Algebraic Topology

based on notes by Andrew Ranicki

2019–20

Contents

0.1 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Lecture 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1 Recap on connectedness (non-examinable) 51.1 Connected spaces . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Paths and path components π0(X) . . . . . . . . . . . . . . . 6

Lecture 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Homotopy theory 112.1 Nonexaminable appendix on function spaces . . . . . . . . . . 12

Lecture 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Lecture 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Some quotient spaces of I2 and ∂I2 213.1 The figure eight . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2 The cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.3 The Mobius band . . . . . . . . . . . . . . . . . . . . . . . . . 243.4 The torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.5 The Klein bottle . . . . . . . . . . . . . . . . . . . . . . . . . 25

Lecture 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.6 The projective plane . . . . . . . . . . . . . . . . . . . . . . . 27

4 Cutting and pasting paths 27Lecture 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

5 The fundamental group π1(X) 32Lecture 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Lecture 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

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5.1 Nonexaminable appendix on the higher homotopy groups . . . 43Lecture 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6 Compact surfaces and their classification 45Lecture 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

7 The fundamental group of Sn is π1(Sn) = 1 for n > 2 50

Lecture 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

8 The fundamental group of the circle is π1(S1) = Z 54

Lecture 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58Lecture 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Lecture 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

9 The n-dimensional projective space RPn 68Lecture 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72Lecture 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

10 Fixed points and non-retraction 77Lecture 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

11 Covering spaces 82Lecture 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85Lecture 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89Lecture 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93Lecture 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

12 Fundamental groups of surfaces and van Kampen’s theorem— non-examinable 9812.1 Fundamental groups via van Kampen’s theorem . . . . . . . . 101

Lecture 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

13 A lightning introduction to homology (non-examinable) 103

Introduction

The aim of this course is to develop the basic notions of algebraic topology,such as homotopy and the fundamental group. We shall relate these notionsto other areas of mathematics — geometry, analysis and algebra, and seeapplications.

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DRPS entry

This course will introduce students to essential notions in algebraic topol-ogy, such as compact surfaces, homotopies, fundamental groups and coveringspaces.

Intended learning outcomes:

1. Construct homotopies and prove homotopy equivalence for simple ex-amples.

2. Calculate fundamental groups of simple topological spaces, using gen-erators and relations or covering spaces as necessary.

3. Calculate simple homotopy invariants, such as degrees and windingnumbers.

4. State and prove standard results about homotopy, and decide whethera simple unseen statement about them is true, providing a proof orcounterexample as appropriate.

5. Provide an elementary example illustrating specified behaviour in rela-tion to a given combination of basic definitions and key theorems acrossthe course.

Intuitively, two spaces are homotopy equivalent if they can be made home-omorphic by shrinking and expanding. For example, a solid ball is homotopyequivalent to a point; a Mobius strip is homotopy equivalent to a standardstrip, since both are homotopy equivalent to a circle.

Recommended books

There are many good books on topology in the Library covering the materialin this course. Amongst them are:

J.R. Munkres, Topology, a first course (Prentice-Hall) 1975, esp. Ch. 8,which deals with the fundamental group. [In the second edition, 2000, thisis Ch. 9.]

C. Kosniowski, A First Course in Algebraic Topology, (CUP) 1980, esp.Chs. 12–17.

K. Janich, Topology (Springer UTM) 1984, esp., Chs. 8 and 10. (Verygood pictures but few exercises).

Beware that other books titled “Algebraic topology” will often focus onmore advanced material than we have time to cover. For instance, we don’t

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cover anything beyond chapter 1 of Hatcher’s Algebraic Topology (and notall of that), which is an affordable text recommended for anyone wishing toread further material.

There are also many relevant online resources, such as Harpreet Bedi’sexcellent youtube lectures on Elementary Homotopy, which cover much ofthe course.

Lecture Notes

Lecture notes are posted on the course website, and lectures will be given onthe assumption that you have tried to look at them in advance. It helps ifyou can come armed with questions on the notes. I might make some changesto the notes as the course progresses, but if so, this should be at least a weekbefore the affected lecture. Also feel free to email me any questions you haveon the notes (but bear in mind I might not have seen your email in time forthe lecture).

0.1 Glossary

We will sometimes use different notation or terminology for concepts en-countered in General Topology. They will be listed here as they come tolight.

1. “map” = “continuous function”

2. I = [0, 1], the closed unit interval with subspace topology in R

3. Let (X, T ) and (Y,U) be topological spaces. Their disjoint union (orcoproduct as in GT notes) X t Y is the set (X ×0)∪ (Y ×1) withthe topology consisting of all sets of the form

(T × 0) ∪ (U × 1) such that T ∈ T , U ∈ U

Example 0.1. If Z = X∪Y for open subspaces X, Y ⊂ Z with X∩Y =∅, then Z is homeomorphic to X t Y .

4. Dn = x ∈ Rn | 0 ≤ ‖x‖ 6 1, the closed unit n-ball.

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Lecture 1

1 Recap on connectedness (non-examinable)

1.1 Connected spaces

Definition 1.1. (i) A topological space X is connected if it cannot bewritten as a union

X = A ∪Bwhere A and B are disjoint non-empty open subsets of X.(ii) A topological space X is disconnected if it is not connected, i.e. if itcan be expressed as a union

X = A ∪B

where A and B are disjoint non-empty open subsets of X.

Remark 1.2. X is connected if and only if the only subsets of X which areboth open and closed are ∅ and X.

Theorem 1.3. R with the usual topology is connected.

Proof. Omitted.

Theorem 1.4. If f : X → Y is continuous and X is connected, then f(X)(with the subspace topology) is connected.

Proof. If f(X) is disconnected then f(X) = (A ∩ f(X)) ∪ (B ∩ f(X))for some open subsets A,B ⊆ Y such that A ∩ B ⊆ Y \f(X). The in-verse images f−1(A), f−1(B) ⊆ X are disjoint open subsets such that X =f−1(A) ∪ f−1(B), in contradiction to the connectedness of X. Hence f(X)is connected.

Corollary 1.5. Connectedness is a topological property: if X, Y are homeo-morphic spaces then X is connected if and only if Y is connected.

Proposition 1.6. Let A be a connected subset of a topological space X andsuppose A ⊆ B ⊆ A. Then B is connected.

Proof. If not, then there exist open sets U, V ⊆ X such that B ∩ U andB ∩ V are disjoint and nonempty and B ⊆ U ∪ V . Then A ∩ U and A ∩ Vare disjoint open subsets of A whose union is A. Since A is connected, eitherA∩U or A∩V must be empty; suppose that A∩U = ∅ so that A ⊆ V . Butsince B ⊆ A, every open set meeting B also meets A; in particular U , (withU ∩B 6= ∅ !) meets A, a contradiction.

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Since nonempty open intervals are connected we immediately obtain:

Corollary 1.7. Every nonempty interval I ⊆ R is connected.

1.2 Paths and path components π0(X)

Definition 1.8. A path in a topological space X is a continuous map α :I = [0, 1] → X. Its initial point is α(0) ∈ X and its terminal point isα(1) ∈ X.

X

α(I)

•α(1)

•α(0)

Example 1.9. By definition, a subset X ⊆ Rn is convex if for any x0, x1 ∈ Xthe segment

[x0, x1] = (1− t)x0 + tx1 | 0 6 t 6 1is wholly contained in X. The path defined by the straight line

α : I → X ; t 7→ (1− t)x0 + tx1

starts at α(0) = x0 and ends at α(1) = x1.

X

α(I)

•α(1)

•α(0)

Definition 1.10. A topological space X is path-connected if for any twopoints x0, x1 ∈ X there exists a path α : I → X from α(0) = x0 to α(1) = x1.

Especially when X is a subspace of some larger space, it’s very importantthat α(t) lies in X for all 0 ≤ t ≤ 1.

Example 1.11. By Example 1.9 a convex subspace X ⊆ Rn is path-connected.I = [0, 1], Dn, Rn, Rn \ 0 (when n > 2) and Sn (when n > 1) are all path-connected spaces.

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Theorem 1.12. If a topological space X is path-connected, then it is alsoconnected.

Proof. Suppose that X is path-connected, but not connected, so that X =A ∪ B for some disjoint nonempty open subsets A,B ⊆ X. Choose pointsa ∈ A, b ∈ B and let α : I → X be a path such that α(0) = a, α(1) = b.The sets

U = α−1(A) , V = α−1(B)

are disjoint nonempty open subsets of I and are such that I = U ∪ V .Since I is connected this is a contradiction. Therefore X must have beenconnected.

Example 1.13. The Euclidean space Rn, the n-sphere Sn (when n > 1), torus,Mobius band, real projective plane, and Klein bottle are path-connected andhence connected.

Theorem 1.14. If n > 2, the spaces Rn and R are not homeomorphic.

Proof. If f : Rn → R were a homeomorphism, then Rn \ 0 and R \ f(0)would also be homeomorphic. But the former is path-connected, hence con-nected, while the latter is disconnected.

Remark 1.15. A connected space need not be path-connected. Indeed, thetopologist’s sine curve (pictured below) defined by

X = (0, y) | − 1 6 y 6 1 ∪ (x, sin 1

x) | 0 < x 6 0.5 ⊆ R2

is connected but not path-connected.

Theorem 1.16. Any connected open subset Ω ⊆ Rn is also path-connected.

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Proof. Fix x ∈ Ω and let

U = y ∈ Ω | there exists a path α : I → Ω from α(0) = x to α(1) = y

be the set of points in Ω which can be joined to x by a path in Ω. Clearlyx ∈ U , so that U 6= ∅, and we shall prove U is both open and closed inΩ, so that U = Ω by the connectedness of Ω. This establishes that Ω ispath-connected.

First let us show that U is open. Let y ∈ U . Since Ω is open in Rn, thereexists r > 0 such that B(y, r) ⊆ Ω. So we can join x to any z ∈ B(y, r) byfirst going from x to y and then going along the straight line from y to z(remaining all the time within Ω). So B(y, r) ⊆ U and therefore U is open.

Now let us see that Rn \ U is open. Let y /∈ U . Since Ω is open in Rn,there exists r > 0 such that B(y, r) ⊆ Ω. So if we could join some x to somez ∈ B(y, r) via a path contained in Ω, we could also join x to y by first goingfrom x to z and then going along the straight line from z to y (remaining allthe time within Ω). But since y /∈ U , we cannot do this, and hence for noz ∈ B(y, r) can we join x to z via a path remaining in Ω. So B(y, r) ⊆ Rn\Uand therefore Rn \ U is open.

Theorem 1.17. Suppose f : X → Y is a continuous map between topologicalspaces and that X is path-connected. Then f(X) is path-connected as asubspace of Y .

Proof. Pick y0 and y1 in f(X). So there are x0, x1 ∈ X such that f(x0) = y0and f(x1) = y1. Let α : [0, 1]→ X be a continuous map with α(0) = x0 andα(1) = x1. Then β = f α is a path in f(X) joining y0 to y1.

Corollary 1.18. Path-connectedness is a topological property: if X, Y arehomeomorphic spaces then X is path-connected if and only if Y is path-connected.

In particular, a path-connected space cannot be homeomorphic to a spacewhich is not path-connected.

Proposition 1.19. For any equivalence relation ∼ on a path-connected spaceX the quotient space Y = X/∼ is path-connected.

Proof. Since the projection p : X → Y is continuous and surjective, thisfollows immediately from Theorem 1.17.

Definition 1.20. The path relation on X is

x0 ∼ x1 if there exists a path α : I → X from α(0) = x0 ∈ X to α(1) = x1 ∈ X.

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The path relation is an equivalence relation. To see this, we proceed asfollows:

Definition 1.21. The constant path at x ∈ X is the path

αx : I → X ; t 7→ x

from αx(0) = x ∈ X to αx(1) = x ∈ X.

Definition 1.22. The reverse of a path α : I → X is the path

−α : I → X ; t 7→ α(1− t)

retracing α backwards, with

•−α(0) = α(1) −α

•−α(1) = α(0)

Definition 1.23. The concatenation of paths α : I → X, β : I → X with

α(1) = β(0) ∈ X

is the path

α • β : I → X ; t 7→

α(2t) if 0 6 t 6 1

2

β(2t− 1) if 126 t 6 1

which starts at α(0), follows along α at twice the speed in the first half,switching at α(1) = β(0) (at half-time) to follow β at twice the speed in thesecond half.

•α • β(0) = α(0) α

•α(1) = β(0) β

•β(1) = α • β(1)

Proposition 1.24. The path relation defined on a space X by x0 ∼ x1 ifthere exists a path α : I → X from α(0) = x0 to α(1) = x1 is an equivalencerelation.

Proof. The path relation is reflexive by 1.21, symmetric by 1.22 and transitiveby 1.23.

Definition 1.25. Let X be a topological space.(i) The path components of X are the equivalence classes of the pathequivalence relation ∼, i.e. the subspaces

[x] = y ∈ X | y ∼ x= y ∈ X | there exists a path α : I → X from α(0) = x to α(1) = y

(ii) The set of path components (which may be infinite) is denoted by

X/∼ = π0(X) .

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The function

X → π0(X) ; x 7→ [x] = equivalence class of x

is surjective.

Proposition 1.26. The cardinality of set of path components π0(X) of atopological space X is a topological invariant, i.e. if X and Y are homeo-morphic, there is a bijection between π0(X) and π0(Y ).

Remark 1.27. However, a bijection between π0(X) and π0(Y ) does not nec-essarily imply that X and Y are homeomorphic.

Proof. This is “obvious” as ultimately the whole structure of the set of pathcomponents of X depends only on the open set structure of X. (Or, if Xand Y are homeomorphic, so are their respective quotient spaces π0(X) andπ0(Y ).) But to prepare for the sorts of arguments which will be regularlyused in this course, we also give a “functorial” proof.

A continuous map f : X → Y induces a function

f∗ : π0(X)→ π0(Y ) ; [x] 7→ [f(x)] .

The composition of maps f : X → Y , g : Y → Z is a map g f : X → Zwhich induces the composition

(g f)∗ = g∗ f∗ : π0(X)→ π0(Z) .

The identity map 1 : X → X induces the identity function

1∗ = 1 : π0(X)→ π0(X) .

A homeomorphism f : X → Y induces a bijection f∗ : π0(X)→ π0(Y ) withinverse

(f∗)−1 = (f−1)∗ : π0(Y )→ π0(X) .

Example 1.28. (i) A topological space X is path-connected if and only ifπ0(X) has a single element.(ii) The topologist’s sine curve of Example 1.15 is compact (as it is closed andbounded) but is homeomorphic to neither a closed interval nor the disjointunion of two closed intervals, as it has one connected component but twopath components.

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Lecture 2

2 Homotopy theory

Definition 2.1. A homotopy between maps f, g : X → Y is a map h :X × I → Y such that

h(x, 0) = f(x) , h(x, 1) = g(x) ∈ Y (x ∈ X) ,

The maps f, g are homotopic, and the homotopy is denoted by

h : f ' g : X → Y .

Example 2.2. If X = x is a space with one element x, a map f : X → Yis the same as an element f(x) ∈ Y . A homotopy h : f ' g : X → Y is thesame as a path h : I → Y with initial point h(0) = f(x) ∈ Y and terminalpoint h(1) = g(x) ∈ Y . A homotopy h : f ' f : X → Y is the same as aclosed path h : I → Y .

A homotopy h : f ' g : X → Y deforms the map f continuously to g inthe space Y X of maps from X to Y , with the ‘compact-open’ topology onY X (defined in the non-examinable appendix to this lecture). For each t ∈ Ithere is defined a map

ht : X → Y ; x 7→ ht(x) = h(x, t)

with h0 = f , h1 = g. Equivalently, there is a continuous choice of path foreach x ∈ X

h(x,−) : I → Y ; t 7→ h(x,−)(t) = h(x, t)

which starts at h(x, 0) = f(x) and ends at h(x, 1) = g(x).

Y

• •h(x, 0) = f(x) h(x, 1) = g(x)h(x, t)

Proposition 2.3. For fixed X, Y the notion of homotopy is an equivalencerelation on the set of maps f : X → Y .

Proof. (i) For every map f : X → Y define the constant homotopy h :f ' f : X → Y by

h : X × I → Y ; (x, t) 7→ f(x)

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(ii) Given a homotopy h : f ' g : X → Y define the reverse homotopy−h : g ' f : X → Y by

−h : X × I → Y ; (x, t) 7→ h(x, 1− t) .

(iii) Given homotopies h1 : f1 ' f2 : X → Y and h2 : f2 ' f3 : X → Ydefine the concatenation homotopy h1 • h2 : f1 ' f3 : X → Y by

h1 • h2 : X × I → Y ; (x, t) 7→

h1(x, 2t) if 0 6 t 6 1

2

h2(x, 2t− 1) if 126 t 6 1 .

•h1 • h2(x, 0) = f1(x)

h1(x,−) at twice the speed•

h1 • h2(x, 12) = f2(x)

h2(x,−) at twice the speed•

h1 • h2(x, 1) = f3(x)

2.1 Nonexaminable appendix on function spaces

Definition 2.4. (i) For any spaces X, Y let Y X be the set of maps f : X →Y .

(ii) The compact-open topology on Y X has basis the subspaces

B(K,V ) = f | f(K) ⊆ V ⊆ Y X

defined for compact K ⊆ X and open V ⊆ Y .

Remark 2.5. (pp. 286-289 of Munkres “Topology”).(i) The compact-open topology has the key property that the function

defined for any spaces X, Y, Z by

Y X×Z → (Y X)Z ; (h : X × Z → Y ) 7→ (H : Z → Y X ; z 7→ (x 7→ h(x, z)))

is continuous. If X is compact Hausdorff this function is a homeomorphism.The special case Z = I identifies homotopies h : f ' g : X → Y with pathsH : I → Y X from H(0) = f to H(1) = g.

(ii) IfX is compact and (Y, d) is a metric space the compact-open topologyon Y X is the topology determined by the metric

d(f, g) = sup d(f(x), g(x)) |x ∈ X (f, g ∈ Y X) .

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Constructing homotopies

The construction of homotopies depends on being able to construct paths,and this is particularly easy in convex subspaces of Rn, by the constructionof straight line paths. (Recall that a subspace Y ⊆ Rn is convex if the linesegment in Rn joining any two points in Y is wholly in Y ).

Proposition 2.6. Any two maps f, g : X → Y to a convex subspace Y ⊆ Rn

are homotopic.

Proof. The map

h : X × I → Y ; (x, t) 7→ (1− t)f(x) + tg(x)

defines a homotopy h : f ' g : X → Y .

N.B. Rn is a convex subspace of Rn.

Example 2.7. (i) Any two paths f, g : I → Rn are homotopic, with

h : I × I → Rn ; (s, t) 7→ (1− t)f(s) + tg(s)

a homotopy h : f ' g.(ii) Given a path f : I → Rn define the straight line path

g : I → Rn ; s 7→ (1− s)f(0) + sf(1)

with g(0) = f(0) and g(1) = f(1). In this case the homotopy h : f ' g of (i)

h : I × I → Rn ; (s, t) 7→ (1− t)f(s) + t((1− s)f(0) + sf(1))

is such that the paths

ht : I → Rn ; s 7→ h(s, t) (0 6 t 6 1)

‘slide’ continuously from the ‘curved’ path h0 = f : I → Rn to the ‘straight’path h1 = g : I → Rn, keeping the endpoints fixed:

ht(0) = f(0) = g(0) , ht(1) = f(1) = g(1) ∈ Rn (0 6 t 6 1) .

In general, geometry is used to construct homotopies, and algebra is usedto show that homotopies with certain properties cannot exist.

Definition 2.8. Two spaces X, Y are homotopy equivalent if there existmaps f : X → Y , g : Y → X and homotopies

h : gf ' 1X : X → X , k : fg ' 1Y : Y → Y .

A map f : X → Y is a homotopy equivalence if there exist such g, h, k.The maps f, g are inverse homotopy equivalences.

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The relation defined on the set of topological spaces by

X ' Y if X is homotopy equivalent to Y

is an equivalence relation.

Definition 2.9. A space X is contractible if it is homotopy equivalent tothe space 0 with one point.

A contractible space X has a point x0 ∈ X and maps

f : X → 0 ; x 7→ 0 , g : 0 → X ; 0 7→ x0

as well as homotopies

h : gf ' 1X : X → X , k : fg ' 10 : 0 → 0 .

The essential feature is the continuous choice of paths

h(x,−) : I → X ; t 7→ h(x, t) (x ∈ X)

from h(x, 0) = gf(x) = x0 to h(x, 1) = x.

Proposition 2.10. The unit line I = [0, 1] is contractible.

Proof. The maps

f : I → 0 ; t 7→ 0 , g : 0 → I ; 0 7→ 0

are related by the homotopies h : gf ' 1I , k : fg ' 10 defined by

h : I × I → I ; (s, t) 7→ st ,

k : 0 × I → 0 ; (0, t)→ 0 .

Remark 2.11. In Proposition 2.6 it was proved that for a convex subspaceY ⊆ Rn any two maps f, g : X → Y are homotopic. In fact, for anycontractible space Y any two maps f, g : X → Y are homotopic.

A non-compact space can be homotopy equivalent to a compact space:

Example 2.12. The non-compact space Dn\0 = x ∈ Rn | 0 < ‖x‖ 6 1 ishomotopy equivalent to the compact space Sn−1 = x ∈ Rn | ‖x‖ = 1, sincethe maps

f : Dn\0 → Sn−1 ; x 7→ x/‖x‖ ,g : Sn−1 → Dn\0 ; x 7→ x

are such that fg = 1 : Sn−1 → Sn−1, and

h : Dn\0 × I → Dn\0 ; (x, t) 7→ tx+ (1− t)x/‖x‖

defines a homotopy h : gf ' 1 : Dn\0 × I → Dn\0.

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Proposition 2.13. Homeomorphic spaces are homotopy equivalent.

Proof. If f : X → Y is a homeomorphism then the actual inverse g = f−1 :Y → X is also a homotopy inverse, with the maps

h : X × I → X ; (x, t) 7→ x ,

k : Y × I → Y ; (y, t) 7→ y

defining the constant homotopies h : gf ' 1X : X → X, k : fg ' 1Y : Y →Y .

Homotopy equivalent spaces have the same homotopy classes of maps.More precisely, if F : X ′ → X and G : Y → Y ′ are both homotopy equiva-lences then the function

homotopy classes of maps f : X → Y → homotopy classes of maps f ′ : X ′ → Y ′ ;

(f : X → Y ) 7→ (GfF : X ′ → Y ′)

is a bijection. Homotopy theory regards homotopy equivalent spaces asbeing isomorphic.

Proposition 2.14. A homotopy equivalence f : X → Y induces a bijectionof the path-component sets

f∗ : π0(X)→ π0(Y ) ; [x] 7→ [f(x)] .

Proof. Let g : Y → X be a homotopy inverse for f , so that there existhomotopies

h : gf ' 1X : X → X , k : fg ' 1Y : Y → Y .

For any x ∈ X there exists a path in X from gf(x) to x

h(x,−) : I → X ; t 7→ h(x, t) ,

so that [x] = [gf(x)] ∈ π0(X). Similarly, for any y ∈ Y there exists a pathin Y from fg(y) to y

k(y,−) : I → Y ; t 7→ k(y, t) ,

so that [y] = [fg(y)] ∈ π0(Y ). It follows that the function

g∗ : π0(Y )→ π0(X) ; [y] 7→ [g(y)]

is an inverse of f∗, so that f∗ is a bijection.

Example 2.15. It is immediate from 2.14 that if X, Y are spaces such thatX is path-connected and Y is not path-connected then X is not homotopyequivalent to Y . Simplest example: X = 0 and Y = 0, 1 with thediscrete topology (= every subset is open). Indeed, two finite spaces X, Ywith the discrete topology are homotopy equivalent if and only if they havethe same number of elements.

15

Lecture 3Definition 2.16. (i) A retraction of a topological space X onto a subspaceY ⊆ X is a map r : X → Y such that

r(y) = y ∈ Y for all y ∈ Y .

The subspace Y is a retract of X.(ii) A deformation retraction of a topological space X onto a subspace

Y ⊆ X is a map h : X × I → X such that

h(x, 0) ∈ Y , h(x, 1) = x ∈ X , for all x ∈ X ,

h(y, 0) = y ∈ Y ⊆ X for all y ∈ Y .

The subspace Y is a deformation retract of X.

Note that for a deformation retraction h of X onto Y the maps

i = inclusion : Y → X , j : X → Y ; x 7→ h(x, 0)

are such thatji = 1Y : Y → Y ,

so that j is a retraction of X onto Y . Moreover, i, j are inverse homotopyequivalences, with

c : Y × I → Y ; (y, t) 7→ y

and h defining homotopies

c : ji ' 1Y : Y → Y ,

h : ij ' 1X : X → X .

Example 2.17. 0 is a deformation retract of [0, 1], with deformation retrac-tion

h : I × I → I ; (x, t) 7→ tx .

While every deformation retract is a retract, not every retract is a defor-mation retract:

Example 2.18. Let X be any non-empty space which is not contractible, forexample 1, 2 with the discrete topology (every subset is open). For anyx0 ∈ X the subspace Y = 1 is a retract which is not a deformation retract.

16

Example 2.19. If X is a non-empty space with the trivial topology ∅, Xthen for any x0 ∈ X the one-point subspace x0 ⊆ X is a deformationretract, with deformation retraction

h : X × I → X ; (x, t) 7→

x0 if t = 0

x if 0 < t 6 1 .

(Recall that any function Y → X to a space with the trivial topology iscontinuous). In particular, X is contractible.

Definition 2.20. A subspace X ⊆ Rn is star-shaped at x ∈ X if for eachy ∈ X the straight line segment

[x, y] = (1− t)x+ ty | 0 6 t 6 1 ⊂ Rn

is wholly contained in X, [x, y] ⊆ X.

Remark 2.21. (i) A subspace X ⊆ Rn is convex if it is star-shaped at everyx ∈ X.

(ii) Construct an example of a star-shaped subspace of Rn which is notconvex!

Proposition 2.22. If X ⊆ Rn is star-shaped at x0 ∈ X then x0 ⊆ X is adeformation retract. In particular, X is contractible.

Proof. The map

h : X × I → X ; (x, t) 7→ x0 + t(x− x0)

is a deformation retraction.

Example 2.23. (i) A non-empty path-connected subspace X ⊆ R is either Ritself or x (x ∈ R) or one of the intervals [a,∞), (a,∞), (−∞, a], (∞, a),[a, b], [a, b), (a, b], (a, b) (a < b ∈ R). In each case X is convex, and hencecontractible.

(ii) For any n > 1 the subspaces Rn, Dn ⊆ Rn are convex, and hencecontractible.

Definition 2.24. The cone on a non-empty space X is the quotient spaceobtained from X × I by collapsing X × 0 to a point

CX = (X × I)/(X × 0) .

Explicitly, CX = (X × I)/ ∼ where (x, 0) ∼ (y, 0) for all x, y ∈ X.

17

The point c = [X × 0] ∈ CX is the cone point, and the subspace

X = X × 1 ⊂ CX

is the cone base.

c

CX

X × 1

For any point [x, s] ∈ CX in a cone there is an obvious path

α : I → CX ; t 7→ [x, st]

along the ray from the cone point α(0) = c to α(1) = [x, s]. Here, obviousmeans that the path varies continuously with [x, s].

This is a very special property for CX to have — by contrast, try tofind an obvious choice of path α : I → S1 from x0 = (1, 0) ∈ S1 to anyx ∈ S1. If x 6= −x0, you can just choose the shorter of the two arcs from x0to x. But for x = −x0 the two arcs are of equal length. It is a non-trivialtheorem that there is no obvious (= continuous) choice of path from x0 tox for all x ∈ S1. [In general, if (X, d) is a metric space such that betweenany two points x, y ∈ X there is a unique path of shortest length, then X iscontractible.]

Proposition 2.25. (i) For any space X the cone point c ⊂ CX is adeformation retract of the cone CX, so that CX is contractible.

(ii) A map f : X → Y is homotopic to a constant map g : X → Y ;x 7→ y0(for some y0 ∈ Y ) if and only if there exists a map F : CX → Y such that

F [x, 1] = f(x) , F [x, 0] = y0 ∈ Y (x ∈ X) .

Proof. (i) The paths along rays from the cone point c define a deformationretraction of CX onto c ⊂ CX

h : CX × I → CX ; ([x, s], t) 7→ [x, st] .

so that the mapsf : CX → 0 ; [x, s] 7→ 0 ,

g : 0 → CX ; 0 7→ [x, 0]

18

are inverse homotopy equivalences.(ii) Given a homotopy h : f ' g : X → Y define a map

F : CX = (X × I)/(X × 0)→ Y ; [x, t] 7→ h(x, 1− t)

such that

F [x, 1] = h(x, 0) = f(x) , F [x, 0] = h(x, 1) = g(x) = y0 ∈ Y (x ∈ X) .

Conversely, given a map F : CX → Y such that F [x, 1] = f(x) let y0 =F [x, 0] ∈ Y , and define a homotopy h : f ' g : X → Y by

h : X × I → Y ; (x, t) 7→ F [x, 1− t] .

Example 2.26. For any n > 0 there is a homeomorphism

CSn → Dn+1 ; [x, t] 7→ tx

sending the cone point c to 0 ∈ Dn+1 and the cone base Sn = Sn × 1 toSn ⊂ Dn+1 by the identity. The inverse homeomorphism is given by

y 7→

[y/‖y‖, ‖y‖] y 6= 0

c y = 0.

By 2.25 (ii) a map f : Sn → Y is homotopic to a constant map if andonly if there exists a map F : Dn+1 → Y such that

F (x) = f(x) ∈ Y (x ∈ Sn) .

Removing the cone point c = [X × 0] ∈ CX from a cone results in aspace CX\c for which there is a homeomorphism

X × (0, 1]→ CX\c ; (x, t) 7→ [x, t] .

In particular, X = X × 1 ⊂ CX\c is a deformation retract.

Proposition 2.27. Extend an equivalence relation ∼ on a space X to anequivalence relation ∼ on the cone CX by the identity, i.e.

(x, s) ∼ (y, t) if (x, s) = (y, t) or s = t = 0 or (s = t = 1 and x ∼ y) .

Then X/∼= (X × 1)/∼ is a deformation retract of Y = (CX\c)/∼,and Y is homotopy equivalent to the quotient space X/∼.

19

Proof. The mapsi : X/∼ → Y ; [x] 7→ [x, 1] ,

j : Y → X/∼ ; [x, t] 7→ [x]

are such that ji = 1 : X/∼ → X/∼ and ij([x, t]) = [x, 1]. The map

h : Y × I → Y ; ([x, t], s) 7→ [x, 1− s+ st]

is a deformation retraction, defining a homotopy h : ij ' 1 : Y → Y whichis constant on X/∼ ⊂ Y (where t = 1).

20

Lecture 4

3 Some quotient spaces of I2 and ∂I2

We shall now deal with the scissors and paper constructions of Workshop 3last semester in a more abstract way.

The unit square

I2 = I × I = (x, y) ∈ R2 | 0 6 x 6 1 and 0 6 y 6 1

has boundary

∂I2 = (x, y) ∈ R2 | either x = 0 or x = 1 or y = 0 or y = 1 .

I2 ∂I2

In this chapter we shall consider the cone of X = ∂I2, noting that themap

β : CX → I2 ; [(x, y), t] 7→ (v, w) = (1− t)(12, 12) + t(x, y)

is a homeomorphism sending the cone point c ∈ CX to the midpoint β(c) =(12, 12) ∈ I2 (similar to the homeomorphism CS1 ∼= D2 of 2.26, which sends

the cone point to (0, 0) ∈ D2, but for I2 the inverse is fiddly to write down).

I2

∂I2

•(x, y)

•(12, 12)

•(v, w)

The straight line ‘ray’ from the cone point c to a point [(x, y), 1] ∈ X×1is sent by β to the line segment in I2 from β(c) = (1

2, 12) to (x, y) ∈ X. Make

sure you understand this homeomorphism, before proceeding further!

By Proposition 2.27, for any equivalence relation ∼ on ∂I2 extended bythe identity to an equivalence relation ∼ on I2 the space obtained from I2/∼by puncturing (i.e. removing) (1

2, 12)

Y = (I2 \ (12, 12))/∼

21

contains the quotient space ∂I2/∼ as a deformation retract.Specifically, we have a deformation retraction

h : Y × I → Y ; ([v, w], s) 7→ β[(x, y), 1− s+ st]

with [(x, y), t] = β−1([v, w]) ∈ C∂I2. The maps

f : Y → ∂I2/∼ ; [v, w] 7→ [x, y] ,

g : ∂I2/∼ → Y ; [x, y] 7→ [x, y]

are inverse homotopy equivalences, such that h : gf ' 1, fg = 1. For any(v, w) 6= (1

2, 12) ∈ I2 we have gf [v, w] = [x, y] with

I → Y ; s 7→ h([v, w], s) = β[(x, y), 1− s+ st]

a path along the straight line segment from h([v, w], 0) = [x, y] to h([v, w], 1) =[v, w].

3.1 The figure eight

Terminology: the one-point union X∨Y of two spaces X, Y with base pointsx0 ∈ X, y0 ∈ Y is the quotient space (X tY )/∼ of the disjoint union X tY ,with x0 ∼ y0. Equivalently

X ∨ Y = (X × y0) ∪ (x0 × Y ) ⊆ X × Y .

The figure eight is the one-point union of two circles

8 = S1 ∨ S1

Define an equivalence relation ∼ on ∂I2 by

(x, 0) ∼ (x, 1) , (0, y) ∼ (1, y) for all x, y ∈ I

as well as the standard (x, y) ∼ (x, y).

22

•

•

••

(x, 1)

(0, y) (1, y)

(x, 0)The quotient space is the figure 8, in the precise sense that the function

∂I2/∼→ 8 ;

[x, 0] = [x, 1] 7→ e2πix in first S1

[0, y] = [1, y] 7→ e2πiy in second S1

is a homeomorphism.Later on, we shall use without proof the following facts:

(i) The space Υ obtained by joining two circles by a line

is homotopy equivalent to 8 – the projection Υ → 8 collapsing thediameter to a point is a homotopy equivalence.

(ii) The space Θ defined by a circle with a diameter

is homotopy equivalent to 8 – the projection Θ → 8 collapsing thediameter to a point is a homotopy equivalence.

3.2 The cylinder

The cylinder is the quotient space

I × S1 = I2/∼ , (x, 0) ∼ (x, 1)

23

,

•

•

(x, 1)

(x, 0)The cylinder I × S1 contains the circle S1 as a deformation retract, with

the inclusionS1 → I × S1 ; z 7→ (0, z)

a homotopy equivalence. (Exercise: prove this!)The quotient space ∂I2/∼ is homeomorphic to Υ (prove this!). The

punctured cylinder (I × S1) \ (12, 12) contains Υ as a deformation retract,

with the inclusion

Υ = ∂I2/∼→ (I × S1) \ (12, 12)

a homotopy equivalence. The punctured cylinder is thus homotopy equivalentto the figure 8.

3.3 The Mobius band

The Mobius band is the quotient space

M = I2/∼ , (x, 0) ∼ (1− x, 1)

•

•

(1− x, 1)

(x, 0)The Mobius band M contains the circle S1 = I/(0 ∼ 1) as a deformation

retract, with the inclusion as the midpoints of the lines joining [0, y] to [1, y]

S1 →M ; [y] 7→ [12, y]

a homotopy equivalence. (Exercise: prove this!). Note that this is *not* theboundary

∂M = (0, 1 × I)/(0, 0) ∼ (1, 1), (1, 0) ∼ (0, 1) ⊂M

24

(draw a picture). The boundary is homeomorphic to S1 (prove this!), but theinclusion ∂M = S1 → M is not a homotopy equivalence, as will be provedlater on in the course using the fundamental group.

The quotient space ∂I2/∼ is homeomorphic to Θ (prove this!). The punc-tured Mobius band contains Θ as a deformation retract, with the inclusion

Θ = ∂I2/∼→M \ (12, 12)

a homotopy equivalence. The punctured Mobius band is thus homotopyequivalent to the figure 8.

3.4 The torus

The torus is the quotient space

S1 × S1 = I2/∼ , (x, 0) ∼ (x, 1) , (0, y) ∼ (1, y)

•

•

••

(x, 1)

(0, y) (1, y)

(x, 0)The quotient space ∂I2/∼ is homeomorphic to the figure 8 (prove this!).

The punctured torus thus contains 8 as a deformation retract, with the in-clusion

∂I2/∼= 8→ (S1 × S1) \ (12, 12)

a homotopy equivalence.

3.5 The Klein bottle

The Klein bottle is the quotient space

K = I2/∼ , (x, 0) ∼ (x, 1) , (0, y) ∼ (1, 1− y)

•

•

•

•

(x, 1)

(0, y)

(1, 1− y)

(x, 0)

25

The quotient space ∂I2/∼ is homeomorphic to the figure 8 (prove this!).The punctured Klein bottle thus contains 8 as a deformation retract, withthe inclusion

∂I2/∼= 8→ K \ (12, 12)

a homotopy equivalence.

26

Lecture 5

3.6 The projective plane

The projective plane is the quotient space

RP2 = I2/∼ , (x, 0) ∼ (1− x, 1) , (0, y) ∼ (1, 1− y)

•

•

•

•

(1− x, 1)

(0, y)

(1, 1− y)

(x, 0)There are four equivalent (i.e. homeomorphic) ways of defining the pro-

jective plane RP2, as a quotient space of I2, S2, D2, R3 :

(i) RP2 = I2/∼, as above.

(ii) RP2 = S2/∼ with (x, y, z) ∼ (−x,−y,−z).

(iii) RP2 = D2/∼ with (x, y) ∼ (−x,−y) for (x, y) ∈ S1.

(iv) RP2 = (R3\(0, 0, 0))/∼ with (x, y, z) ∼ (λx, λy, λz) for λ 6= 0 ∈ R.

The quotient space ∂I2/∼ is homeomorphic to the circle S1, and is adeformation retract of RP2 \(1

2, 12) (prove this!). The punctured projective

plane is thus homotopy equivalent to a circle:

RP2 \ (12, 12) ' ∂I2/∼ = S1 .

Remark 3.1. The circle, the figure eight, the torus and the Klein bottle arenot homotopy equivalent to each other: the proofs require the ‘fundamen-tal group’ which will be developed later in the course. Homotopy equiv-alent spaces have isomorphic fundamental groups, and these spaces havenon-isomorphic fundamental groups.

4 Cutting and pasting paths

The fundamental group π1(X, x0) will be defined in section 5 for any spaceX and base point x0 ∈ X to be the set of ‘rel 0, 1 homotopy classes’ ofpaths α : [0, 1]→ X such that

α(0) = α(1) = x0 ∈ X ,

27

with appropriate group law and inversion. For this purpose it is necessaryto be able to paste together paths by concatenation. For nontrivial applica-tions it is also necessary to improve paths, by cutting them into convenientlysmaller pieces and if possible straightening each piece!

What does ‘rel 0, 1’ mean? Keeping the endpoints α(0), α(1) ∈ X of apath α : I → X fixed during the homotopy.

Definition 4.1. If f, g : A → X are maps and B ⊆ A is a subspace suchthat

f(b) = g(b) ∈ X (b ∈ B)

then a homotopy rel B (or relative to B) is a homotopy h : f ' g : A→ Xsuch that

h(b, t) = f(b) = g(b) ∈ X (b ∈ B, t ∈ I) .

We shall be particularly concerned with the special case

(A,B) = (I, 0, 1)

Example 4.2. A homotopy rel 0, 1 of two paths α0, α1 : I → X with thesame endpoints

α0(0) = α1(0) , α0(1) = α1(1) ∈ X

is a collection of paths ht : I → X (0 6 t 6 1) with the same endpoints

ht(0) = α0(0) = α1(0) , ht(1) = α0(1) = α1(1)

such that h0 = α0, h1 = α1 and such that the function

h : I × I → X ; (s, t) 7→ ht(s)

is continuous.

It will turn out that any two paths α0, α1 : I → X with the same end-points are homotopic rel 0, 1 for X = Rn or Sn for n > 2, but not neces-sarily so for X = S1: while this may be intuitively obvious, it is quite hardto prove.

As was already seen in General Topology it is convenient to be able toglue together paths, using the following construction:

Definition 4.3. The concatenation at λ ∈ (0, 1) of paths α, β : I → Xsuch that α(1) = β(0) ∈ X is the path

α •λ β : I → X ; s 7→

α(s

λ) if 0 6 s 6 λ

β(s− λ1− λ

) if λ 6 s 6 1 .

28

with

α •λ β(0) = α(0) , α •λ β(λ) = α(1) = β(0) , α •λ β(1) = β(1) .

Note that the image of a concatenation is the union of the images

(α •λ β)(I) = α(I) ∪ β(I) ⊆ X .

Example 4.4. (i) The concatenation of two paths α : I → X, β : I → X withα(1) = β(0) ∈ X defined in Definition 1.23 is the concatenation of 4.3 withλ = 1

2

α • β = α •12β : I → X ; t 7→

α(2t) if 0 6 t 6 1

2

β(2t− 1) if 126 t 6 1

which starts at α(0), follows along α at twice the speed in the first half,switching at α(1) = β(0) (at half-time) to follow β at twice the speed in thesecond half.

(ii) The concatenation of paths as in (i) was used in Proposition 2.3 toprove that homotopy is a transitive relation.

•α • β(0) = α(0) α

•α(1) = β(0) β

•β(1) = α • β(1)

Proposition 4.5. The rel 0, 1 homotopy class of α •λ β : I → X isindependent of λ.

Proof. For any λ, µ ∈ (0, 1) the function

ν : I → (0, 1) ; t 7→ ν(t) = (1− t)λ+ tµ

is such thath : I × I → X ; (s, t) 7→ (α •ν(t) β)(s)

defines a rel 0, 1 homotopy

h : α •λ β ' α •µ β : I → X ,

withh(s, 0) = (α •λ β)(s) , h(s, 1) = (α •µ β)(s) .

29

s

t

α

β

•(λ, 0)

•

0 6 s 6 ν(t)

ν(t) 6 s 6 1

(µ, 1)

α •µ β

α •λ βThis is a picture of the unit square I2 in the (s, t)-plane subdivided accordingto how h is defined: the line joining (λ, 0) to (µ, 1) is s = ν(t) and

h(s, t) = (α •ν(t) β)(s) =

α(

s

ν(t)) if 0 6 s 6 ν(t)

β(s− ν(t)

1− ν(t)) if ν(t) 6 s 6 1 .

(The picture is perhaps more memorable than the formula.)

More generally:

Definition 4.6. Suppose given N paths α1, α2, . . . , αN : I → X which adjoineach other, so that

α1(1) = α2(0) , α2(1) = α3(0) , . . . , αN−1(1) = αN(0) ∈ X ,

and denote this collection of paths by α. Suppose given also a partition λ ofI as a union of intervals

I = [λ0, λ1] ∪ [λ1, λ2] ∪ · · · ∪ [λN−1, λN ]

withλ0 = 0 < λ1 < λ2 < · · · < λN−1 < λN = 1 .

The N-tuple concatenation path is

αλ = α1 •λ1 α2 • · · · •λN−1αN : I → X ; s 7→ αi(

s− λi−1λi − λi−1

) if s ∈ [λi−1, λi]

from αλ(0) = α1(0) to αλ(1) = αN(1) ∈ X.

30

Proposition 4.7. The rel 0, 1 homotopy class of an N-tuple concatenationαλ : I → X is independent of λ.

Proof. For any partitions λ, µ of I the map

h : I × I → X ; (s, t) 7→ (α(1−t)λ+tµ)(s)

defines a rel 0, 1 homotopy

h : αλ ' αµ : I → X .

Example 4.8. (i) Proposition 4.5 is the special case of Proposition 4.7 withN = 2.

(ii) Consider the special case N = 3, writing α1, α2, α3 as α, β, γ. Letthen α, β, γ : I → X be three adjoining paths, such that

α(1) = β(0) , β(1) = γ(0) ∈ X .

The concatenation of α, β, γ for 0 < λ < λ′ < 1 is the path

δ = α •λ β •λ′ γ : I → X ; s 7→

α(s/λ) if 0 6 s 6 λ

β((s− λ)/(λ′ − λ)) if λ 6 s 6 λ′

γ((s− λ′)/(1− λ′)) if λ′ 6 s 6 1 .

such thatδ(0) = α(0) , δ(1) = γ(1) ∈ X .

• • • •0 α λ β λ′ γ 1

The path starts by going along α at 1/λ the speed on [0, λ], followed by goingalong β at 1/(λ′ − λ) the speed on [λ, λ′], and finishes by going along γ at1/(1− λ′) the speed on [λ′, 1]. The homotopy class rel 0, 1 of δ : I → X isindependent of the choice of λ, λ′.

31

Lecture 6

5 The fundamental group π1(X)

The fundamental group is the fundamental algebraic object associated to atopological space! It was first defined by Poincare around 1900, and is thekey to using algebra to prove theorems in topology. Why should one wantto do this? Roughly speaking, geometric constructions tell us what can bedone in topology, while algebraic obstructions tell us what cannot be done.Homeomorphic spaces have isomorphic groups. Given two spaces X, Y itmay be possible to construct a homeomorphism X → Y by geometry, or itmay be possible to prove algebraically that their fundamental groups are notisomorphic, so that X, Y are not homeomorphic.

The fundamental group π1(X, x0) of a space X at a base point x0 ∈ Xis a geometrically defined group of the ‘based homotopy classes’ of maps ω :S1 → X such that ω(1, 0) = x0. The isomorphism class of π1(X, x0) dependsonly on the homotopy type of the path-component of the base point x0 ∈ X.Homotopy equivalent path-connected spaces have isomorphic fundamentalgroups. The fundamental group is used to distinguish non-homeomorphicspaces: path-connected spaces with non-isomorphic fundamental groups arenot homotopy equivalent, and (a fortiori) not homeomorphic.

Definition 5.1. (i) A based space (X, x0) is a space with a base pointx0 ∈ X.

(ii) A based map f : (X, x0) → (Y, y0) is a map f : X → Y such thatf(x0) = y0 ∈ Y .

(iii) A based homotopy h : f ' g : (X, x0) → (Y, y0) is a homotopyh : f ' g : X → Y such that h(x0, t) = y0 ∈ Y (t ∈ I).

Proposition 5.2. For any based spaces (X, x0), (Y, y0) based homotopy isan equivalence relation on the set of based maps f : (X, x0)→ (Y, y0).

Proof. Exactly as for unbased homotopy.

Definition 5.3. A based loop is a based map ω : (S1, 1)→ (X, x0) where1 = (1, 0) ∈ S1.

32

ω(1) = x0•

X

ω(S1)

In view of the homeomorphism

I/0 ∼ 1 → S1 ; [t] 7→ (cos 2πt, sin 2πt)

a loop ω : S1 → X at ω(1) = x0 ∈ X is essentially the same as a closed pathα : I → X such that

α(0) = α(1) = x0 ∈ X ,

with α and ω related by

α(t) = ω(cos 2πt, sin 2πt) ∈ X (t ∈ I)

The closed path α is the composite

α : Iprojection // S1 ω // X .

The rel 0, 1 homotopy classes of closed paths α : I → X such that α(0) =α(1) = x0 ∈ X are in one-one correspondence with the rel 1 homotopyclasses of loops ω : S1 → X with ω(1) = x0 ∈ X.

Homotopy theory uses the topological properties of closed paths I → Xand loops S1 → X and the algebraic properties of groups to decide whethertopological spaces are homotopy equivalent. Since I is contractible (2.10)two paths α, β : I → X are homotopic if and only if α(I), β(I) ⊆ X arein the same path component of X. In order to use the homotopy classesof paths I → X to detect more than just the path components of X, it isnecessary to keep the endpoints fixed!

Definition 5.4. The fundamental group π1(X, x0) at a base point x0 ∈ Xis the set of rel 0, 1 homotopy classes [α] of closed paths α : I → X suchthat

α(0) = α(1) = x0 ∈ Xwith multiplication by the concatenation of paths (4.3)

π1(X, x0)× π1(X, x0)→ π1(X, x0) ; ([α], [β]) 7→ [α][β] = [α • β] ,

33

and inverses by path reversal (1.22)

π1(X, x0)→ π1(X, x0) ; [α] 7→ [α]−1 = [−α]

and neutral element [ex0 ] ∈ π1(X, x0) the class of the constant path

ex0 : I → X ; t 7→ x0 .

It is of course also possible to regard π1(X, x0) as the set of rel 1homotopy classes [ω] of loops ω : S1 → X such that ω(1) = x0 ∈ X. Thepath formulation is more convenient for algebra, while the loops are moregeometric.

Theorem 5.5. The fundamental group π1(X, x0) is a group.

Proof. I. [ex0 ] is a unit: [α][ex0 ] = [α] ∈ π1(X, x0).Define a rel 0, 1 homotopy

h : α • ex0 ' α : I → X

by

h : I × I → X ; (s, t) 7→

α(2s/(1 + t)) if s 6 (1 + t)/2

x0 if s > (1 + t)/2 .

To make sense of this formula draw the unit square in the (s, t)-plane andjoin the point (1

2, 0) to the point (1, 1) by the line s = (1 + t)/2. Think what

happens at each time t ∈ I: the path

ht = α •(1+t)/2 ex0 : I → X ; s 7→ ht(s) = h(s, t)

starts by going along α at 2/(1+ t) the speed on [0, (1+ t)/2], and then staysput at x0 on [(1 + t)/2, 1]. The homotopy h starts at h0 = α • ex0 and endsat h1 = α.

s =(1 + t)/2

α

α ex0t = 0

t = 1

s = 0 s = 12 s = 1

(Work out the corresponding formula for [ex0 ][α] = [α] ∈ π1(X, x0).)

34

II. Inverses: [α][−α] = [ex0 ] ∈ π1(X, x0).Define a rel 0, 1 homotopy

h : α • −α ' ex0 : I → X

by

h : I × I → X ; (s, t) 7→

x0 if 0 6 s 6 t/2

α(2s− t) if t/2 6 s 6 12

α(2− 2s− t) if 126 s 6 1− t/2

x0 if 1− t/2 6 s 6 1 .

s =t/2

s =1− t/2

−αα

ex0

t = 0

t = 1

s = 0 s = 12 s = 1

Again, think what happens at each time t ∈ I: the path

ht : I → X ; s→ ht(s) = h(s, t)

is constant on [0, t/2], goes along the restriction α| : [0, 1−t]→ X (i.e. usingonly a part of α) at twice the speed on [t/2, 1

2], then along the restriction

−α| : [t, 1] → X at twice the speed on [12, 1 − t/2], and stays constant on

[1 − t/2, 1]. Note that α(1 − t) = −α(t) is essential for continuity. Thehomotopy h starts at h0 = α • −α and ends at h1 = ex0 . (Work out thecorresponding formula for [−α][α] = [ex0 ].)

III. Associativity of multiplication: ([α][β])[γ] = [α]([β][γ]) ∈ π1(X, x0).Let α, β, γ : I → X be paths which send each endpoint to x0 ∈ X. For

0 < λ < µ < 1 let

δ(λ, µ) = α •λ β •µ γ : I → X

be the triple concatenation of 4.6. From the definitions

([α][β])[γ] = δ(1/4, 12) : I → X ; s 7→

α(4s) if 0 6 s 6 1/4

β(4s− 1) if 1/4 6 s 6 12

γ(2s− 1) if 126 s 6 1

35

and

[α]([β][γ]) = δ(12, 3/4) : I → X ; s 7→

α(2s) if 0 6 s 6 1

2

β(4s− 2) if 126 s 6 3/4

γ(4s− 3) if 3/4 6 s 6 1 .

α

α

β

β

γ

γt = 0

t = 1

s = 0 s = 1/4 s = 12

s = 12

s = 3/4

s = 1

α β γ

s = (1 + t)/4

s = (2 + t)/4

Finally, construct a homotopy rel 0, 1

h : ([α][β])[γ] ' [α]([β][γ]) : I → X

byht = δ((1− t)/4 + t/2, (1− t)/2 + t(3/4))

= δ((1 + t)/4, (2 + t)/4) : I → X

with h0 = δ(1/4, 12), h1 = δ(1

2, 3/4).

Remark 5.6. The following table gives the fundamental groups of some path-connected spaces (which will be calculated in subsequent chapters):

X π1(X)Rn 1S1 Z

Sn (n > 2) 1RPn (n > 2) Z2

S1 × S1 Z⊕ Z8 Z ∗ Z

36

(i) π1(Rn) = 1 by the convexity of Rn.(ii) The fundamental group π1(S

1) = Z is generated by the identity loop1 : S1 → S1. The homotopy class of a loop f : S1 → S1 is the degree of f ,the number deg(f) ∈ Z introduced in Workshop 2.

(iii) The proof of π1(Sn) = 1 for n > 2 is easy once it is known that

every loop ω : S1 → Sn is homotopic to a non-surjective one, i.e. that aspace-filling closed path in Sn is homotopic to one which misses at least onepoint of Sn.

(iv) The n-dimensional projective space RPn is the quotient of Sn by theequivalence relation

x ∼ y if either x = y or x = −y ;

for n > 2 π1(RPn) = Z2 is the cyclic group of order 2, generated by thesquare root loop

σ : S1 → RPn ; (cos2πt, sin2πt) 7→ [cosπt, sinπt, 0, . . . , 0] .

(v) The fundamental group of the torus S1 × S1 is the free abelian groupon two generators π1(S

1 × S1) = Z⊕ Z. The generators (1, 0), (0, 1) are thebased homotopy classes of the meridian and longitude based loops:

(1, 0) : S1 → S1 × S1 ; x 7→ (x, 1) ,

(0, 1) : S1 → S1 × S1 ; y 7→ (1, y) .

(vi) The figure eight space 8 is the one-point union of two copies of S1.The fundamental group π1(8) = Z ∗ Z is the nonabelian free group on twogenerators. Elements of this group are all words on the alphabet a, b, a−1b−1,but cancelling expressions like aa−1. In this group, note that ab 6= ba, andaba−1b−1 6= 1. The generators of π1(8) are the based homotopy classes of thetwo based loops defined by the images of the two circles.

37

Lecture 7The fundamental group π1(X, x0) of a space X at a base point x0 ∈ X is

defined geometrically, in terms of paths α : I → X such that α(0) = α(1) =x0, or equivalently in terms of loops ω : S1 → X such that ω(1) = x0 ∈ X.A calculation of π1(X, x0) is an algebraic description. In general, it is quitedifficult to compute π1(X, x0), unless there is a geometric reason for it to bethe trivial group 1.

Definition 5.7. A based space (X, x0) is simply-connected if X is path-connected and π1(X, x0) = 1.

Example 5.8. The space X = x0 with a single point x0 ∈ X is simply-connected.

In Theorem 5.15 below it will be proved that if X, Y are homotopy equiva-lent path-connected spaces then the fundamental groups π1(X, x0), π1(Y, y0)are isomorphic, for any x0 ∈ X, y0 ∈ Y . In particular, the fundamental groupπ1(X, x0) of a contractible space X is trivial, so that X is simply-connected.

In dealing with the fundamental group π1(X, x0) of a path-connectedspace X it is usual to just write π1(X), since the the isomorphism class ofπ1(X, x0) is independent of the base point x0 ∈ X. (This is the special caseof Theorem 5.15 below with X = Y ).

A space determines a group. (In fact, every group arises as the fundamen-tal group of a space). A map of spaces determines a group homomorphism.

Proposition 5.9. (i) A map f : X → Y induces a group homomorphism

f∗ : π1(X, x0)→ π1(Y, f(x0)) ; [α] 7→ [fα]

for any base point x0 ∈ X.(ii) The identity map 1 : X → X induces the identity homomorphism

1∗ = 1 : π1(X, x0)→ π1(X, x0) .

(iii) The composite gf : X → Z of maps f : X → Y , g : Y → Z induces thecomposite group homomorphism

(gf)∗ = g∗f∗ : π1(X, x0)→ π1(Y, f(x0))→ π1(Z, gf(x0)) .

Proof. Easy consequences of the definitions!

Definition 5.10. Let X be a space with a base point x0 ∈ X. A mapf : X → Y is a homotopy equivalence rel x0 if there exists a mapg : Y → X such that g(f(x0)) = x0, a homotopy rel x0 denoted h : gf '1X : X → X (with h(x0, t) = x0 for t ∈ I) and a homotopy rel f(x0)denoted K : fg ' 1Y : Y → Y (with K(f(x0), t) = f(x0) for t ∈ I).

38

Proposition 5.11. (i) If f1, f2 : X → Y are maps which are related by a relx0 homotopy h : f1 ' f2 : X → Y then

(f1)∗ = (f2)∗ : π1(X, x0)→ π1(Y, f1(x0)) .

(ii) If f : X → Y is a homotopy equivalence rel x0 then f∗ is an isomor-phism, with inverse

(f∗)−1 = g∗ : π1(Y, f(x0))→ π1(X, x0) ,

for g as in Definition 5.10.

Proof. Easy consequences of the definitions!

In fact, the isomorphism class of the fundamental group π1(X, x0) is in-dependent of the choice of the base point x0 ∈ X within its path component.(Recall that the path component of x0 ∈ X is the set of all x ∈ X suchthat there exists a path γ : I → X from γ(0) = x0 to γ(1) = x.) Also,if f : X → Y is any homotopy equivalence (not necessarily rel x0) thenf∗ : π1(X, x0)→ π1(Y, f(x0)) is an isomorphism.

Proposition 5.12. (i) A path γ : I → X determines an isomorphism ofgroups

γ# : π1(X, γ(0))→ π1(X, γ(1)) ; [α] 7→ [−γ • α • γ]

with −γ : I → X; t 7→ γ(1 − t) the reverse path from −γ(0) = γ(1) to−γ(1) = γ(0). The reverse path −γ determines the inverse isomorphism

(−γ)# = (γ#)−1 : π1(X, γ(1))→ π1(X, γ(0)) .

(ii) If γ = ex0 is the constant path at x0 ∈ X (with γ(t) = x0) then[γ] = [ex0 ] ∈ π1(X, x0) is the unit element and

γ# = 1 : π1(X, x0)→ π1(X, x0) ; [α] 7→ [α]

is the identity automorphism.(iii) The isomorphism γ# depends only on the rel 0, 1 homotopy class

of γ.(iv) If γ = γ1 • γ2 : I → X is the concatenation of paths γ1, γ2 : I → X

with γ1(1) = γ2(0) ∈ X then

γ# = (γ2)#(γ1)# : π1(X, γ1(0))→ π1(X, γ2(1)) .

39

Proof. (i) Define a homotopy rel 0, 1

h : γ • −γ ' eγ(0) : I → X

(with ex0 the constant path at x0) by

h(s, t) =

γ(0) if 0 6 s 6 t/2

γ(2s− t) if t/2 6 s 6 12

γ(2− 2s− t) if 126 s 6 1− t/2

γ(0) if 1− t/2 6 s 6 1 .

It follows that if α, β : I → X are closed paths at γ(0) (i.e. such thatα(0) = α(1) = β(0) = β(1) = γ(0) ∈ X) then

γ#([α][β]) = γ#([α • β])

= [−γ • (α • β) • γ]

= [(−γ • α • γ) • (−γ • β • γ)]

= γ#([α])γ#([β]) ∈ π1(X, γ(1))

so that γ# preserves multiplications. Also

γ#[eγ(0)] = [−γ • eγ(0) • γ] = [eγ(1)] ∈ π1(X, γ(1)) ,

so γ# preserves the units. Therefore γ# is a group homomorphism.It follows from the existence of rel 0, 1 homotopies γ • −γ ' eγ(0) and

−γ • γ ' eγ(1) that γ#, −γ# are inverse isomorphisms.(ii) If γ is constant there exists a rel 0, 1 homotopy −γ • α • γ ' α.(iii) A rel 0, 1 homotopy δ : γ ' γ′ : I → X determines a rel 0, 1

homotopy

−δ • 1 • δ : − γ • α • γ ' γ′ • α • γ′ : I → X .

(iv) There exists a rel 0, 1 homotopy −γ ' −γ2•−γ1 and hence a rel 0, 1homotopy

−γ • α • γ ' −γ2 • (−γ1 • α • γ1) • γ2 : I → X .

Example 5.13. A closed path γ : I → X with γ(0) = γ(1) = x0 ∈ Xdetermines the conjugation automorphism

γ# : π1(X, x0)→ π1(X, x0) ; [α] 7→ [−γ • α • γ] = [γ]−1[α][γ]

(which is the identity if π1(X, x0) is abelian).

40

Lecture 8Proposition 5.14. Given maps F,G : X → Y , a homotopy H : F ' G :X → Y and a base point x0 ∈ X define a path

γ : I → Y ; t 7→ H(x0, t)

from γ(0) = F (x0) to γ(1) = G(x0), so that there is defined an isomorphism

γ# : π1(Y, F (x0))→ π1(Y,G(x0)) .

The induced homomorphisms of fundamental groups

F∗ : π1(X, x0)→ π1(Y, F (x0)) , G∗ : π1(X, x0)→ π1(Y,G(x0))

are such that

G∗ = γ#F∗ : π1(X, x0)→ π1(Y,G(x0)) .

Proof. For t ∈ I define the path γt : I → Y ; s → H(x0, st) from γt(0) =F (x0) to γt(1) = H(x0, t) = γ(t). For any closed path α : I → X withα(0) = α(1) = x0 ∈ X the triple concatenations

γt •1/3 (Htα) •2/3 (−γt) : I → Y (t ∈ I)

define a rel 0, 1 homotopy (eFx0)#F∗(α) ' (−γ)#G∗(α). By the inverseand identity relations, we deduce that G∗ ' γ#F∗

Theorem 5.15. If f : X → Y is a homotopy equivalence then

f∗ : π1(X, x0)→ π1(Y, f(x0))

is an isomorphism of groups for any base point x0 ∈ X.

Proof. Let g : Y → X be a homotopy inverse of f , so that there existhomotopies

h : 1X ' gf : X → X , h′ : 1Y ' fg : Y → Y .

Consider the group homomorphisms

π1(X, x0)f∗ // π1(Y, f(x0))

g∗ // π1(X, gf(x0))f ′∗ // π1(Y, fgf(x0))

41

where f ′ is also induced by f . Define the path γ : I → X; t→ h(x0, t) fromγ(0) = x0 to γ(1) = gf(x0). By Proposition 5.14 applied to the homotopyH = h : F = 1X ' G = gf : X → X

G∗ = g∗f∗ = γ# : π1(X, x0)→ π1(X, gf(x0))

is an isomorphism, so that f∗ : π1(X, x0)→ π1(Y, f(x0)) is one-one. Similarly,it follows from h′ : fg ' 1Y that f ′∗g∗ is an isomorphism, so that g∗ is alsoone-one. It follows from g∗f∗ being onto that g∗ is also onto, and hence anisomorphism. Finally, note that the composite

e = (g∗f∗)−1g∗ : π1(Y, f(x0))→ π1(X, gf(x0))→ π1(X, x0)

is an isomorphism such that ef∗ = 1 : π1(X, x0) → π1(X, x0), so that f∗ isan isomorphism with inverse (f∗)

−1 = e.

Example 5.16. A contractible space X is simply-connected; π1(X, x0) = 1for any base point x0 ∈ X.

Example 5.17. (i) If Y ⊆ X is a deformation retract (Definition 2.16), i.e.if the inclusion i : Y → X is a homotopy equivalence, then i∗ : π1(Y, y0) →π1(X, iy0) is an isomorphism for any y0 ∈ Y .

(ii) The maps

i = inclusion : Sn−1 → Rn\0 ,

j : Rn\0 → Sn−1 ; x 7→ x

‖x‖

are inverse homotopy equivalences, with ji = 1 : Sn−1 → Sn−1 and

h : Rn\0 × I → Rn\0 ; (x, t) 7→ (1− t) x‖x‖ + tx

defining a homotopy h : ij ' 1 : Rn\0 → Rn\0. Thus i induces anisomorphism

i∗ : π1(Sn−1) ∼= π1(Rn\0) .

Remark 5.18 (Non-examinable). For any spaceX the cone CX is contractible(and in particular path-connected) so π1(CX) = 1. For any equivalencerelation ∼ on X let p : X → X/∼ be the projection, let ∼ be the extendedequivalence relation on CX as in Proposition 2.27, with (x, 1) ∼ (x′, 1) ifx ∼ x′. The quotient space Y = CX/∼ is path-connected, but not in generalsimply-connected.

42

The van Kampen theorem (which we will see briefly and non-examinablyat the end of the course) shows that for path-connected X, the fundamentalgroup of Y is the quotient group

π1(Y ) = π1(X/∼)/N

with N the normal subgroup generated by im(p∗ : π1(X)→ π1(X/∼)).It should be clear that a loop ω : S1 → X × 1

2 ⊂ Y is homotopic to

the constant loop at the cone point of Y . This expression for π1(Y ) can beused to compute the fundamental groups of all the quotient spaces I2/∼ inChapter 8, with X = ∂I2 ∼= S1, CX ∼= I2 ∼= D2 (and indeed for all cellcomplexes). We shall use a different method in the course. The first step forboth methods is to compute π1(S

1) = Z.

5.1 Nonexaminable appendix on the higher homotopygroups

Remark 5.19. Let X be a space with a base point x0 ∈ X.(i) Let Ωx0(X) ⊂ XS1

be the subspace consisting of the maps ω : S1 → Xsuch that ω(1) = x0 ∈ X with 1 = (1, 0) ∈ S1, called the based loops atx0. The path-component set of Ωx0(X) is in bijective correspondence withthe fundamental group π1(X, x0) of homotopy classes of based loops

π0(Ωx0(X)) = π1(X, x0)

(about which more later) since a path I → Ωx0(X) is the same as a rel 1homotopy of based loops in X.

(ii) More generally, for any n > 1 let Ωx0(Sn, X) ⊂ XSn

be the subspaceof the maps ω : Sn → X such that ω(1) = x0 ∈ X with 1 = (1, 0, . . . , 0) ∈ Sn.The set

πn(X, x0) = π0(Ωx0(Sn, X))

can be identified with the set of homotopy classes of such maps ω keepingthe image of 1 ∈ Sn fixed at x0 ∈ X. The set πn(X, x0) can be given a groupstructure, which is abelian for n > 2; a map of based spaces f : (X, x0) →(Y, y0) induces group homomorphisms f∗ : πn(X, x0) → πn(Y, y0) (n > 2)which are isomorphisms if f : X → Y is a homotopy equivalence, just likefor n = 1.

(iii) The standard proof of Brouwer’s theorem on the topological invari-ance of dimension (that Rm is homeomorphic to Rn if and only if m = n)uses the one-point compactification (Rm)∞ = Sm. It is immediate from the

43

computation

πn(Sm, 1) =

Z if n = m

0 if n < m

that Sm is homeomorphic to Sn if and only if m = n. (The computation ofπn(Sm, 1) for n 6 m = 1 will be worked out in the course, using the Z-valueddegree of maps f : S1 → S1 for n = m = 1. The general case n < m, m > 2is not much more difficult, using the Z-valued degree of maps f : Sn → Sn forn = m). If n < m and Sn were homeomorphic to Sm then the infinite cyclicgroup πn(Sn, 1) = Z would be isomorphic to the zero group πn(Sm, 1) = 0 –a contradiction!

44

Lecture 9

6 Compact surfaces and their classification

We will now look at more examples of topological spaces with interestingfundamental groups, namely compact surfaces. Although we are not in aposition to compute the fundamental groups yet, by taking them on trustwe’ll see that they can be used to distinguish between different surfaces.

Definition 6.1. (i) A Hausdorff topological spaceM is called an n-dimensional(topological) manifold if it admits a countable cover by open subsets U ⊆Msuch that each U is homeomorphic to the Euclidean n-space Rn.

(ii) A 2-dimensional manifold is called a surface.

It is true, but not obvious, that if m 6= n then an m-dimensional manifoldcannot be homeomorphic to an n-dimensional manifold. This follows fromthe invariance of dimension: the Euclidean spaces Rm,Rn are homeomorphicif and only if m = n.

Example 6.2. (i) Rn is an n-dimensional manifold: take U = Rn !(ii) Sn is an n-dimensional manifold.(iii) The torus T 2 and the Klein bottle K are surfaces.

Definition 6.3. 1. A surface M is orientable if there is no subspaceN ⊂M which is homeomorphic to a Mobius band.

2. A surface M is nonorientable if there is a subspace N ⊂M which ishomeomorphic to a Mobius band.

Example 6.4. 1. The plane R2, the sphere S2 and the torus T 2 are ori-entable. (This is not obvious).

2. The Klein bottle K and the projective plane RP2 are not orientable.(Although not exactly obvious, it is much easier to verify that there isa subspace which is homeomorphic to a Mobius band than that thereis no such subspace).

Remark 6.5. A surface with boundary (M,∂M) is a space M togetherwith a subspace ∂M (the boundary) such that M\∂M is a surface (theinterior), and every x ∈ ∂M has an open neighbourhood U ⊂M such thatU is homeomorphic to the upper half-plane R2

+ = (y, z) ∈ R2 | z ≥ 0, withU ∩ ∂M homeomorphic to the real line.

For example, (D2, S1) is a surface with boundary. Also, The Mobius bandM defines a surface with boundary (M,S1).

45

Definition 6.6. Let g ≥ 0 be an integer. The sphere with g handles H(g)(or the g-holed torus) is the orientable surface obtained from the 2-sphereS2 by punching out 2g disjoint copies of D2 and joining up the 2g holes byg handles, each a copy of the cylinder S1 × [0, 1].

Example 6.7. 1. H(0) = S2.

2. H(1) = T 2 = S1 × S1, the 2-torus.

3. Here are two pictures of H(2) (taken from the books of Armstrong andStillwell):

4. Here are the orientable surfaces H(g) of genus g = 0, 1, 2, 3, . . . :

Poincare’s fundamental polygons give us a way to study surfaces, bydescribing them as quotient spaces of polygons with respect to an equivalencerelation on the boundary. Here are the diagrams for H(1), H(2), H(3), takenfrom Hatcher’s book:

46

Here are pictures, taken from Stillwell’s book, showing how the corre-spondence works for H(2):

There is an algebraic convention for summarising the equivalence rela-tions. Starting from the dot, the expressions for H(1) and H(2) as aba−1b−1

and aba−1b−1cdc−1d−1.In this vein, each space H(g) can be obtained as a quotient space of a

(4g)-gon, where the equivalence relation on the boundary corresponds to the

47

expressiona1b1a

−11 b−11 a2b2a

−12 b−12 · · · agbga−1g b−1g .

These descriptions of the equivalence relations for the fundamental polygonsare very closely related to the fundamental groups of these surfaces!

If we puncture H(g), can you see what the result is homotopy equivalentto?

Definition 6.8. Let g ≥ 1 be an integer. The sphere with g cross-capsM(g) is the nonorientable surface obtained from the 2-sphere S2 by punchingout g disjoint copies of D2 and replacing each of the g holes by glueing in gcopies of the Mobius band.

There also exist fundamental polygons for these non-orientable surfaces.The space M(g) can be obtained as a quotient space of a (2g)-gon, with theequivalence relation on the boundary corresponding to the expression

a1a1a2a2 · · · ag.

If we puncture M(g), can you see what the result is homotopy equivalentto?

Theorem 6.9 (Classification theorem for compact surfaces (non-examinable)).

1. Every connected compact orientable surface M is homeomorphic toH(g) for a unique g ≥ 0.

2. Every connected compact nonorientable surface M is homeomorphic toM(g) for a unique g ≥ 1.

3. Connected compact surfaces M,M ′ are homeomorphic if and only ifthere exists a group isomorphism π1(M) ∼= π1(M

′).

Proof. See Armstrong (pp. 16–18, Chapter 7).

Example 6.10.

1. A connected compact surface M is homeomorphic to S2 if and only ifπ1(M) = 1.

2. A connected compact surface M is homeomorphic to RP2 if and onlyif π1(M) = Z2.

Every connected compact surface M is homeomorphic to either H(g)for some g ≥ 0 (if M is orientable) or to M(g) for some g ≥ 1 (if M isnonorientable).

48

Definition 6.11. The number g is the genus of M .

Roughly speaking, g is the number of ‘holes’ in M .How does the fundamental group π1(M) determine the genus?

Definition 6.12. 1. The commutator of any two elements a, b ∈ G ina group G is

[a, b] = aba−1b−1 ∈ G .

2. The abelianisation of a group G is the abelian quotient group

Gab = G/[G,G] ,

with [G,G] / G the normal subgroup generated by the commutators[a, b] (a, b ∈ G).

The appendix of Armstrong (pp. 241–243) is a brief account of the pre-sentation of groups in terms of generators and relations. For a more detailedaccount see pp.40–51 of Stillwell’s Classical topology and combinatorialgroup theory (51.57 Sti).

Proposition 6.13 (Non-examinable). We have:

1. The abelianisation of π1(H(g)) is the free abelian group of rank 2g

π1(H(g))ab = Z2g .

2. The abelianisation of π1(M(g)) is the quotient of the free abelian groupZg by the cyclic subgroup generated by (2, 2, . . . , 2)

π1(M(g))ab = Zg/(2, 2, . . . , 2) ∼= Zg−1 ⊕ Z2 .

Proof. See Armstrong p.168 and Stillwell, pp.82–84.

Note that Proposition 6.13 only describes the abelianisations of the fun-damental groups. We will not describe the fundamental groups themselvesuntil the end of the course.

49

Lecture 10

7 The fundamental group of Sn is π1(Sn) = 1

for n > 2

For any n > 1, a path α : I → Sn can be space-filling, i.e. onto, withα(I) = Sn (Peano, 1890). We will show that every path α : I → Sn ishomotopic rel 0, 1 to the concatenation of paths which are not onto. Forn > 2 it follows that α itself is homotopic rel 0, 1 to a path which is notonto, which suffices to prove that π1(S

n) = 1 for n > 2. This is definitelynot true for X = S1, but at least every path α : I → S1 is homotopicrel 0, 1 to the concatenation of paths which are not onto, allowing thecomputation π1(S

1) = Z (in the next section).

Lemma 7.1. The spaces

U+ = Sn − (0, . . . , 0, 1) = (x1, x2, . . . , xn+1) ∈ Sn |xn+1 6= 1,U− = Sn − (0, . . . , 0,−1) = (x1, x2, . . . , xn+1) ∈ Sn |xn+1 6= −1 .

are both homeomorphic to Rn.

Proof (non-examinable). Explicit homeomorphisms are given by the stereo-graphic projection maps

φ+ : U+ → Rn ; x = (x1, x2, . . . , xn+1) 7→ φ+(x) =1

1− xn+1

(x1, x2, . . . , xn) ,

φ− : U− → Rn ; x = (x1, x2, . . . , xn+1) 7→ φ−(x) =1

1 + xn+1

(x1, x2, . . . , xn) .

(The formula for φ+(x) is obtained by vector analysis, from the requirementthat the vectors (0, . . . , 0, 1), x, (φ+(x), 0) ∈ Rn+1 be collinear:

(φ+(x), 0) = sx+ (1− s)(0, . . . , 0, 1) with s =2

1− xn+1

.

Similarly for φ−(x).) Their inverses are

(φ+)−1(a1, . . . , an) = (2a1, 2a2, . . . , 2an,∑a2i − 1)/(1 +

∑i a

2i )

(φ−)−1(a1, . . . , an) = (2a1, 2a2, . . . , 2an, 1−∑a2i )/(1 +

∑i a

2i ).

50

We now introduce a useful result in General Topology, the Lebesgue Cov-ering Lemma:

Lemma 7.2. Let X be a compact metric space and Uλ | λ ∈ Λ an opencover of X. Then there exists a positive number δ > 0 (the “Lebesgue num-ber” of the cover) such that for all x ∈ X, B(x, δ) lies entirely inside somesingle Uλ.

Proof (Non-examinable). By compactness there is a finite subcover Uλ1 , . . . , Uλn .Let

fj(x) = d(x,X \ Uλj).Then fj is a continuous real-valued function which, since X \ Uλj is closed,is zero if and only if x /∈ Uλj . Since the Uλj cover X, for every x at least onefj(x) > 0. So if we set

f(x) = max16j6n

fj(x)

we have that f is continuous and f(x) > 0 on X.Thus the function x 7→ f(x)−1 is continous, hence has an upper bound

M by compactness of X, so f(x) ≥ M−1 for all x ∈ X. This means thatfor all x ∈ X there exists a Uλj such that d(x,X \ Uλj) ≥ M−1, and soB(x,M−1) ⊆ Uλj .

Note that compactness is necessary for the Lebesgue Covering Lemma:it does not hold for (0, 1) with cover (0, 1).

Proposition 7.3. (i) Every map α : [s0, s1] → Rn is homotopic rel s0, s1to a linear map

β : [s0, s1]→ Rn ; s 7→ (s1 − s)α(s0) + (s− s0)α(s1)

s1 − s0.

with image the straight line segment from α(s0) to α(s1) ∈ Rn

β([s0, s1]) = [α(s0), α(s1)] ⊂ Rn .

(ii) For n > 2 every path α : I → Sn is homotopic rel 0, 1 to a pathβ : I → Sn which is not onto.

Proof (details non-examinable). (i) Straightforward: define a rel 0, 1 ho-motopy

γ : α ' β : I → Rn

by a straight line homotopy

γ : I × I → Rn ; (s, t) 7→ (1− t)α(s) + tβ(s) .

51

(ii) Let U = U+, U− be the open cover of Sn

Sn = U+ ∪ U−

defined by the complements of two antipodal points.The unit line I = [0, 1] is a compact metric space and the open subsets

α−1(U+), α−1(U−) define an open cover F of I. For any integer N > 0define a sequence t0, . . . , tN by ti = i

Nso the intervals [ti, ti+1] have diameter

1/N and subdivide I.By the Lebesgue Covering Lemma there is an N > 0 so large that each

[ti, ti+1] is contained in some U ∈ F , and so α[ti, ti+1] is contained in eitherU+ or U− (possibly both).

If α[ti, ti+1] ⊂ U+, then as in (i), the composite

φ+ α|[ti,ti+1] : [ti, ti+1]→ Rn

with the stereographic projection map is homotopic rel ti, ti+1 to a linearmap.

Applying φ−1+ , this gives us a homotopy γi rel ti, ti+1 between α|[ti,ti+1]

and a map βi : [ti, ti+1]→ U+ for which φ+ βi is linear.On the other hand, if α[ti, ti+1] 6⊂ U+, then α[ti, ti+1] ⊂ U−, and there is

a homotopy γi rel ti, ti+1 between α|[ti,ti+1] and a path βi which becomes astraight line segment under the stereographic projection φ−.

Concatenating these gives a ‘piecewise-linear’ path

β : I → Sn ; s 7→ βi(s) if ti 6 s 6 ti+1

and a ‘piecewise-linear’ homotopy rel 0, 1

γ : I × I → Sn ; (s, t) 7→ γi(s, t) (if ti 6 t 6 ti+1) .

between α and β.It remains to show that β is not onto. Consider the union of the inverse

stereographic projections

f = (φ+)−1 ∪ (φ−)−1 : Rn ∪ Rn → Sn .

For n > 2, Sn is not the image f(C1 ∪C2) with C1, C2 ⊂ Rn subsets definedby the union of a finite collection of line segments. (One way to see this isthat C1 ∪ φ+(φ−)−1(C2 \ 0) is a finite union of lines and circles.)

Note that the proof of Proposition 7.3 (ii) is quite false for n = 1: S1 isa union of two doubly-overlapping arcs, so the map

f = (φ+)−1 ∪ (φ−)−1 : R ∪ R→ S1

can send a union of two line segments (one in each R) onto S1.

52

Theorem 7.4. π1(Sn) = 1 for n > 2.

Proof. Since Sn is path-connected there is no need to specify a base point,although it is conventional to select 1 = (1, 0, . . . , 0) ∈ Sn as the base point.Represent an arbitrary element [α] ∈ π1(Sn, 1) by a path α : I → Sn suchthat

α(0) = α(1) = 1 ∈ Sn .

By Proposition 7.3(ii), α is homotopic rel 0, 1 to a path β : I → Sn

which is not onto. Choose x ∈ Sn\β(I), noting that x 6= 1 ∈ Sn andβ(I) ⊆ Sn\x. The inclusion i : Sn\x → Sn induces a homomorphism ofgroups

i∗ : π1(Sn\x, 1)→ π1(S

n, 1)

such thati∗[β] = [iβ] = [α] ∈ π1(Sn, 1) .

But Sn\x is homeomorphic to Rn, so it is contractible (= homotopyequivalent to a point) and

π1(Sn\x, 1) = 1 .

Thus [β] = 1 ∈ π1(Sn\x, 1) and

[α] = i∗(1) = 1 ∈ π1(Sn, 1) ,

so that π1(Sn, 1) = 1.

Corollary 7.5. For n > 3

π1(Rn\0) = 1 .

Proof. By Example 5.17(ii) π1(Rn\0) ∼= π1(Sn−1), for any n > 1.

53

Lecture 11

8 The fundamental group of the circle is π1(S1) =

ZWe shall prove that π1(S

1) = Z by defining and studying the homotopytheoretic properties of the degree of a loop ω : S1 → S1, which is an integerdegree(ω) ∈ Z counting the number of anticlockwise turns of ω.

Regard the circle as the space of complex numbers of unit length

S1 = z ∈ C | |z| = 1 .

Proposition 8.1. For each N ∈ Z let

ωN : S1 → S1 ; z 7→ zN

be the map winding the circle around itself N times in the anticlockwisedirection, with ωN(1) = 1. The function

Z→ π1(S1) ; N 7→ [ωN ]

is a homomorphism of groups.

Proof. It suffices to show that for any N ∈ Z we have [ωN ] = [ω1]N , since

that automatically implies [ωM+N ] = [ω1]M · [ω1]

N .The composite of the projection

p : I → S1 ; t 7→ e2πit = cos(2πt) + isin(2πt)

and ωN is the closed path

αN = ωN p : I → S1 ; t 7→ e2πiNt = cos(2πNt) + isin(2πNt)

with αN(0) = αN(1) = 1.For N > 1,

[αN ] = α1 •1/N α1 •2/N α1 · · · •(N − 1)/N α1 : I → S1

in the terminology defined above, corresponding to the partition of I into N(equal) subintervals

I = [0, 1/N ] ∪ [1/N, 2/N ] ∪ · · · ∪ [(N − 1)/N, 1] .

54

By Proposition 4.7, the rel 0, 1 homotopy class of a concatenation isindependent of the choice of partition of I = [0, 1], so

[ωN ] = [αN ] = [α1]N = (ω1)

N ∈ π1(S1) .

To generalise this to negative integers, we note that α−N is the reverse ofαN

α−N = αN : I → S1 ; t 7→ e−2πiNt = e2πiN(1− t),

so[ω−N ] = [ωN ]−1 = ([ω1]

N)−1 = [ω1]−N .

We shall prove that Z→ π1(S1);N 7→ [ωN ] is an isomorphism of groups,

using the topology of ‘angle maps’ to define the inverse isomorphism

degree : π1(S1)→ Z ; ω 7→ degree(ω)

which counts the number of times a loop ω : S1 → S1 winds around S1 in theanticlockwise sense. For differentiable ω the degree can be computed quiteeasily by counting (with signs) the number of points in ω−1(z) for any z ∈ S1

with ω−1(z) finite (see Remark 8.9 below).In general ω is not differentiable, although it is homotopic rel 1 to a

differentiable loop, so we take a different approach. Our definition of degreewill make use of the projection

p : R→ S1 ; x 7→ e2πix

and its key properties:

(i) For every x ∈ R the open subset U = (x, x + 1) ⊂ R is such that prestricts to a homeomorphism

p| : U → p(U) ; u 7→ p(u)

with p(U) = S1\e2πix.

(ii) p is onto, with the inverse image of y = p(x) ∈ S1 given by

p−1(y) = x+ n |n ∈ Z ⊂ R .

Any real numbers x0, x1 ∈ R with p(x0) = p(x1) ∈ S1 differ by aninteger x0 − x1 ∈ Z ⊂ R.

55

We shall show that for every path α : I → S1 there exists a ‘lift’ pathθ : I → R such that α = pθ : I → S1. If α is closed

pθ(0) = α(0) = α(1) = pθ(1) ∈ S1 ,

allowing the degree of α to be defined by

degree(α) = θ(1)− θ(0) ∈ Z .

Definition 8.2. An angle map or lift of a map α : [t0, t1] → S1 is a mapθ : [t0, t1]→ R such that

α(t) = e2πiθ(t) ∈ S1 (t ∈ [t0, t1]) ,

with θ(t0) = θ0 ∈ R the initial angle and θ(t1) = θ1 ∈ R the terminalangle.

It is an essential ingredient of the definition that the angle map θ :[t0, t1] → R be continuous, i.e. that θ be a map. The angle map fits into acommutative diagram

Rp

[t0, t1]

θ<<

α // S1

Proposition 8.3. Given a map α : [t0, t1]→ S1 and θ0 ∈ R such that

α(t0) = e2πiθ0 ∈ S1

there is a unique angle map θ : [t0, t1]→ R for α with initial angle

θ(t0) = θ0 ∈ R .

Proof (details non-examinable). Consider first the special case when α is notonto. Choose an x ∈ R such that

e2πix ∈ S1\α([t0, t1]) , θ0 ∈ (x, x+ 1) ,

and note that the restriction of p defines a homeomorphism

q = p| : (x, x+ 1)→ S1\e2πix ; y 7→ e2πiy

such thatq(θ0) = e2πiθ0 = α(t0) ∈ S1 .

56

The composite

θ : [t0, t1]α // S1\e2πix

q−1 // (x, x+ 1) incl. // R

is an angle map θ with θ(t0) = θ0 ∈ R.If θ′ : [t0, t1] → R is another angle map such that θ′(t0) = θ0 ∈ R then

θ′([t0, t1]) ⊂ (x, x+ 1) and it is immediate from the commutative triangle

(x, x+ 1)

q

[t0, t1]

θ′88

α // S1\e2πix

thatθ′ = q−1 α = θ : [t0, t1]→ R .

Next, consider the general case of a map α : [t0, t1]→ S1 which may notbe onto, with t0 < t1. As in the proof of Proposition 7.3 (ii) define an opencover U = U+, U− of S1 by

U+ = S1\1 , U− = S1\−1 ,

so that α−1U = α−1(U+), α−1(U−) is an open cover of the compact metricspace [t0, t1].

By the Lebesgue Covering Lemma there exists an integer N > 1 so largethat for

sj := t0 + j(t1−t0)N

,

the N subintervals [sj, sj+1] ⊂ [t0, t1] have diameter (t1− t0)/N so small that

[sj, sj+1] ⊂ α−1(U+) or α−1(U−) (0 6 j 6 N − 1) .

None of the restrictions α|[sj ,sj+1] is onto. Apply the special case to obtainan angle map θ|[s0,s1] : [s0, s1] → R for α|[s0,s1] with initial angle θ0. Next,apply the special case to obtain an angle map θ|[s1,s2] : [s1, s2] → R forα|[s1,s2] with initial angle θ|[s0,s1](s1) ∈ R the previous terminal angle. Andso on, until each α|[sj ,sj+1] has an angle map θ|[sj ,sj+1] → R with initial angleθ|[sj ,sj+1](sj) = θ|[sj−1,sj ](sj) ∈ R the previous terminal angle.

Piecing together the angle maps θ|[sj ,sj+1] for α|[sj ,sj+1], we obtain an anglemap

θ : [t0, t1]→ R ; t 7→ θ|[sj ,sj+1](t) if sj 6 t 6 sj+1

for α with initial angle θ0.

57

Lecture 12Definition 8.4. (i) The degree of a closed path α : I → S1 is the differencebetween the terminal and initial angles in any angle map θ : I = [0, 1] → Rfor α

degree(α) = θ(1)− θ(0) ∈ Z ,

using α(0) = α(1) ∈ S1 to ensure that this is an integer.(ii) The degree of a loop ω : S1 → S1 is

degree(ω) = degree(α) ∈ Z

with α the closed path

α : I → S1 ; t 7→ ω(e2πit) .

Example 8.5. The closed path defined for N ∈ Z by

αN : I → S1 ; t 7→ e2πiNt

has angle mapθN : I → R ; t 7→ Nt

so that the degree is

degree(αN) = θN(1)− θN(0) = N ∈ Z ⊂ R .

The loop

ωN : S1 → S1 ; z = e2πit 7→ zN = e2πiNt

thus hasdegree(ωN) = N ∈ Z .

Example 8.6. The antipodal map is a loop

ω : S1 → S1 ; z 7→ −z .

The corresponding closed path

α : I → S1 ; t 7→ (cos (2π(t+ 12)), sin (2π(t+ 1

2)))

has angle mapθ : I → R ; t 7→ t+ 1

2

sodegree(ω) = degree(α) = θ(1)− θ(0) = 1 ∈ Z .

58

We shall also need angle maps for homotopies:

Proposition 8.7. (i) Let α, β : [t0, t1]→ S1 be paths related by a homotopyγ : α ' β : [t0, t1]→ S1. Given an initial angle θ0 for α there exists a uniquehomotopy ψ : θ ' φ : [t0, t1]→ R such that

γ(s, t) = e2πiψ(s, t) ∈ S1 , ψ(t0, 0) = θ0 ∈ R (s ∈ [t0, t1], t ∈ I) ,

with

θ : [t0, t1]→ R ; s 7→ ψ(s, 0) , φ : [t0, t1]→ R ; s 7→ ψ(s, 1)

angle maps for α, β

α(s) = e2πiθ(s) , β(s) = e2πiφ(s) (s ∈ [t0, t1]) .

(ii) Ifα(t0) = β(t0) , α(t1) = β(t1) ∈ S1

and γ : α ' β is a homotopy rel t0, t1 then ψ is a homotopy rel t0, t1.

Proof (details non-examinable). (i) As for angle maps in Proposition 8.3.First consider the special case when γ is not onto: choose x ∈ R such that

e2πix ∈ S1\γ([t0, t1]× I) , θ0 ∈ (x, x+ 1) ,

and define ψ = q−1 γ to fit into a commutative triangle

(x, x+ 1)

q

[t0, t1]× I

ψ77

γ // S1\e2πix

with ψ(t0, 0) = θ0 ∈ R.Next, consider the general case of a homotopy γ : [t0, t1]× I → S1 which

may not be onto, with t0 < t1. Define an open cover U = U+, U− of S1 by

U+ = S1\1 , U− = S1\−1 ,

so that γ−1U = γ−1(U+), γ−1(U−) is an open cover of the compact metricspace [t0, t1]× I.

By the Lebesgue Covering Lemma there exists a Lebesgue number N > 1so large that the N2 subrectangles

[sj, sj+1]× [ kN, k+1N

] ⊂ [t0, t1]× I,

59

for sj := t0 +j(t1 − t0)

Nhave diameter

√1 + (t1 − t0)2/N so small that

[sj, sj+1]× [ kN, k+1N

] ⊂ α−1(U+) or α−1(U−) (0 6 j, k 6 N − 1) .

None of the restrictions

γ|[sj ,sj+1]×[

kN,k+1N

]: [sj, sj+1]× [ k

N, k+1N

]→ S1 (0 6 j, k 6 N − 1)

is onto, so we may apply the special case N2 times and piece them togetherto obtain ψ.

(ii) A special case of (i), noting that the functions

I → Z ; t 7→ ψ(ti, t)− ψ(ti, 0) (i = 0, 1)

are continuous, taking the value 0 at t = 0, so that

ψ(ti, t) = ψ(ti, 0) ∈ R (i = 0, 1, t ∈ [0, 1])

and ψ is a homotopy rel t0, t1.

Theorem 8.8. The fundamental group of the circle S1 is the infinite cyclicgroup, with inverse group homomorphisms

Z→ π1(S1) ; N 7→ ωN ,

degree : π1(S1)→ Z ; ω 7→ degree(ω) .

Proof. In the first instance, let us check that degree(ω) ∈ Z depends only onthe rel 1 homotopy class of a loop ω : S1 → S1 based at 1 ∈ S1, i.e. thatπ1(S

1)→ Z is well-defined.Suppose that α, β : I → S1 are closed paths such that

α(0) = β(0) = α(1) = β(1) = 1 ∈ S1

and[α] = [β] = [ω] ∈ π1(S1) .

There exists a rel 0, 1 homotopy γ : I × I → S1 with the closed paths

γt : I → S1 ; s 7→ γ(s, t) (s ∈ I)

such thatγt(0) = γt(1) = 1 , γ0 = α , γ1 = β .

60

By Proposition 8.7 there exists an angle map ψ : I × I → R for γ suchthat

γt(s) = γ(s, t) = e2πiψ(s, t) ∈ S1 (s, t ∈ I)

with ψ(0, 0) = 0. Thus for each t ∈ I there is defined an angle map

ψt : I → R ; s 7→ ψ(s, t)

for γt, anddegree(γt) = ψt(1)− ψt(0) ∈ Z .

(In fact, ψt(0) = 0). The degree function

I → Z ; t 7→ degree(γt)

is continuous, I is connected, and Z is disconnected. The degree functionmust therefore be constant, and

degree(α) = degree(γ0) = degree(γ1) = degree(β) ∈ Z .

Next, we verify that degree : π1(S1) → Z is a group homomorphism: if

α, β : I → S1 are closed paths with angle maps θ, φ : I → R such thatθ(0) = φ(0) = 0 then

degree(α) = θ(1) , degree(β) = φ(1) ∈ Z .

The concatenation α • β : I → S1 has angle map

λ : I → R ; t 7→

θ(2t) if 0 6 t 6 1

2

θ(1) + φ(2t− 1) if 126 t 6 1.

The degree of α • β is

degree(α • β) = λ(1)

= θ(1) + φ(1)

= degree(α) + degree(β) ∈ Z ,

so degree is a group homomorphism. For each n ∈ Z the standard path αndefined above has degree(αn) = n, so degree is surjective.

It remains to prove that degree is injective. Let then α : I → S1 be aclosed path with α(0) = α(1) = 1, such that degree(α) = 0 ∈ Z. There existsan angle map θ : I → R for α with θ(0) = θ(1) = 0 ∈ Z, and the map

H : I × I → S1 ; (s, t) 7→ e2πiθ(s)t

61

defines a rel 0, 1 homotopy H : α0 ' α : I → S1, with

H(s, 0) = e2πi0 = 1 = α0(s) , H(s, 1) = e2πiθ(s) = α(s) ,

H(0, t) = e2πiθ(0)t = 1 , H(1, t) = e2πiθ(1)t = 1 .

So [α] = [α0] = [e1] = 1 ∈ π1(S1) is the identity element, and degree is

injective.

Remark 8.9 (Non-examinable). It is possible to compute degree(α) ∈ Z fora differentiable closed path

α : I → S1 ; t 7→ (α1(t), α2(t))

as follows. The tangent vector at t ∈ I

α′(t) = (α′1(t), α′2(t)) ∈ R2

is orthogonal to α(t), as is clear geometrically, or by differentiating

α1(t)2 + α2(t)

2 = 1

to obtain

2(α1(t)α′1(t) + α2(t)α

′2(t)) = 2α(t) · α′(t) = 0 ∈ R .

Define a sign

ε(t) =

+1

−1

0

if

α1(t)α

′2(t) > α2(t)α

′1(t)

α1(t)α′2(t) < α2(t)α

′1(t)

α1(t)α′2(t) = α2(t)α

′1(t)

according to whether α′(t) is pointing anticlockwise/clockwise, or is 0. Thenfor any z ∈ S1 such that α−1(z) ⊂ I is finite

degree(α) =∑

t∈α−1(z)

ε(t) ∈ Z .

62

Lecture 13Example 8.10. Consider the Mobius band

M = I × I/(0, t) ∼ (1, 1− t) .

(i) The inclusion of the middle circle

f1 : S1 →M ; e2πis 7→ [s, 12]

is a homotopy equivalence, inducing an isomorphism of groups (f1)∗ : π1(S1)→

π1(M) with inverse (f1)−1∗ = g∗ induced by the homotopy inverse

g : M → S1 ; [s, t] 7→ e2πis .

(ii) The inclusion of the boundary circle

f2 : S1 →M ; e2πis 7→

(2s, 0) if 0 6 s 6 1

2

(2s− 1, 1) if 126 s 6 1

is not a homotopy equivalence: the composite

gf2 : S1 → S1 ; z = e2πis 7→ z2 = e4πis

has degree 2, so that the induced homomorphism of groups

(f1)−1∗ (f2)∗ = (gf2)∗ = 2 : π1(S

1) = Z→ π1(S1) = Z

is not an isomorphism.

Remark 8.11. (i) The definition of degree(ω) ∈ Z works for any loop ω :S1 → S1 (or closed path α : I → S1), without reference to a base point inS1. The proof of Theorem 8.8 is easily extended to also prove that the degreedepends only on the homotopy class of ω, and that two loops ω, ω′ : S1 → S1

are homotopic if and only if

degree(ω) = degree(ω′) ∈ Z .

(ii) In particular, if a loop ω : S1 → S1 extends to a map δω : D2 → S1 thendegree(ω) = 0 ∈ Z.

Example 8.12. For any map f : C→ C\0 the loop

ω : S1 → S1 ; z 7→ f(z)/‖f(z)‖

extends to the map

δω : D2 → S1 ; z 7→ f(z)/‖f(z)‖ ,

identifying D2 = z ∈ C | ‖z‖ 6 1. It follows from Remark 8.11 (ii) thatdegree(ω) = 0 ∈ Z.

63

Definition 8.13. The winding number around 0, denoted W (α) ∈ Z, ofa closed path

α : I → C\0 (α(0) = α(1))

is defined to be the degree of the loop

ω : S1 → S1 ; e2πit 7→ α(t)|α(t)| ,

that isW (α) = degree(ω : S1 → S1) = θ(1)− θ(0) ∈ Z

for any angle map θ : I → R such that

α(t) = |α(t)|e2πiθ(t) ∈ C\0 (t ∈ I) .

Remark 8.14 (Non-examinable). (i) For analytic α the winding number canbe computed as the complex contour integral in Cauchy’s theorem

W (α) = 12πi

∮ω

dzz.

It is possible to prove directly that the winding number function

π1(C\0, 1)→ Z ; [α] 7→ W (α)

is an isomorphism.However, a better way would be to use the calculation π1(S

1) = Z ofTheorem 8.8 and the isomorphism π1(S

1) ∼= π1(R2\0) given by Example5.17(ii). The inclusion S1 → R2\0 = C\0 is a homotopy equivalencewith homotopy inverse

j : C\0 → S1 ; z 7→ z|z|

inducing an isomorphism

j∗ : π1(C\0)→ π1(S1) = Z ; α 7→ W (α) .

(ii) The winding number W (α) ∈ Z around 0 of a differentiable closedpath α : I → C\0 can be calculated by looking at each crossing pointof the curve and a radius drawn from 0 ∈ C to which the curve is nottangent, assigning +1 (resp. −1) where the curve is turning anticlockwise(resp. clockwise) and adding up all these +1’s and −1’s, in the manner ofRemark 8.9.

64

Theorem 8.15. (The Fundamental Theorem of Algebra) For n > 1, a degreen polynomial with complex coefficients

p(z) =n∑k=0

akzk (ak ∈ C, an 6= 0)

has at least one root, an element z ∈ C such that p(z) = 0 ∈ C.

Proof (details non-examinable). Replacing p(z) by p(z)/an we can take an =1. Assume that p(z) 6= 0 for all z ∈ C. The loop

p : S1 → C\0 ; z 7→ p(z)

extends to a map

δp : D2 = z ∈ C | |z| 6 1 → C\0 ; z 7→ p(z)

so that W (p) = 0 (by Example 8.12). We shall obtain a contradiction to theassumption that p(z) 6= 0 for all z ∈ C by proving that W (p) = n 6= 0. Forany r > 0 define

qr : C\0 → C\0 ; z 7→ rz .

The following analysis ensures that for sufficiently large r > 0 the map

h : S1 × I → C ; (z, t) 7→ (1− t)p(rz) + t(rz)n

never takes the value 0, in which case it defines a homotopy

h1 : pqr ' (qr)n : S1 → C\0

with(qr)

n : S1 → C\0 ; z 7→ (rz)n .

The maps defined for any r > 0 by

h2 : S1 × I → C\0 ; (z, t) 7→ p((1− t+ rt)z) ,

h3 : S1 × I → C\0 ; (z, t) 7→ ((1− t+ rt)z)n

are homotopies

h2 : p ' pqr , h3 : ωn ' (qr)n : S1 → C\0 ,

so that there is a concatenation homotopy

h2 • h1 • −h3 : p ' ωn : S1 → C\0

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andW (p) = W (ωn) = n ∈ Z .

Now, for r > 0 sufficiently large and any z ∈ S1

|(rz)n − p(rz)| = |n−1∑k=0

ak(rz)k| 6n−1∑k=0

|ak|rk < rn = |(rz)n|

and by the triangle inequality

0 < |(rz)n| − (1− t)|(rz)n − p(rz)|6 |(1− t)(p(rz)− (rz)n) + (rz)n| = |(1− t)p(rz) + t(rz)n| .

Remark 8.16 (Non-examinable). The image of an injective map i : S1 → R2

is a closed curveK = i(S1) ⊂ R2 .

The Jordan curve theorem (Google it!) proves that the complement R2\Khas two path-components, a one inside K with compact closure, and oneoutside K with non-compact closure.

The question of whether x ∈ R2\K is inside K or outside K is decidedby the degree of the loop

ωx : S1 → S1 ; y 7→ i(y)− x‖i(y)− x‖

which sends y ∈ S1 to the unit vector ωx(y) ∈ S1 ⊂ R2 parallel to thedirected straight line segment joining x to i(y) ∈ K.

[x, i(y)] = (1− t)x+ ti(y) | t ∈ I ⊂ R2 .

Specifically,

|degree(ωx)| =

1 if x is inside K

0 if x is outside K .

For any z ∈ S1 define the ray from x in the direction of z

[x, z( = x+ tz | t > 0 ⊂ R2 ,

so that(ωx)

−1(z) = y ∈ S1 | i(y) ∈ [x, z(

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is the set of points of K which can be ”seen” from x ∈ R2\K by looking inthe direction z. If i is differentiable and [x, z( is not tangent to K, or if K isa polygon and [x, z( does not contain any edges of K, then the intersection[x, z(∩K is finite, and

|degree(ωx)| =

1 if [x, z(∩K has an odd number of points

0 if [x, z(∩K has an even number of points .

In particular, if there exists z ∈ S1 such that [x, z(∩K is empty (i.e. youcannot see any point of K in the direction z from x) then x is outside K.

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Lecture 14

9 The n-dimensional projective space RPn

We shall now define an interesting family of n-dimensional manifolds.

Definition 9.1. The n-dimensional real projective space is the quotient spaceof the n-sphere

RPn = Sn/∼

with ∼ the antipodal equivalence relation

v ∼ w if either v = w or v = − w ∈ Sn .

Let p : Sn → RPn; v 7→ [v] be the natural projection. For every x ∈ RPnthe inverse image is the equivalence class

p−1(x) = x,−x ⊂ Sn

consisting of an unordered pair of antipodal points in Sn. By the definitionof a quotient space a subset U ⊆ RPn is open if and only if p−1(U) ⊆ Sn isopen.

A point x ∈ RPn can be viewed in several different ways:

(i) as a subset x,−x ⊂ Sn, with x = p(x) = p(−x) ∈ RPn,

(ii) as a line L ⊂ Rn+1 through the origin 0, with x = p(v

‖v‖) ∈ RPn for

any v 6= 0 ∈ L,

(iii) as an equivalence class of non-zero vectors v 6= 0 ∈ Rn+1 with

v ∼ v′ if v′ = λv for some λ 6= 0 ∈ R ,

and x = p(v

‖v‖) = p(

−v‖v‖

) ∈ RPn for any v in the class.

Remark 9.2. The 1-dimensional real projective space RP1 is homeomorphicto S1 via the ‘square root’ homeomorphism

q : S1 → RP1 ; z = eiθ →√z = eiθ/2,−eiθ/2

such that p = qr : S1 → RP1 with r : S1 → S1; z = eiθ 7→ z2 = e2iθ.

The 2-dimensional real projective space RP2 is also known as the projec-tive plane.

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Exercise 9.3. (i) Prove that RP2 is homeomorphic to the quotient space I2/∼with

(x, 0) ∼ (1− x, 1) , (0, y) ∼ (1, 1− y) (0 6 x, y 6 1) .

(ii) Prove that RP2 is homeomorphic to the quotient space of the Mobiusband obtained by identifying all the points of the boundary circle to a singlepoint.

Exercise 9.4. Prove that RPn is the space obtained from X1 = Rn by attach-ing an (n − 1)-sphere X2 = Sn−1 ‘at infinity’, with antipodal points on the(n− 1)-sphere identified. )

The n-dimensional projective space RPn is a quotient space of a compactpath-connected space Sn, so it is also compact and is path-connected We canthus write π1(RPn) without specifying a base point.

We shall prove in Theorem 9.12 below that π1(RPn) is isomorphic to theinfinite cyclic group Z for n = 1, and to the cyclic group Z2 of order 2 forn > 2.

Definition 9.5. The homogeneous coordinates of a point in RPn is theequivalence class [x0, x1, . . . , xn] ∈ RPn of non-zero vectors (x0, x1, . . . , xn) 6=(0, 0, . . . , 0) ∈ Rn+1, with

(x0, x1, . . . , xn) ∼ (y0, y1, . . . , yn)

if (y0, y1, . . . , yn) = λ(x0, x1, . . . , xn) ∈ Rn+1 for some λ 6= 0 ∈ R

The projection

p : Sn → RPn ; x = (x0, x1, . . . , xn) 7→ [x] = [x0, x1, . . . , xn]

has the key property that for any x, y ∈ Sn with p(x) = p(y) ∈ RPn eitherx = y or x = −y.

Every point on the unit circle

z = cos(2πt) + isin(2πt) ∈ S1

(regarded as a complex number) has two square roots

√z = ± (cos(πt) + isin(πt)) ∈ S1 .

Definition 9.6. The square root loop in RPn is

σ : S1 → RPn ; z = cos(2πt)+isin(2πt) 7→√z = [cos(πt), sin(πt), 0, . . . , 0] .

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It is convenient to write the cyclic group of 2 elements as

Z2 = −1,+1 ,

with the group operation given by multiplication.

Proposition 9.7. (i) For n = 1 the square root loop

σ : S1 → RP1 ; z = e2πit = cos(2πt) + isin(2πt) 7→√z = [cos πt, sin πt]

is a homeomorphism, so

π1(RP1) = π1(S1) = Z .

(ii) The square of the square root σ ∈ π1(RPn) is

[σ]2 = [qp] ∈ π1(RPn)

with p : S1 → RP1 the projection and

q : RP1 → RPn ; [x1, x2] 7→ [x1, x2, 0, . . . , 0]

the inclusion.(iii) For n > 2

[σ]2 = 1 ∈ π1(RPn)

and there is defined a group homomorphism

[σ] : Z2 → π1(RPn) ; − 1 7→ [σ] .

Proof. (i) By construction.(ii) The paths

σ1 : I → Sn ; t 7→ (cos(πt), sin(πt), 0, . . . , 0) ,

σ2 : I → Sn ; t 7→ (cos(π(t+ 1)), sin(π(t+ 1)), 0, . . . , 0)

are such that

σ1(0) = σ2(1) = (1, 0, . . . , 0) , σ1(1) = σ2(0) = (−1, 0, . . . , 0) ∈ Sn ,σ1 • σ2 : I → Sn ; t 7→ [cos(2πt), sin(2πt), 0, . . . , 0] ,

pσ1(t) = pσ2(t) = σ(e2πit) ∈ RPn .

The concatenation of the square root loop σ with itself is thus the compositeof the projection p : S1 → RP1 and the inclusion RP1 ⊂ RPn

σ • σ = qp : S1 → RPn ; (x1, x2) 7→ [x1, x2, 0, . . . , 0] .

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(iii) As n > 2 it is possible to extend σ • σ to the map

D2 → RPn ; (x1, x2) 7→ [x1, x2,√

1− (x1)2 − (x2)2, 0, . . . , 0]

and π1(D2) = 1, so

[σ]2 ∈ im(π1(D2)→ π1(RPn)) = 1 .

We shall use topology to define a group homomorphism

sign : π1(RPn)→ Z2

which will be shown in Theorem 9.12 below to be an isomorphism for n > 2.While RP1 is homeomorphic to S1, RPn is not homeomorphic to Sn for n > 2,since π1(S

n) = 1 (Theorem 7.4) and π1(RPn) = Z2.

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Lecture 15Definition 9.8. A path-connected subset U ⊂ RPn is small if p−1(U) ⊂ Sn

has 2 path-components U+, U− and the restrictions

p+ = p| : U+ → U , p− = p| : U− → U

are homeomorphisms.

Example 9.9. For any x ∈ Sn and ε < 1 the open subset

U = p(Bε(x)) ⊂ RPn

(homeomorphic to Rn) is small, with

Bε(x) = y ∈ Sn | ‖x− y‖ < ε , p−1(U) = Bε(x) ∪ −Bε(x) ⊂ Sn .

If y ∈ Bε(x) ∩ −Bε(x) then y,−y ∈ Bε(x) and

‖y − (−y)‖ = 2‖y‖ = 2 6 ‖x− y‖+ ‖x+ y‖ < 2ε ,

a contradiction.

Definition 9.10. A lift of a map α : [t0, t1]→ RPn is a map θ : [t0, t1]→ Sn

such thatα(t) = p(θ(t)) ∈ RPn (t ∈ [t0, t1]) .

Sn

p

[t0, t1]

θ;;

α // RPn

If α is closedpθ(0) = α(0) = α(1) = pθ(1) ∈ Sn ,

allowing the sign of α to be defined by

sign(α) =

+1 if θ(0) = θ(1)

−1 if θ(0) = −θ(1) .

Proposition 9.11. Given a map α : [t0, t1]→ RPn and θ0 ∈ Sn such that

α(t0) = p(θ0) ∈ RPn

there is a unique lift θ : [t0, t1]→ Sn for α with initial point

θ(t0) = θ0 ∈ Sn .

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Proof. The method of proof is essentially the same as for Proposition 8.3,which proved the existence of an angle map for each path in S1. Considerfirst the special case when α([t0, t1]) ⊂ U for a small open subset U ⊂ RPn,so that

θ0 ∈ p−1(U) = U+ ∪ U− .

Let ε ∈ +,− be the unique sign such that θ0 ∈ Uε. The composite

θ = (pε)−1 α : [t0, t1]→ U → (pε)

−1(Uε) ⊂ Sn

is the unique lift of α with θ(t0) = θ0.Next, consider the general case of a map α : [t0, t1] → RPn such that

α([t0, t1]) may not be contained in a single small open subset U ⊂ RPn.Define an open cover U = U of RPn consisting of all the small open subsetsU ⊂ RPn. By the Lebesgue Covering Lemma there exists an integer N > 1so large that the N subintervals in the partition

[t0, t1] =N−1⋃j=0

[sj, sj+1]

with

s0 = t0 < s1 = t0 +(t1− t0)/N < · · · < sj = t0 +j(t1− t0)/N < · · · < sN = t1

have diameter (t1 − t0)/N so small that

α([sj, sj+1]) ⊂ Uj ⊂ RPn (0 6 j 6 N − 1)

for small open Uj ⊂ RPn. Let

α(j) = α| : [sj, sj+1]→ RPn .

Apply the special case to construct the unique lift θ(0) : [s0, s1] → Sn forα(0) with θ(0)(s0) = θ0. Next, apply the special case to construct the uniquelift θ(1) : [s1, s2] → Sn for α(1) with θ(1)(s1) = θ(0)(s1) ∈ RPn. And soon . . . , until each α(j) has a lift θ(j) : [sj, sj+1] → Sn. Piecing together thelifts θ(0), θ(1), . . . , θ(N−1) for α(0), α(1), . . . , α(N−1) there is obtained theunique lift

θ = θ(0) ∪ · · · ∪ θ(N − 1) : [t0, t1]→ Sn ; t→ θ(j)(t) if sj 6 t 6 sj+1

for α = α(0) ∪ · · · ∪ α(N − 1) with α(t0) = θ0.

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For a closed path α : I → RPn and for a choice of θ0 ∈ p−1(α(0)) ⊂ Sn

the endpoints θ(0) = θ0, θ(1) ∈ Sn of the lift θ : I → Sn are such that

p(θ(0)) = α(0) = α(1) = p(θ(1)) ∈ RPn .

There are two mutually exclusive possibilities:

either θ(0) = θ(1) ∈ Sn or θ(0) = −θ(1) ∈ Sn

or equivalently

the lift θ is/is not a closed path in Sn.

The possibilities are independent of the choice of θ0.

Theorem 9.12. (i) The function

sign : π1(RPn)→ Z2 = +1,−1 ; [α] 7→

+1 if θ(0) = θ(1)

−1 if θ(0) = −θ(1)

is a surjective group homomorphism which takes the value −1 on the squareroot loop σ ∈ π1(RPn).

(ii) For n = 1

sign : π1(RP1) = Z→ Z2 ; N 7→

+1 if N is even

−1 if N is odd

is the surjection of the infinite cyclic group onto the cyclic group of order 2.(iii) For n > 2 sign : π1(RPn)→ Z2 is an isomorphism of groups inverse

to σ : Z2 → π1(RPn).

Proof. The method of proof is essentially the same as for Theorem 8.8.(i) Adapting Proposition 8.7, we can use the Lebesgue Covering Lemma

to lift homotopies, and thus show that sign(ω) ∈ Z depends only on the rel1 homotopy class of a loop ω : S1 → RPn.

The function sign : π1(RPn) → Z2 is a group homomorphism, since aconcatenation of closed paths lifts to the concatenation of paths. The squareroot loop σ : S1 → RPn lifts to the path

θ : I → Sn ; t 7→ (cos(πt), sin(πt), 0, . . . , 0)

withθ(0) = (1, 0, . . . , 0) 6= θ(1) = (−1, 0, . . . , 0) ∈ Sn

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so that sign(σ) = −1 ∈ Z2.(ii) There is only one surjective group homomorphism Z→ Z2.(iii) The sign function is injective for n > 2 since for any closed path

α : I → RPn at α(0) = α(1) with a lift θ : I → Sn to a closed path atθ(0) = θ(1)

[α] = p∗[θ] = 1 ∈ π1(RPn)

on account of [θ] = 1 ∈ π1(Sn) = 1 (Theorem 7.4).

Example 9.13. The projection p : S1 → RP1 corresponds to the closed path

α : I → RP1 ; t 7→ p(e2πit) = [cos (2πt), sin (2πt)] .

The lift of αθ : I → S1 ; t 7→ e2πit

has θ(0) = θ(1) = (1, 0), so

sign(p) = + 1 ∈ Z2 .

This is in agreement with Theorem 9.12 (ii), since composition of p with theinverse of the square root homeomorphism σ : S1 → RP1 is the squaringfunction

σ−1 p : S1 → S1 ; z 7→ z2 ,

so that

degree(p) := degree(σ−1 p) = 2 ∈ π1(RP1) = Z .

Example 9.14. For any a, b ∈ S2 with a · b = 0 ∈ R the path

θ : I → S2 ; t 7→ b cosπt+ (a× b) sinπt

traces out half the great circle a⊥ ⊂ S2 from θ(0) = b to θ(1) = −b. Thenα = pθ : I → RP2 is a closed path tracing out the projective line

L(a) = p(c) | c ∈ S2, a · c = 0 ∈ R ⊂ RP2

with [α] = −1 ∈ π1(RP2) = Z2. Any path β : I → S2 with

β(0) = b , β(1) = − b ∈ S2

is homotopic rel 0, 1 to θ, for any a ∈ b⊥. Thus β = pβ : I → RP2 isa closed path with β(0) = β(1) = p(b) ∈ RP2 such that β is homotopic rel0, 1 to α and

[β] = [α] = − 1 ∈ π1(RP2, b) = Z2 .

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Remark 9.15. The sign of a loop ω : S1 → RPn is +1 or −1, depending onwhether the space

S(ω) = (x, y) ∈ S1 × Sn |ω(x) = p(y) ∈ RPn

is homeomorphic to S1 t S1 or S1.

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Lecture 16

10 Fixed points and non-retraction

Definition 10.1. A fixed point of a map f : X → X is point x ∈ X suchthat f(x) = x.

Topology is very good at predicting which maps f : X → X have fixedpoints, although not very good at actually finding them! The fixed pointtheorems of topology have applications not only in mathematics (e.g. dif-ferential equations, dynamical systems, analysis, . . . ) but also in economicsand game theory. An equilibrium of some system is a fixed point of somemap f ; in order to actually find such fixed points it is necessary to employnon-topological methods such as analysis or linear programming, which areadapted to f .

Example 10.2. (i) Every x ∈ X is a fixed point of the identity map

f = 1 : X → X ; x 7→ x .

(ii) The rotation of R2 through an angle θ with 0 < θ < 2π

Rθ =

(cos θ −sin θsin θ cos θ

): R2 → R2

has only (0, 0) as fixed point.(iii) The fixed points of the reflection of R2 through the line y = xtanφ

Sφ =

(cos 2φ sin 2φsin 2φ −cos 2φ

): R2 → R2

are precisely the points on the line.(iv) A translation of Rn by a 6= 0 ∈ Rn

T : Rn → Rn ; x 7→ x+ a

does not have fixed points.(v) The antipodal map

T : Sn → Sn ; x 7→ −x

does not have fixed points.

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For certain spaces X every map f : X → X can be shown to have a fixedpoint. The Brouwer fixed point theorem (dating from 1910) is that everymap f : Dn → Dn has a fixed point, for every n > 1. This is a consequenceof the non-retraction theorem: there is no map g : Dn → Sn−1 whichis the identity on Sn−1 ⊂ Dn. We shall only consider the cases n = 0, 1,although the method extends to all n > 2.

Definition 10.3. Let i : A→ X be the inclusion of a subspace A ⊂ X. Thesubspace A is a retract of X if there exists a map j : X → A such thatj i = identity : A→ A. The map j is a retraction of X onto A.

Example 10.4. (i) For every a ∈ X the subspace A = a ⊂ X is a retract.(ii) Let A = a, b ⊆ X, with a 6= b, and X Hausdorff. There exist

disjoint open subsets U, V ⊂ X such that a ∈ U , b ∈ V so that A has thediscrete topology, with both a = A ∩ U and b = A ∩ V open in A. Thesubspace A is a retract of X if and only if a, b are in different components ofX.

(iii) For any n > 1 Sn−1 ⊂ Dn\0 is a retract, with

j : Dn\0 → Sn−1 ; x 7→ x

‖x‖

a retraction of Dn\0 onto Sn−1.

Proposition 10.5. Let X be a path-connected space, and let i : A → X bethe inclusion of a path-connected subspace A.

(i) If A is a retract of X, i : A→ X is the inclusion and j : X → A is aretraction the induced group homomorphisms

i∗ : π1(A)→ π1(X) , j∗ : π1(X)→ π1(A)

are such that

j∗ i∗ = identity : π1(A)i∗ // π1(X)

j∗ // π1(A) .

Thus i∗ : π1(A)→ π1(X) is injective, and j∗ : π1(X)→ π1(A) is surjective.(ii) If π1(A) 6= 1 and π1(X) = 1 then A cannot be a retract of X.

Proof. (i) Immediate from

(j i)∗ = j∗ i∗ , identity∗ = identity

and the fact that homotopic maps induce the same homomorphisms of thefundamental groups.

(ii) Immediate from (i).

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Remark 10.6. If X is a connected space and A ⊂ X is a retract then A mustbe connected: Proposition 10.5 is the π1-analogue.

Proposition 10.7. If there exists a map f : Dn → Dn without fixed pointsthen the inclusion i : Sn−1 → Dn admits a retraction.

Proof. Define a retraction of i

j : Dn → Sn−1 ; x 7→ j(x)

by sending x ∈ Dn to the unique point

j(x) = (1− t)f(x) + tx ∈ Sn−1

with t > 1. Explicitly,

t = 1 +x·(f(x)−x)+

√(x·(f(x)−x))2−4‖f(x)−x‖2(‖x‖2−1)

2‖f(x)−x‖2

If x ∈ Dn\Sn−1 then f(x), x, j(x) ∈ Dn are distinct collinear points. Ifx ∈ Sn−1 then j(x) = x.

•f(x)

•x

•j(x)

Theorem 10.8. (Non-retraction for n = 1, 2)(i) S0 ⊂ D1 is not a retract.(ii) S1 ⊂ D2 is not a retract.

Proof. (i) Suppose j : D1 → S0 is a retraction of the inclusion i : S0 → D1,so that

ji = 1 : S0 → S0 .

Now j is onto, D1 is connected, S0 is disconnected : a contradiction!(ii) Apply Proposition 10.5: suppose j : D2 → S1 is a retraction of the

inclusion i : S1 → D2, so that

ji = 1 : S1 → S1 .

induces

(ji)∗ = j∗i∗ = 1 = 0 : π1(S1) = Z→ π1(D

2) = 0 → π1(S1) = Z

- a contradiction!

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Corollary 10.9. (Brouwer fixed point theorem for n = 1, 2)Every map f : Dn → Dn has a fixed point.

Proof. Immediate from Proposition 10.7 and Theorem 10.8.

Remark 10.10 (Non-examinable). (i) The non-retraction Theorem 10.8 forn = 1 is the intermediate value theorem in disguise.

(ii) The non-retraction Theorem 10.8 generalizes to all n > 3: Sn−1 ⊂ Dn

is not a retract. The standard proof uses the computations

πn−1(Sn−1) = Z , πn−1(D

n) = 0

of the higher homotopy groups (cf. Remark 5.19). The proof of the Brouwerfixed point theorem for n > 3 follows as in the cases n = 1, 2.

(iii) The proof of the Brouwer fixed point theorem is non-constructive,i.e. does not provide an algorithm for actually finding a fixed point. Thisis very ironic, since Brouwer subsequently developed ‘intuitionistic logic’ inwhich only constructive proofs are allowed!

What about the fixed points of maps f : Sn → Sn ?

Proposition 10.11. A map f : Sn → Sn without fixed points is homotopicto the antipodal map a : Sn → Sn;x 7→ −x.

Proof. The map

h : Sn × I → Sn ; (x, t) 7→ (1−t)f(x)−tx‖(1−t)f(x)−tx‖

defines a homotopy f ' a. (If (1− t)f(x)− tx = 0 ∈ Rn+1 the unit vectorsf(x) 6= x ∈ Sn are parallel, so that f(x) = −x and 0 = (1 − t)f(x) − tx =f(x), a contradiction).

Example 10.12. The rotation of S1 through an angle θ ∈ [0, 2π)

Rθ : S1 → S1 ; z 7→ zeiθ

has fixed points if θ = 0 (since R0 = 1) and no fixed points if θ > 0, sinceeiθ 6= 1. The map

S1 × I → S1 ; (z, t) 7→ zetθi

defines a homotopy R0 = 1 ' Rθ, so

degree(Rθ) = degree(1) = 1 ∈ Z .

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Lecture 17Proposition 10.13. (i) A map f : S1 → S1 without fixed points has degree1.

(ii) A map f : S1 → S1 with degree 6= 1 has a fixed point.

Proof. (i) By Proposition 10.11 f is homotopic to the antipodal map

a = Rπ : S1 → S1 ; z 7→ −z .

By Example 10.12

degree(f) = degree(a) = degree(1) = 1 ∈ Z .

(ii) This is just the contrapositive of (i).

Remark 10.14 (Non-examinable). (i) It can be shown that the antipodal mapa : Sn → Sn is homotopic to the identity if and only if n is odd. (This maybe deduced for example from the higher homotopy group computation

a∗ = (−1)n+1 : πn(Sn) = Z→ πn(Sn) = Z

also due to Brouwer).(ii) A vector field on Sn is a map v : Sn → Rn+1 such that

x · v(x) = 0 ∈ R for all x ∈ Sn ,

so that v(x) is tangent to Sn ⊂ Rn+1 at x ∈ Sn.If v(x) 6= 0 for all x ∈ Sn, the maps

f : Sn → Sn ; x 7→ v(x)

‖v(x)‖,

h : Sn × I → Sn ; (x, t) 7→ (1− t)v(x) + tx

‖(1− t)v(x) + tx‖

are such that h : f ' 1 : Sn → Sn, so that f is homotopic to the identity.It follows from Proposition 10.11 and (i) that for even n = 2m f must

have a fixed point x = f(x) ∈ Sn. But for such a fixed point v(x) 6= 0 ∈ Rn+1

would be both parallel and orthogonal to x. So there is no such v. This is theHairy Ball Theorem (first proved by Poincare for m = 1, but not calledthat by him): every vector field v on S2m has at least one x ∈ S2m withv(x) = 0 ∈ R2m+1.

On the other hand, for odd n = 2m+ 1 there exist non-zero vector fieldson S2m+1, for example

v : S2m+1 → R2m+2\0 ;

(x1, x2, . . . , x2m+2) 7→ (−xm+1,−xm+2, . . . ,−x2m+2, x1, x2, . . . , xm) .

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11 Covering spaces

It is practically impossible to actually compute the fundamental group π1(X, x0)directly from the definition! However, in many cases it is possible to computeit geometrically, as the ‘group of covering translations’ of a ‘covering space’ ofX. For any topological space X the set Homeo(X) of all homeomorphisms

h : X → X is a group, with the composition of homeomorphisms as thegroup law.

In favourable circumstances it is possible to compute π1(X, x0) for a path-

connected space X using the universal covering projection p : X → Xwith X simply-connected (= path connected and π1(X) = 0), such that

each inverse image p−1(x) ⊂ X (x ∈ X) is a copy of π1(X, x0) as a discretesubspace.

The fundamental group π1(X, x0) is isomorphic to the subgroup Homeop(X)

of Homeo(X) consisting of the homeomorphisms h : X → X such that

ph = p : X → X, called the covering translations. This correspondence(which will take us a few lectures) generalises the degree homomorphism onπ1(S

1): we lift loops α : S1 → X to paths α : I → X, and find that thehomotopy class [α] is determined by the difference between α(0) and α(1).

Recall that a topological space F is discrete if it has the topology in whichevery subset is open.

Definition 11.1. A covering space of a space X with fibre the discrete spaceF is a space X with a covering projection map p : X → X such that foreach x ∈ X there exists an open subset U ⊆ X with x ∈ U , and with ahomeomorphism

φ : F × U → p−1(U)

such thatpφ(a, u) = u ∈ U ⊆ X (a ∈ F, u ∈ U) .

In particular, for each x ∈ X, p−1(x) is homeomorphic to F .

A covering projection p : X → X is a ‘local homeomorphism’: for eachx ∈ X there exists an open subset U ⊆ X such that x ∈ U and

U → p(U) ; u 7→ p(u)

is a homeomorphism, with p(U) ⊆ X an open subset.

Definition 11.2. Given a covering projection p : X → X let Homeop(X) be

the subgroup of Homeo(X) consisting of the homeomorphisms h : X → Xsuch that

ph = p : X → X ,

82

i.e. such that the diagram

X

p

h // X

pX

is commutative. Such h are called covering translations.

Example 11.3. A map p : X → X is a homeomorphism if and only if it is acovering projection with fibre F = 1.

Definition 11.4. A covering projection p : X → X with fibre F is trivial ifthere exists a homeomorphism

φ : F ×X → X

such thatpφ(a, x) = x ∈ X (a ∈ F, x ∈ X) .

A particular choice of φ is a trivialisation of p.

Example 11.5. For any space X and discrete space F the covering projection

p : X = F ×X → X ; (a, x) 7→ x

is trivial, with the identity trivialization φ = 1 : F ×X → X.

Example 11.6. Let p : X → X be a covering projection, with fibre F .(i) For any subset U ⊆ X the restriction p| : p−1(U) → U is a covering

projection, with the same fibre F . By Definition 11.1, X has an open coverU ⊂ X such that the restrictions p| : p−1(U)→ U are trivial. But p itselfneed not be trivial – see Example 11.7 below.

(ii) If p is trivial and φ1, φ2 : F ×X → X are two trivialisations there isa unique function

σ : X → Homeo(F )

such that

φ2(a, x) = φ1(σ(x)(a), x) ∈ X (a ∈ F, x ∈ X) .

Both F and Homeo(F ) are given the discrete topology, and the homeomor-phisms (= bijections in this case) σ(x) : F → F vary continuously withx ∈ X, so the function σ is constant on each path-component of X.

83

(iii) If p is trivial and φ is a trivialization, a covering translation h : X →X is necessarily of the form

h : X → X ; φ(a, x) 7→ φ(σ(x)(a), x)

with the bijections σ(x) : F → F varying continuously with x ∈ X as in (ii).For path-connected X the σ(x)’s are all the same, and the group of coveringtranslations is

Homeop(X) = Homeo(F )

the group of all bijections F → F .

Example 11.7. (i) The map

p : R→ S1 ; x 7→ e2πix

is a covering projection with fibre Z. Note that p is not trivial, since R is nothomeomorphic to Z × S1. The group of covering translations is the infinitecyclic group Z (see Example 11.9 below for details).

(ii) For any n 6= 0 ∈ Z the map

pn : S1 → S1 ; z 7→ zn

is a covering projection.If |n| = 1 then pn : S1 → S1 is a homeomorphism (= the identity map

for n = 1, the reflection in the real axis for n = −1), so it has fibre F = 1and group of covering translations Homeopn(S1) = 1.

If |n| > 2 then pn : S1 → S1 has fibre F = 1, 2, . . . , |n|. The coveringtranslation defined by anticlockwise rotation through the angle 2π/n (= theantipodal map for |n| = 2)

hn : S1 → S1 ; z 7→ ze2πi/n

generates the group of covering translations

Homeopn(S1) = (hn)j | 0 6 j 6 |n| − 1 = Z|n| ⊂ Homeo(S1) ,

a cyclic group of order |n|.

84

Lecture 18Here is a very general construction of covering projections:

Theorem 11.8. Given a space X and a subgroup G ⊆ Homeo(X) define an

equivalence relation ∼ on X by

x1 ∼ x2 if there exists g ∈ G such that x2 = g(x1)

and write

X = X/∼ = X/G ,

with quotient map p : X → X.

Suppose that for each x ∈ X there exists an open neighbourhood U of xsuch that

g(U) ∩ U = ∅ for g ∈ G \ 1 .

Then p : X → X is a covering projection with fibre G. Furthermore, if X ispath-connected then so is X, and the group of covering translations of p is

Homeop(X) = G ⊂ Homeo(X) .

Proof. If we have g1u1 = g2u2 ∈ g1(U)∩ g2(U), then u2 = gu1 ∈ g(U)∩U forg = (g2)

−1g1, so g = 1 ∈ G and g1 = g2. Thus the subsets g(U) ⊂ p−1p(U)are disjoint and hence clopen, so

p−1p(U) = gu | g ∈ G, u ∈ U =⊔g∈G

g(U) ⊆ X

is a disjoint union and the surjective map

G× U → p−1p(U) ; (g, u) 7→ gu

is a homeomorphism.From the description of p−1p(U) above, it follows that p is an open map,

and the surjective open map p|U : U → p(U) is injective by the argumentabove, hence a homeomorphism, so we may combine these maps to give therequired homeomorphism

φ : G× p(U)→ p−1p(U) ; (g, p(u)) 7→ gu.

If h ∈ Homeop(X), then for any x ∈ X there is a unique gx ∈ G suchthat

h(x) = gx(x) ∈ X .

85

For the open subset W = h−1(gxU)∩U , we have h(W )∩g(W ) = ∅ if g 6= gx,so we must also have gy = gx for all y ∈ W . Thus the function

X → G ; x→ gx

is continuous. If X is path-connected, the function is constant (since G isdiscrete), so h = gx and thus

Homeop(X) = G .

Example 11.9. For each n ∈ Z the translation of R by n units to the rightdefines a homeomorphism

hn : R→ R ; x 7→ x+ n

with hnhm = hm+n. The infinite cyclic subgroup

G = hn |n ∈ Z ⊂ Homeo(R)

satisfies the hypothesis of Theorem 11.8, so that

p : R→ R/G = R/Z = S1 ; x 7→ e2πix

is a covering projection with fibre G = Z and group of covering translations

Homeop(R) = G = Z ⊂ Homeo(R) .

Example 11.10. Let p : X → X be a covering projection. A lift of a mapf : Y → X is a map f : Y → X such that

p(f(y)) = f(y) ∈ X (y ∈ Y )

so that there is defined a commutative diagram

X

p

Y

f??

f // X

Example 11.11. For the trivial covering projection p : X = F × X → X ofExample 11.5 define a lift of any map f : Y → X by choosing a point a ∈ Fand setting

fa : Y → X = F ×X ; y 7→ (a, f(y)) .

If Y is path-connected every lift of f is of this type, and the function a→ fadefines a one-one correspondence between the points a ∈ F and the lifts fof f .

86

Theorem 11.12. (Path lifting property) Let p : X → X be a covering

projection with fibre F . Let x0 ∈ X, x0 ∈ X be such that p(x0) = x0 ∈ X.(i) Every path α : I → X with α(0) = x0 ∈ X has a unique lift to a path

α : I → X such that α(0) = x0 ∈ X.(ii) Let α, β : I → X be paths with

α(0) = β(0) = x0 ∈ X , α(1) = β(1) ∈ X .

and let α, β : I → X be the lifts with α(0) = β(0) = x0 ∈ X given by (i).Every rel 0, 1 homotopy

h : α ' β : I → X

has a unique lift to a rel 0, 1 homotopy

h : α ' β : I → X

and in particular

α(1) = h(1, t) = β(1) ∈ X (t ∈ I) .

Proof. (Nonexaminable) (i) For each s ∈ I there exists an open subset Us ⊆X with α(s) ∈ Us, and with a homeomorphism φs : F × Us → p−1(Us) suchthat

pφs(a, u) = u ∈ Us ⊆ X (a ∈ F, u ∈ Us) .

By the Lebesgue Covering Lemma there exists an integerN > 1 and s1, s2, . . . , sN ∈I such that

[ kN, k+1N

] ⊂ α−1(Usk) (0 6 k < N) .

If the lift α has already been defined on [0, kN

] and

(φsk)−1α( kN

) = (fk, α( kN

)) ∈ F × Usk

there is a unique extension to a lift on [0, k+1N

] by

α : [ kN, k+1N

]→ X ; s→ φsk(fk, α(s)) .

with(φsk+1

)−1α(k+1N

) = (fk+1, α(k+1N

)) ∈ F × Usk+1.

(ii) For each t ∈ I there is defined a path

ht : I → X ; s→ h(s, t)

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withh0(s) = α(s) , h1(s) = β(s) ,

ht(0) = h(s, 0) = x0 , ht(1) = α(1) = β(1) ∈ X .

Let ht : I → X be the unique lift of ht with ht(0) = x0 ∈ X given by (i), andset

h : I × I → X ; (s, t) 7→ ht(s) .

Definition 11.13. Given a covering projection p : X → X and a pathα : I → X define the fibre transport bijection

α# : p−1(α(0))→ p−1(α(1)) ; x 7→ αx(1)

where αx : I → X is the unique lift of α given by Theorem 11.12 (i) with

αx(0) = x ∈ X .

Remark 11.14. In a covering p : X → X, the inverse image p−1(x) ⊆ X ofany x ∈ X is homeomorphic to F , but not in a particular way: a choice ofopen subset U ⊆ X and a homeomorphism φ : F × U → p−1(u) with x ∈ Uand pφ(a, u) = p(u) determine a homeomorphism

F → p−1(x) ; (a, u) 7→ φ(a, u)

but different choices of U, φ will give a different homeomorphism F ∼= p−1(x).If x, y ∈ X are joined by a particular path α : I → X with α(0) = x,

α(1) = y ∈ X the fibre transport is a particular bijection α# : p−1(x) →p−1(y). By Theorem 11.12 (ii) α# depends only on the rel 0, 1 homotopyclass of α. If α : I → X is a closed path with α(0) = α(1) = x ∈ X then

α# : p−1(x)→ p−1(x)

is a self-homeomorphism (= permutation) of p−1(x).

88

Lecture 19Recall that a space X is simply-connected if it is path-connected and

π1(X) = 1 .

Proposition 11.15. Every covering projection p : X → X of a simply-connected space X is trivial.

Proof. (Nonexaminable) Let F be the fibre. Choose a base point x0 ∈ X,and an open neighbourhood U0 ⊆ X of x0 with a trivialisation

φ0 : F × U0 → p−1(U0)

of p| : p−1(U0)→ U0, i.e. a homeomorphism such that

pφ0(a, u) = u ∈ X (a ∈ F, u ∈ U0) .

In particular, there is defined a bijection

F → p−1(x0) ; a 7→ φ0(a, x0) .

For each x ∈ X choose a path αx : I → X from αx(0) = x0 to αx(1) = x,and use fibre transport to define a homeomorphism

φ : F ×X → X ; (a, x) 7→ (αx)#(φ0(a, x0)) .

The condition π1(X) = 1 is needed to prove that φ is independent of thechoices of paths αx.

Example 11.16. (i) By Proposition 11.15 every covering p : I → I is trivial,

with a homeomorphism φ : F × I → I such that pφ(a, x) = x.(ii) For any discrete space F and bijection σ : F → F define a covering

p : S1 → S1 with fibre F by

p : S1 = F × I/(x, 0) ∼ (σ(x), 1) → S1 = I/0 ∼ 1 ; [x, t] 7→ [t] .

In fact, every covering p : S1 → S1 arises in this way: define the closed path

α : I → S1 ; t 7→ e2πit

with α(0) = α(1) = 1 ∈ S1, and note that the fibre transport is a bijection

α# : F = p−1(1)→ F = p−1(1)

89

such that

p : S1 = F × I/(x, 0) ∼ (α#(x), 1) → S1 = I/0 ∼ 1 ; [x, t] 7→ [t] .

(iii) For the covering projection p : R → S1;x 7→ e2πix of Example 11.7 (i)the fibre transport bijection in (ii) is just the shift map

σ : F = Z→ F = Z ; x 7→ x+ 1 ,

with a homeomorphism

Z× I/(x, 0) ' (x+ 1, 1) → R ; [y, t] 7→ y − t .

(iv) For the covering projection pn : S1 → S1; z 7→ zn of Example 11.7 (ii)with n > 1 the fibre transport bijection is the cyclic permutation

α# = (1 2 . . . n) : F = 1, 2, . . . , n → F = 1, 2, . . . , n ; k 7→ k+1 ( mod n) .

(v) A permutation σ ∈ Σn determines a covering map

p : S1 = 1, 2, . . . , n×I/(x, 0) ∼ (σ(x), 1) → S1 = I/0 ∼ 1 ; (x, t) 7→ t

with fibre F = 1, 2, . . . , n and fibre transport bijection α# = σ. Exercise:

prove that S1 has one path-component for each cycle (i1 i2 . . . im) in the cycle

decomposition of σ, and that Homeop(S1) is isomorphic to the subgroup of Σn

consisting of the permutations τ such that τσ = στ , called the centralizerof σ.

Definition 11.17. A covering projection p : X → X of a path-connectedspace X is universal if X is simply-connected.

Example 11.18. (i) The covering projection p : R→ S1 is universal.(ii) The covering projection p : Sn → RPn is universal for n > 2 .

Recall that a space X is locally path connected if for each x ∈ X and foreach open subset U ⊆ X with x ∈ U there is a path-connected open subsetV ⊆ U with x ∈ V . (Main example: open subsets of Rn).

Theorem 11.19. Let X be a path-connected, locally path-connected spacewith a universal covering projection p : X → X. Let x0 ∈ X, x0 ∈ X be basepoints such that p(x0) = x0.(i) The function

π1(X, x0)→ p−1(x0) ; [α]→ α#(x0)

90

is a bijection.(ii) For each y ∈ p−1(x0) there is a unique covering translation hy ∈ Homeop(X)such that

hy(x0) = y ∈ X .

The functionp−1(x0)→ Homeop(X) ; y → hy

is a bijection, with inverse h→ h(x0). The composite bijection

π1(X, x0)→ p−1(x0)→ Homeop(X)

is an isomorphism of groups.

Proof. (Nonexaminable) (i) The function π1(X, x0)→ p−1(x0) is well-defined,by the rel 0, 1 homotopy invariance (11.14) of the fibre transport α#. Since

X is simply-connected, for every x1 ∈ p−1(x0) there is a unique rel 0, 1homotopy class of paths α : I → X with

α(0) = x0 , α(1) = x1 ∈ X

so thatx1 = (pα)#(x0) ∈ p−1(x0)

is the image of pα ∈ π1(X, x0), and π1(X, x0)→ p−1(x0) is surjective. Injec-tivity now follows because α is unique up to rel 0, 1 homotopy.

(ii) By the path-connectedness of X for each z ∈ X there exists a path

β : I → X from β(0) = x0 to β(1) = z.

X

•x0

•z

•y

•hy(z)

X

p

•x0

•p(z)

β

α

α

There is a unique lift of α = pβ : I → X to a path α : I → X such thatα(0) = y; define

hy(z) = α#(y) = α(1) ∈ X

91

withphy(z) = pα(1) = α(1) = pβ(1) = p(z) ∈ X .

The construction of hy(z) is independent of the choice of β, by the simple-

connectedness of X. The function

hy : X → X ; z → hy(z)

is continuous because X (and hence X) is locally path-connected. For every

t ∈ I the restriction α| : [0, t] → X is the unique lift of the restrictionα| : [0, t]→ X such that α(0) = y, so that

α(t) = hy(β(t)) ∈ X

and in particular

y = α(0) = hy(β(0)) = hy(x0) ∈ X .

Example 11.20. It is possible to apply Theorem 11.19 to the covering projec-tion p : R→ S1 of Example 11.9, since R is simply-connected. It is preciselythe path-lifting Theorem 11.12 (i) which is used to prove that for every closedpath α : I → S1 with α(0) = 1, there exists a lift α : I → R with α(0) = 0,allowing the degree to be defined by

degree(α) = α#(0) = α(1)− α(0) ∈ Z ,

with Theorem 11.12 (ii) showing that the degree depends only on the rel0, 1 homotopy class of α. The bijection of Theorem 11.19 (i) is given bythe degree map

π1(S1, 1)→ p−1(1) = Z ; α→ degree(α) = α(1) ,

since α(0) = 0. The bijection of Theorem 11.19 (ii) is given by

p−1(1) = Z→ Homeop(R) ; n→ hn

with hn : R → R;x → x + n the unique covering translation such thathn(0) = n. The isomorphism of Theorem 11.19 (iii) is given by

π1(S1, 1)→ Homeop(R) ; α→ hn

with α(1) = n. In particular, we have that π1(S1) = Z is the infinite cyclic

group generated by 1 : S1 → S1.

92

Lecture 20Example 11.21. Define homeomorphisms A,B : R2 → R2 by

A(x, y) = (x+ 1, y) , B(x, y) = (x, y + 1)

such that

AB = BA : R2 → R2 ; (x, y) 7→ (x+ 1, y + 1)

andArBs(x, y) = (x+ r, y + s) for any r, s ∈ Z .

Let G ⊂ Homeo(R2) be the subgroup generated by A,B, that is

G = ArBs | r, s ∈ Z ⊂ Homeo(R2)

with multiplication by

(ArBs)(AtBu) = Ar+tBs+u ∈ G .

The quotient space X = X/G is obtained from the unit square I2 = I × Iby identifying

(0, y) ∼ A(0, y) = (1, y) and (x, 0) ∼ B(x, 0) = (x, 1) .

The quotient space

•(x, 0)

•(x, 1)

•(0, y) •(1, y)

is a torusX = X/G = I2/∼ = T = S1 × S1 .

93

Theorems 11.8, 11.19 apply to the universal covering projection

p : X = R2 → X = X/G = T .

The fundamental group of the torus is thus the free abelian group on 2generators

π1(T ) = Homeop(X) = G = Z⊕ Z .

Example 11.22. Define homeomorphisms A,B : R2 → R2 by

A(x, y) = (x+ 1, y) , B(x, y) = (−x+ 1, y + 1)

such that

AB = BA−1 : R2 → R2 ; (x, y) 7→ (−x+ 2, y + 1)

and

ArBs(x, y) = ((−1)sx+ r +1 + (−1)s+1

2, y + s) for any r, s ∈ Z .

Let G ⊂ Homeo(R2) be the (nonabelian) subgroup generated by A,B, thatis

G = ArBs | r, s ∈ Z ⊂ Homeo(R2)

with multiplication by

(ArBs)(AtBu) = Ar+tBs(−1)t+u ∈ G .

The quotient space X = X/G is obtained from the unit square I2 by identi-fying

(0, y) ∼ A(0, y) = (1, y) and (x, 0) ∼ B(x, 0) = (1− x, 1) .

The quotient space

•(x, 0)

•(1− x, 1)

•(0, y) •(1, y)

is a Klein bottleX = X/G = I2/∼ = K .

94

Theorems 11.8, 11.19 apply to the universal covering projection

p : X = R2 → X = X/G = K .

The fundamental group of the Klein bottle is thus

π1(K) = Homeop(X) = G .

Example 11.23. The figure 8 space

X = S1 × 1 ∪ 1 × S1 ⊂ S1 × S1

Covering Spaces Section 1.3 57

As our general theory will show, these examples for n ≥ 1 together with the

helix example exhaust all the connected coverings spaces of S1 . There are many

other disconnected covering spaces of S1 , such as n disjoint circles each mapped

homeomorphically onto S1 , but these disconnected covering spaces are just disjoint

unions of connected ones. We will usually restrict our attention to connected covering

spaces as these contain most of the interesting features of covering spaces.

The covering spaces of S1∨S1 form a remarkably rich family illustrating most of

the general theory very concretely, so let us look at a few of these covering spaces to

get an idea of what is going on. To abbreviate notation, set X = S1∨ S1 . We view this

as a graph with one vertex and two edges. We label the edges

b aa and b and we choose orientations for a and b . Now let

X be any other graph with four edges meeting at each vertex,

and suppose the edges of X have been assigned labels a and b and orientations in

such a way that the local picture near each vertex is the same as in X , so there is an

a edge oriented toward the vertex, an a edge oriented away from the vertex, a b edge

oriented toward the vertex, and a b edge oriented away from the vertex. To give a

name to this structure, let us call X a 2 oriented graph.

The table on the next page shows just a small sample of the infinite variety of

possible examples.

Given a 2 oriented graph X we can construct a map p : X→X sending all vertices

of X to the vertex of X and sending each edge of X to the edge of X with the same

label by a map that is a homeomorphism on the interior of the edge and preserves

orientation. It is clear that the covering space condition is satisfied for p . The con-

verse is also true: Every covering space of X is a graph that inherits a 2 orientation

from X .

As the reader will discover by experimentation, it seems that every graph having

four edges incident at each vertex can be 2 oriented. This can be proved for finite

graphs as follows. A very classical and easily shown fact is that every finite connected

graph with an even number of edges incident at each vertex has an Eulerian circuit,

a loop traversing each edge exactly once. If there are four edges at each vertex, then

labeling the edges of an Eulerian circuit alternately a and b produces a labeling with

two a and two b edges at each vertex. The union of the a edges is then a collection

of disjoint circles, as is the union of the b edges. Choosing orientations for all these

circles gives a 2 orientation.

It is a theorem in graph theory that infinite graphs with four edges incident at each

vertex can also be 2 oriented; see Chapter 13 of [Koenig 1990] for a proof. There is

also a generalization to n oriented graphs, which are covering spaces of the wedge

sum of n circles.

has universal cover X an infinite tree with one vertex for each word in twoletters a, b

w = am1bn1am2bn2 . . . amkbnk (mj, nj ∈ Z, k > 0)

and one edge joining w,w′ if and only if w−1w′ ∈ a, a−1, b, b−1.

95

Covering Spaces Section 1.3 59

A simply-connected covering space of X can be constructed in the following way.

Start with the open intervals (−1,1) in the coordinate

axes of R2 . Next, for a fixed number λ , 0 < λ < 1/2, for

example λ = 1/3, adjoin four open segments of length

2λ , at distance λ from the ends of the previous seg-

ments and perpendicular to them, the new shorter seg-

ments being bisected by the older ones. For the third

stage, add perpendicular open segments of length 2λ2

at distance λ2 from the endpoints of all the previous

segments and bisected by them. The process is now

repeated indefinitely, at the nth stage adding open segments of length 2λn−1 at dis-

tance λn−1 from all the previous endpoints. The union of all these open segments is

a graph, with vertices the intersection points of horizontal and vertical segments, and

edges the subsegments between adjacent vertices. We label all the horizontal edges

a , oriented to the right, and all the vertical edges b , oriented upward.

This covering space is called the universal cover of X because, as our general

theory will show, it is a covering space of every other connected covering space of X .

The covering spaces (1)–(14) in the table are all nonsimply-connected. Their fun-

damental groups are free with bases represented by the loops specified by the listed

words in a and b , starting at the basepoint x0 indicated by the heavily shaded ver-

tex. This can be proved in each case by applying van Kampen’s theorem. One can

also interpret the list of words as generators of the image subgroup p∗(π1(X, x0)

)in π1(X,x0) =

⟨a,b

⟩. A general fact we shall prove about covering spaces is that

the induced map p∗ :π1(X, x0)→π1(X,x0) is always injective. Thus we have the at-

first-glance paradoxical fact that the free group on two generators can contain as a

subgroup a free group on any finite number of generators, or even on a countably

infinite set of generators as in examples (10) and (11).

Changing the basepoint vertex changes the subgroup p∗(π1(X, x0)

)to a conju-

gate subgroup in π1(X,x0) . The conjugating element of π1(X,x0) is represented by

any loop that is the projection of a path in X joining one basepoint to the other. For

example, the covering spaces (3) and (4) differ only in the choice of basepoints, and

the corresponding subgroups of π1(X,x0) differ by conjugation by b .

The main classification theorem for covering spaces says that by associating the

subgroup p∗(π1(X, x0)

)to the covering space p : X→X , we obtain a one-to-one

correspondence between all the different connected covering spaces of X and the

conjugacy classes of subgroups of π1(X,x0) . If one keeps track of the basepoint

vertex x0 ∈ X , then this is a one-to-one correspondence between covering spaces

p : (X, x0)→(X,x0) and actual subgroups of π1(X,x0) , not just conjugacy classes.

Of course, for these statements to make sense one has to have a precise notion of

when two covering spaces are the same, or ‘isomorphic.’ In the case at hand, an iso-

The pictures are taken from pages 57,59 of Hatcher’s Algebraic Topology,which can be downloaded fromhttp://www.math.cornell.edu/ hatcher/AT/ATpage.html

The projection p : X → X sends each vertex to 1 ∈ S1, and maps eachedge onto S1 (one-one on the interior). For each word w let hw : X → X bethe unique covering translation with hw(w′) = ww′, so that hw′w = hw′hw.The fundamental group of X is the free nonabelian group on two generators

π1(X) = Homeop(X) = hw |w = Z ∗ Z ⊂ Homeo(X) .

Remark 11.24. If p : X → X is a universal covering projection satisfying thehypothesis of Theorem 11.19, then for any subgroup

G ⊆ π1(X) = Homeop(X)

there is defined a universal covering projection

q : Y = X → Y = X/G

satisfying the hypothesis of Theorem 11.19, with

π1(Y ) = Homeoq(Y ) = G .

If G ⊆ π1(X) is a normal subgroup the projection r : Y → X is a coveringprojection with

Homeor(Y ) = π1(X)/G .

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Remark 11.25. Theorem 11.19 gives a geometric method for computing thefundamental group of a path-connected space X which admits a universalcovering p : X → X, namely

π1(X, x0) = Homeop(X) = p−1(x0) .

For any path-connected space X and x0 ∈ X, let X be the topologicalspace of equivalence class of paths α : I → X such that α(0) = x0, withα ∼ α′ if there exists a rel 0, 1 homotopy β : α ' α′ : I → X, and

p : X → X ; α 7→ α(1) .

It is a theorem that p is the universal covering projection of X with fibreF = p−1(x0) = π1(X, x0) if X is semi-locally simply-connected, meaning thatfor every x ∈ X there exists an open subset U ⊆ X with x ∈ U such that theinclusion i : U → X induces the trivial homomorphism i∗ = 1 : π1(U, x) →π1(X, x) (in which case p−1(U) is homeomorphic to U × π1(X, x)).

In general, this is too synthetic a construction of the universal cover to beof use in the computation of π1(X). In practice, a geometrically interesting

space X has a geometrically interesting universal cover X, and this can beused to compute π1(X).

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Lecture 21

12 Fundamental groups of surfaces and van

Kampen’s theorem — non-examinable

This section is a brief outline of the application of the fundamental group tothe classification of compact surfaces (see Definition 6.1).

Remark 12.1. A double cover of a space X is a covering projection p : Y →X such that the fibre F = −1,+1 is the discrete space with two points.

A nonorientable surface M has an orientable double cover p : M → M ,

with M an orientable surface. For example, K = T 2, RP2

= S2.

Example 12.2. 1. M(1) = RP2, the projective plane, with orientable dou-

ble cover M(1) = H(0) = S2.

2. M(2) = K, the Klein bottle, with orientable double cover M(2) =H(1) = T 2.

The classification theorem 6.9 states that the following conditions onconnected compact surfaces M,M ′ are equivalent:

1. M ,M ′ are homeomorphic,

2. M and M ′ are both orientable or both nonorientable, and have thesame genus,

3. the fundamental groups π1(M), π1(M′) are isomorphic.

Proposition 12.3. If we write FS for the free abelian group generated by aset S, and 〈x〉 for the normal subgroup generated by x, then we have:

1. For the orientable surface H(g) with g ≥ 1, there exists a coveringprojection p : R2 → H(g) with

π1(H(g)) = Homeop(R2) = Fa1, b1, . . . , ag, bg/〈[a1, b1][a2, b2] . . . [ag, bg]〉 .

2. For the nonorientable surface M(g) with g ≥ 2, there exists a coveringprojection q : R2 →M(g) with

π1(M(g)) = Homeoq(R2) = Fc1, c2, . . . , cg/〈(c1)2(c2)2 . . . (cg)2〉 .

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3. The orientable double cover (4.3) of M(g) is

M(g) = H(g − 1) .

Proof. See Armstrong p.168 and Stillwell, pp.82–84.

Surfaces have interesting geometric properties:

Example 12.4. 1. The expression of the torus as an quotient space

H(1) = T 2 = R2/(x, y) ∼ (x+ 1, y) ∼ (x, y + 1)

makes apparent the covering projection p : R2 → H(1) = T . Thehomeomorphisms

a1 : R2 → R2 ; (x, y) → (x+ 1, y) ,

b1 : R2 → R2 ; (x, y) → (x, y + 1)

generate Homeop(R2), with the relation a1b1 = b1a1 (or equivalently[a1, b1] = 1).

2. The expression of the Klein bottle as an quotient space

M(2) = K = R/(x, y) ∼ (x+ 1, y) ∼ (1− x, y + 12)

makes apparent the covering projection q : R2 → M(2) = K. Thehomeomorphisms

c1 : R2 → R2 ; (x, y) → (1− x, y − 12) ,

c2 : R2 → R2 ; (x, y) → (2− x, y + 12)

generate Homeoq(R2), with the relation (c1)2(c2)

2 = 1.

3. Note that q factors as q : R2 p−→ T 2 r−→ K with r : T 2 → K a coveringprojection with fibre Z2, and

a1 = c2c1 , b1 = (c1)−2 : R2 → R2 ,

with π1(T2) = Homeop(R2) / π1(K) = Homeoq(R2) a normal subgroup

of index 2. Thus T 2 = K is the orientable double cover (4.3) of K.

Remark 12.5. See Stillwell’s Geometry of surfaces (51.51 Sti) for a beau-tiful account of the connections between surfaces and hyperbolic geome-try: the groups of homeomorphisms π1(H(g)) = Homeop(R2), π1(M(g)) =Homeoq(R2) in 12.3 for g ≥ 2 are groups of isometries of the hyperbolic

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plane H 2 (= the topological space R2 with a hyperbolic geometry). It ispossible to realize each connected compact surface M of genus g ≥ 2 as anquotient space M = H 2/π1(M) with π1(M) acting on H 2 by symmetries ofa tessellation of H 2. (The projection H 2 →M is a covering as in Definition11.1).

Example 12.6. The Poincare D2-model for the hyperbolic plane H 2 is theopen disc (D2), in which the ‘lines’ are the circular arcs orthogonal to theboundary circle (at infinity) S1. The orientable surface H(2) of genus 2 isthe ‘hyperbolic surface’

H(2) = H 2/π1(H(2))

with π1(H(2)) = a1, b1, a2, b2 | [a1, b1][a2, b2] acting on H 2 by the symme-tries of a tessellation of H 2 by regular (hyperbolic) octagons, as in the firstpicture:

to a neighbour.Here is a picture of H(2) cut open along the loops a1, b1, a2, b2

The pictures are from Stillwell’s books.

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12.1 Fundamental groups via van Kampen’s theorem

If we look at the fundamental polygon associated to a surface, then each edgegives a closed loop on the surface, and hence an element of the fundamentalgroup. The path which traces the whole boundary of the polygon is homo-topic to a constant, by shrinking it to the centre of the polygon. For instance,

for the orientable surfaceH(2) of genus 2,we have [a], [b], [c], [d] ∈ π1H(2), with the class of the boundary path [aba−1b−1cdc−1d−1] =1. We therefore have a group homomorphism

Fa, b, c, d/〈aba−1b−1cdc−1d−1〉 → π1H(2),

and this is precisely how the group isomorphisms of Proposition 12.3 arise.The following theorem establishes isomorphisms of this kind:

Definition 12.7. Given groups Gi = FSi/〈Ti〉 for i = 1, 2, 3, and grouphomomorphisms θ : G2 → G1 and φ : G2 → G3, the pushout G1 ∗G2 G3 isgiven by

F (S1 ∪ S3)/〈T1 ∪ T3 ∪ φ(s)θ(s)−1 : s ∈ S2〉.

This is a very ad hoc definition, which we have not even shown is well-defined (i.e. independent of the choices of generators and relations). Thepushout is characterised by the following universal property, which ensuresit is well-defined up to group isomorphism:

Proposition 12.8. Given groups G1, G2, G3 as above and a group H, theset of group homomorphisms α : G1 ∗G2 G3 → H is isomorphic to the setof pairs

(α1 : G1 → H,α3 : G3 → H)

or group homomorphisms satisfying α1 θ = α3 φ : G2 → H.For the natural group homomorphisms γ1 : G1 → G1 ∗G2 G3 and

γ1 : G3 → G1 ∗G2 G3, the homomorphisms α1, α3 are given by αi = α γi.

Theorem 12.9 (van Kampen (very non-examinable)). Let X be a topologicalspace which is the union of two open and path connected subspaces U , W .Suppose U ∩ W is path connected and nonempty, and let x0 be a point inU ∩W that will be used as the base of all fundamental groups. Then

π1X ∼= (π1U) ∗π1(U∩W ) (π1W ).

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In particular, if W is contractible, then

π1X ∼= (π1U)/〈j∗π1(U ∩W )〉,

for j : U ∩W → U .

Proof. See Theorem 1.20 in Hatcher’s book.

Now given a compact surface X = P/ ∼ for a fundamental 2n-gon P , welet W be the interior of P , and U = (P \ 0)/ ∼. Thus W is contractible,U is homotopy equivalent to n circles joined together at a single point, andU ∩ W is a punctured open polygon, so homotopy equivalent to a circle.Repeatedly applying van Kampen’s theorem, we see that π1U is the freegroup on n generators, and we see that j∗ sends the generator of π1(U ∩W )to the expression corresponding to the equivalence relation on the boundary.

For instance the relation aba−1b−1 for H(1) in Proposition 12.3 exactlycorresponds to our standard representation for the torus, while the descrip-tion of M(2) relates to our usual representation of the Klein bottle by settinga = c1, b = c1c2, so (c1)

2(c2)2 = aba−1b,

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Lecture 22

13 A lightning introduction to homology (non-

examinable)

Although we saw in §5.1 that higher homotopy groups πnX are quite easyto define, they can be very difficult to compute. A major open problem is tocomplete the calculation of the higher homotopy groups of n-spheres.

There is another series of groups, the homology groups Hn(X) (n ≥ 0),which are quite easy to compute, but tricky to define — see for instance Ch.2 of Hatcher’s book. We actually saw the first homology group H1(X) indisguise in Proposition 6.13, since it is the abelianisation of π1X. The 0thhomology group H0X is even easier to describe: it is the free abelian groupgenerated by the set π0X.

Instead of giving a definition of homology groups, I will now explain howthey are computed. We start with four basic properties:

1. The homology groups of the empty space ∅ are given by Hi(∅) = 0 forall i ≥ 0.

2. The homology groups of a point ∗ are given byH0(∗) = Z andHi(∗) = 0for all i > 0.

3. A map f : X → Y induces homomorphisms f∗ : HnX → HnY .

4. If f is a homotopy equivalence, then f∗ is an isomorphism.

On their own, these properties only suffice to describe homology of con-tractible and empty spaces, but there is an analogue of homology of vanKampen’s theorem, called the Mayer–Vietoris theorem. We begin with adefinition.

Definition 13.1. A sequence A → B → C → D → . . . of group homomor-phisms is called exact if the image of each map is the kernel of the next.

Theorem 13.2 (Mayer-Vietoris). Let X be a topological space and A,B betwo subspaces whose interiors cover X (the interiors need not be disjoint).Denote the inclusion maps by I : A → X and j : B → X. Then there isa long exact sequence

. . .→ Hn(A ∩B)(j∗,i∗)−−−→ HnA⊕HnB

i∗−j∗−−−→ HnX → Hn−1(A ∩B)(j∗,i∗)−−−→ . . .

. . .i∗−j∗−−−→ H1X → H0(A ∩B)

(j∗,i∗)−−−→ H0A⊕H0Bi∗−j∗−−−→ H0X → 0.

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Example 13.3. The simplest application of the theorem comes when the sub-spaces A,B are disjoint, so X = AtB. In that case, the long exact sequencebreaks down into exact sequences

0→ HnA⊕HnBi∗−j∗−−−→ HnX → 0,

giving an isomorphism Hn(A tB) ∼= HnA⊕HnB.

Remark 13.4. The Mayer–Vietoris theorem does not necessarily determineH∗X from H∗A and H∗B. This is because there can exist short exact se-quences with the same ends but a different term in the middle, such as:

0 // Z 2 // Z // Z/2Z // 0,

0 // Z // Z⊕ Z/2Z // Z/2Z // 0.

Exercise 13.5. 1. By expressing Sn as a union of contractible open sub-spaces, prove inductively that for n ≥ 1,

Hi(Sn) ∼=

Z i = 0, n

0 else.

2. Now try some other spaces:

For instance, the figure of 8 has homology groups Z,Z2, 0, 0, . . ., and ingeneral S1 ∨ . . . ∨ S1︸ ︷︷ ︸

n

has homology groups Z,Zn, 0, 0, . . ..

The orientable surfaces H(g) have homology groups Z,Z2g,Z, 0, 0, . . .,while the non-orientable surfaces M(g) have homology groups Z,Zg ⊕Z/2Z, 0, 0, . . ..

Exercise 13.6. Can you see how to use homology to show that Rn is nothomeomorphic to Rm for any m 6= n.

Exercise 13.7. Can you see how to use homology to prove the higher-dimensionalnon-retraction theorem: that for n ≥ 1 there is no map g : Dn → Sn−1

which is the identity on Sn−1 ⊂ Dn (cf. Theorem 10.8, which used the fun-damental group). A corollary is the Brouwer fixed point theorem, that everymap f : Dn → Dn has a fixed point.

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