = e0. ⇐⇒ α ∼ constant path atx0⇐⇒ [αcx0] ∈H ⇐⇒ [α] ∈H .

65

Index

Symbols

ε-ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

A

adjunction space . . . . . . . . . . . . . . . . . . . . . 33attaching map . . . . . . . . . . . . . . . . . . . . . . . 33

B

basepoint . . . . . . . . . . . . . . . . . . . . . . . . . . . 18basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

C

category . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21closed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5compact . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12comparable . . . . . . . . . . . . . . . . . . . . . . . . . . 5connected . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9connected component . . . . . . . . . . . . . . . . 12connected sum . . . . . . . . . . . . . . . . . . . . . . 34continuous . . . . . . . . . . . . . . . . . . . . . . . . . . . 7countable topology . . . . . . . . . . . . . . . . . . . . 4coveriant functor . . . . . . . . . . . . . . . . . . . . 21covering map . . . . . . . . . . . . . . . . . . . . . . . . 21covering space . . . . . . . . . . . . . . . . . . . . . . . 21

D

deformation retract . . . . . . . . . . . . . . . . . . 29deformation retraction . . . . . . . . . . . . . . . 29direct sum . . . . . . . . . . . . . . . . . . . . . . . . . . 35

external . . . . . . . . . . . . . . . . . . . . . . . . 36discrete topology . . . . . . . . . . . . . . . . . . . . . 4disjoint union . . . . . . . . . . . . . . . . . . . . . . . 33disjoint union topology . . . . . . . . . . . . . . . 33

E

edges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

embedding . . . . . . . . . . . . . . . . . . . . . . . . . . 22equivalence . . . . . . . . . . . . . . . . . . . . . . . . . 60equivalent . . . . . . . . . . . . . . . . . . . . . . . 53, 60external direct sum . . . . . . . . . . . . . . . . . . 36external free product . . . . . . . . . . . . . . . . . 39

F

nal point . . . . . . . . . . . . . . . . . . . . . . . . . . . 16ner . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5nite topology . . . . . . . . . . . . . . . . . . . . . . . . 4nitely generated . . . . . . . . . . . . . . . . . . . . 43nitely presented . . . . . . . . . . . . . . . . . . . . 43rst homology group . . . . . . . . . . . . . . . . . 50

free abelian group. . . . . . . . . . . . . . . . . . . . 37free group. . . . . . . . . . . . . . . . . . . . . . . . . . 41 ffree product . . . . . . . . . . . . . . . . . . . . . . . . . 38fundamental group . . . . . . . . . . . . . . . . . . 19

G

generate . . . . . . . . . . . . . . . . . . . . . . . . . 35, 38group of the link . . . . . . . . . . . . . . . . . . . . . 54

H

Hausdorff . . . . . . . . . . . . . . . . . . . . . . . . . . . 14homeomorphism . . . . . . . . . . . . . . . . . . . . . 8homeotopic . . . . . . . . . . . . . . . . . . . . . . . . . 16homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . 16homotopy equivalence . . . . . . . . . . . . . . . 30homotopy equivalent . . . . . . . . . . . . . . . . . 30homotopy inverse . . . . . . . . . . . . . . . . . . . . 30homotopy type . . . . . . . . . . . . . . . . . . . . . . 30

I

induced homomorphism . . . . . . . . . . . . . 20initial point . . . . . . . . . . . . . . . . . . . . . . . . . 16interior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

K

knot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

66

INDEX

L

least upper bound property . . . . . . . . . . . 13length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38liing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24liing correspondence . . . . . . . . . . . . . . . . 26link . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53locally connected . . . . . . . . . . . . . . . . . . . . 15locally path connected . . . . . . . . . . . . . . . . 15loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

M

metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6metric topology . . . . . . . . . . . . . . . . . . . . . . 6

N

nullhomotopic . . . . . . . . . . . . . . . . . . . . . . 16

O

open . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4open neighbourhood . . . . . . . . . . . . . . . . . 14order relation . . . . . . . . . . . . . . . . . . . . . . . 12order topology . . . . . . . . . . . . . . . . . . . . . . 13

P

path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11, 16path components . . . . . . . . . . . . . . . . . . . . 12path connected . . . . . . . . . . . . . . . . . . . . . . 11path homotopic . . . . . . . . . . . . . . . . . . . . . 16presentation . . . . . . . . . . . . . . . . . . . . . . . . . 43product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17product topology . . . . . . . . . . . . . . . . . . . . . 6

Q

quotient map . . . . . . . . . . . . . . . . . . . . . . . . . 8quotient topology . . . . . . . . . . . . . . . . . . . . . 9

R

rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38reduced word . . . . . . . . . . . . . . . . . . . . . . . 38relation subgroup . . . . . . . . . . . . . . . . . . . . 43reparametrisation . . . . . . . . . . . . . . . . . . . . 18retract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27retraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

S

semilocally simply connected . . . . . . . . . 63separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9simply connected . . . . . . . . . . . . . . . . . . . . 20standard topology . . . . . . . . . . . . . . . . . . . . 6stereographic projection . . . . . . . . . . . . . . 34strictly ner . . . . . . . . . . . . . . . . . . . . . . . . . . 5subset

closed . . . . . . . . . . . . . . . . . . . . . . . . . . . 4open . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

subspace topology . . . . . . . . . . . . . . . . . . . . 7sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

T

topological property . . . . . . . . . . . . . . . . . . 8topological space . . . . . . . . . . . . . . . . . . . . . 4topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

countable . . . . . . . . . . . . . . . . . . . . . . . . 4discrete . . . . . . . . . . . . . . . . . . . . . . . . . 4nite . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

metric . . . . . . . . . . . . . . . . . . . . . . . . . . 6product . . . . . . . . . . . . . . . . . . . . . . . . . 6subspace . . . . . . . . . . . . . . . . . . . . . . . . 7trivial . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52triangulation . . . . . . . . . . . . . . . . . . . . . . . . 52trivial topology . . . . . . . . . . . . . . . . . . . . . . . 4

U

universal covering space . . . . . . . . . . . . . . 62

V

vertices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

W

wedge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23wedge sum . . . . . . . . . . . . . . . . . . . . . . . . . . 33word . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

67

Documents

Introduction to Algebraic Topologyuser.math.uzh.ch/berner/files/algtop.pdf · Topology Claim: XisconnectediﬀtheonlysubsetsofXthatarebothopenandclosedare∅and X. P