Introduction to Algebraic Topologyuregina.ca/~pso748/AT_notes.pdf · General Topology (a brief...

Introduction to Algebraic Topology

Paul Arnaud Songhafouo Tsopméné

Contents

0 General Topology (a brief recall) 5

0.1 Definition and Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

0.2 Subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

0.3 Continuous function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

0.4 Homeomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

0.5 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

0.6 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1 Homotopy type 9

1.1 Definition and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2 Contractible spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 The fundamental group 13

2.1 Definition of π1(X) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.1 Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.2 Homotopy of paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.3 Loops and the definition of the fundamental group . . . . . . . . . . . . . . . . . . . . 15

2.1.4 The dependence of π1(X,x) on the choice of the basepoint . . . . . . . . . . . . . . . . 16

2.2 The fundamental group of the circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2.1 Four lemmas in order to prove that π1(S1) = Z . . . . . . . . . . . . . . . . . . . . . . 17

2.2.2 Proving that π1(S1) = Z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.3 Induced homomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4 Applications of the fundamental group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.4.1 The Brouwer fixed point theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.4.2 The fundamental theorem of algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5 The Van Kampen theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.5.1 Free products of groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3

4 CONTENTS

2.5.2 Van Kampen’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3 Homology 31

3.1 Chain complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2 Simplicial homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2.1 ∆-complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.2.2 Definition of the simplicial homology and examples . . . . . . . . . . . . . . . . . . . . 36

3.3 Singular homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.3.2 Exact sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.3.3 Relative homology groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Bibliography 41

Chapter 0

General Topology (a brief recall)

In this chapter we recall briefly the following concepts: Topological space, subsapce, continuous function,homeomorphism, compactness, and connectedness.

0.1 Definition and Example

Definition 0.1.1. A topological space or just a space is a couple (X,O(X)) where X is a set, and O(X)

is a family of subsets of X satisfying the following three axioms:

(T1) The emptyset and the set X itself belong to O(X).

(T2) O(X) is closed under finite intersections. That is, for any finite family {Uj}j∈J of elements of O(X),∩j∈JUj ∈ O(X).

(T3) O(X) is closed under unions. That is, for any family {Ui}i∈I of elements of O(X), ∪i∈IUi ∈ O(X).

Elements of O(X) are called open subsets of X, and the complement of elements of O(X) are called closedsubsets of X. The family O(X) is called a topology on X.

Example 0.1.2. Let a ∈ Rn, and let r ≥ 0 be a real number. An open ball in Rn of center a and radius ris the subset B(a, r) defined by

B(a, r) = {x ∈ Rn : ‖x− a‖ < r}.

Here ‖.‖ stands for the standard Euclidean norm in Rn. Note that B(a, r) = ∅ when r = 0. Define O(Rn)

as

O(Rn) = set of unions of open balls.

One can easily check that O(Rn) is a topology on Rn.

Throughout this course, Rn will be equipped with the topology O(Rn) we just defined. We will often writeX for (X,O(X)) when the topology is understood.

5

6 CHAPTER 0. GENERAL TOPOLOGY (A BRIEF RECALL)

0.2 Subspace

Definition 0.2.1. Let (X,O(X)) be a topological space, and A be a subset of X. We say that A is endowedwith the subspace topology if the topology, O(A), on A is of the form:

O(A) = O(X) ∩A = {U ⊆ A : there exists W ∈ O(X) such that U = W ∩A} .

In that case A is called a subspace of X, and O(A) is called the induced topology.

In this course, all subsets of Rn will be endowed with the subspace topology. We will often deal with thefollowing examples of subspaces.

Example 0.2.2. (i) X = Rn+1, and A = Sn is the n-dimensional sphere or just the n-sphere. Recallthat

Sn =

{(x1, · · · , xn+1) ∈ Rn+1 :

n+1∑i=1

x2i = 1

}.

If n = 0 we have S0 = {−1, 1} ⊆ R. So S0 consists of only two elements. The 1-sphere S1 is calledthe circle, and the 2-sphere will be sometimes just called the sphere.

(ii) X = Rn, and A = Dn is the n-dimensional disk or just the n-disk. Recall that

Dn =

{(x1, · · · , xn) ∈ Rn :

n∑i=1

x2i ≤ 1

}.

The space D0 is a single point, while D1 = [−1, 1] ⊆ R. The 2-disk D2 will sometimes just called thedisk. Note that the boundary of Dn is Sn−1.

(iii) X = R, and A = [0, 1] is the unit interval. Here are some examples of open subsets of [0, 1]. For anya ∈ (0, 1), the interval [0, a) is an open subset of [0, 1] because [0, a) = (−1, a)∩ [0, 1] and (−1, a) is anopen subset of R. Similarly, (a, 1] and (a, b), where b ∈ (0, 1) is such that a < b, are open subsets of[0, 1].

0.3 Continuous function

Most of spaces we will encounter in this course are “metrizable”, that is, there is a notion of distance betweentwo points. If one writes x ≈ y to saying that the distance between x and y is small enough (or equivalentlyto saying that x and y are very closed to each other), then roughly speaking a function f : X −→ Y betweentwo topological spaces is continuous if for any x, x′ ∈ X, we have the implication

x ≈ x′ ⇒ f(x) ≈ f(x′).

Explicitly, we have the following definition.

Definition 0.3.1. A function f : (X,O(X)) −→ (Y,O(Y )) is continuous if for every U ∈ O(Y ), f−1(U) ∈O(X).

The word “map” will mean continuous function. Or if we want to emphasize about the continuity, we willsay continuous map.

0.4. HOMEOMORPHISM 7

0.4 Homeomorphism

Roughly speaking, two spaces X and Y are homeomorphic if one can carry X to Y by a continuous deforma-tion and vise versa. By continuous deformation, we mean a deformation in which nearby points stay nearbyduring all the process. So when performing such a deformation, it is allowed to stretch or bend, but it is notallowed to glue or tear. To be more precise, we have the following.

Definition 0.4.1. We say that two topological spaces X and Y are homeomorphic, and we denote X ∼= Y ,if there exists a bijection f : X −→ Y such that both f and f−1 are continuous.

Example 0.4.2. The open interval (0, 1) is homeomorphic to R. Indeed, let f : (0, 1) −→ (−π2 ,π2 ) defined

by f(t) = (1 − t)(−π2 ) + tπ2 , and let g : (−π2 ,π2 ) −→ R defined as g(x) = tan(x). One can easily check that

f is a homeomorphism. So (0, 1) ∼= (−π2 ,π2 ). It is well known that the second map is a homeomorphism as

well. So (−π2 ,π2 ) ∼= R.

The map f from Example 0.4.2 or more generally a map of the form (1 − t)a + tb will be extensively usedin next section, and also in Chapter 2 to define most of homotopies.

0.5 Compactness

Intuitively, a space is said to be compact if it contains no “holes”. For instance, the circle without one pointis not compact.

Definition 0.5.1. A topological space X is said to be compact if for any family {Ui}i∈I of open subsets ofX such that X = ∪i∈IUi, there is a finite subset J of I such that X = ∪j∈JUi.

Such a family {Ui}i∈I is called an open cover of X.

Example 0.5.2. We admit the following theorem due to Heine-Borel. Recall that a subset of Rn is boundedif it is contained in one ball.

Theorem. A subspace of Rn is compact if and only if it is closed and bounded.

As a consequence, we have the following.

(i) All closed and bounded intervals (that is, intervals of the form [a, b]) of R are compact. In particular,the interval [0, 1] is compact.

(ii) For all n ≥ 0, the n-sphere Sn is compact.

(iii) For all n ≥ 1, Rn is not compact because it is not bounded. However R0, which is a single point, iscompact.

0.6 Connectedness

Roughly speaking, a space X is said to be connected if it can be “held by only one hand” or if it consists ofonly one piece. Explicitly, we have the following definition.

Definition 0.6.1. We say that a space X is connected if it cannot be represented as the union of two ormore disjoint nonempty open subsets.

8 CHAPTER 0. GENERAL TOPOLOGY (A BRIEF RECALL)

Example 0.6.2. We admit the following well known topological results.

(i) For all n ≥ 0, Rn and Dn are connected.

(ii) For all n ≥ 1, Sn is connected. But S0 = {−1, 1} is not connected since S0 = {−1} ∪ {1} where both{−1} and {1} are open subsets of S0. Recall that S0 ⊆ R is endowed with the subspace topology as wementioned just after Definition 0.2.1.

Chapter 1

Homotopy type

In Section 0.4, we introduced the notion of homeomorphism between two topological spaces X and Y . Inthis chapter we are going to introduce a weaker notion called homotopy equivalence. By “weaker”, we meanthat two spaces can be homotopy equivalent without being homeomorphic. For instance, X = D2 andY = {(0, 0)} are not homeomorphic since manifestly there is no bijection from X to Y . But they arehomotopy equivalent as we shall see in Example 1.1.9 below. Contrary to the topological word (where twospaces are the same if they are homeomorphic), in the algebraic topology world two spaces are the same ifthey are homotopy equivalent.

From now onI = [0, 1].

Before we introduce the notion of homotopy equivalence, we need to first define the notion of homotopybetween maps.

1.1 Definition and examples

Definition 1.1.1. Let f, g : X −→ Y be continuous maps. We say that f is homotopic to g, and we writef ∼ g, if there exists a continuous map H : X × I −→ Y such that H(x, 0) = f(x) and H(x, 1) = g(x) forall x ∈ X. The map H is called a homotopy from f to g.

Sometimes we will write Ht(x) for H(x, t). One should think t as the time running from 0 to 1. At theinitial time t = 0 we have H0 = f , and at the final time t = 1 we have H1 = g.

Example 1.1.2. The maps f, g : R −→ R defined as f(x) = ex and g(x) = x2 are homotopic. Indeed,one can easily check that the map H : R × I −→ R defined as H(x, t) = (1 − t)ex + tx2 is a homotopyfrom f to g. In fact, any continuous maps f and g from R to R are always homotopic (just define H asH(x, t) = (1− t)f(x) + tg(x)).

Definition 1.1.3. A homotopy H is called linear if it is of the form H(x, t) = (1− t)a(x) + tb(x).

Proposition 1.1.4. Let X and Y be two spaces. The relation ∼ is an equivalence relation in the set ofmaps from X to Y .

Proof. (i) Reflexivity. For any map f : X −→ Y , one has f ∼ f , the homotopy from f to f being theconstant homotopy H : X × I −→ Y,H(x, t) = f(x).

9

10 CHAPTER 1. HOMOTOPY TYPE

(ii) Symmetry. Let f, g : X −→ Y be two maps such that f ∼ g. We want to show that g ∼ f . Let H bea homotopy from f to g, and define K : X × I −→ Y by K(x, t) = H(x, 1− t). We claim that K is ahomotopy from g to f . Indeed, first of all, the map K is continuous since H is continuous. Next at thetime t = 0, we have K(x, 0) = H(x, 1) = g(x), and at the time t = 1 we have K(x, 1) = H(x, 0) = f(x).

(iii) Transitivity. Let f, g, h : X −→ Y be three maps such that f ∼ g and g ∼ h. We want to prove thatf ∼ h. Let H be a homotopy from f to g, and K be a homotopy from g to h. Define L : X × I −→ Y

by

L(x, t) =

{H(x, 2t) if 0 ≤ t ≤ 1

2

K(x, 2t− 1) if 12 ≤ t ≤ 1.

The map L is continuous because at the time t = 12 we have H(x, 2 1

2 ) = H(x, 1) = g(x), and atthe same time we have K(x, 2t − 1) = K(x, 0) = g(x). Moreover L(x, 0) = H(x, 0) = f(x), andL(x, 1) = K(x, 1) = h(x) for all x.

Given maps f : X −→ Y and g : Y −→ Z, the composition g ◦ f will be sometimes just denoted gf .

Proposition 1.1.5. Let f0, f1 : X −→ Y and g0, g1 : Y −→ Z be continuous maps such that f0 ∼ f1 andg0 ∼ g1. Then g0f0 ∼ g1f1.

Proof. Let H : X × I −→ Y be a homotopy from f0 to f1, and K : Y × I −→ Z be a homotopy from g0

to g1. Then for any x ∈ X,H(x, 0) = f0(x) and H(x, 1) = f1(x). And for any y ∈ Y,K(y, 0) = g0(y) andK(y, 1) = g1(y). Define L : X × I −→ Z by L(x, t) = K(H(x, t), t). The map L is continuous since it is acomposition of continuous maps. Furthermore, we have

L(x, 0) = K(H(x, 0), 0) = g0(H(x, 0)) = g0f0(x), and L(x, 1) = K(H(x, 1), 1) = g1(H(x, 1)) = g1f1(x),

which completes the proof.

The following is an immediate consequence of Proposition 1.1.5.

Corollary 1.1.6. (i) Let f0, f1 : X −→ Y and g : Y −→ Z be maps such that f0 ∼ f1. Then gf0 ∼ gf1.

(ii) Let f : X −→ Y and g0, g1 : Y −→ Z such that g0 ∼ g1. Then g0f ∼ g1f .

Definition 1.1.7. A map f : X −→ Y is called a homotopy equivalence if there exists another map g : Y −→X such that gf ∼ idX and fg ∼ idY .

Here idX : X −→ X stands for the identity map. Note that g is just a map (not necessarily the inverse off) from Y to X.

Now we are ready to define the notion of homotopy between topological spaces.

Definition 1.1.8. We say that two topological spaces X and Y have the same homotopy type or are homo-topy equivalent, and we denote X ' Y , if there exists a homotopy equivalence X −→ Y .

Before we see some examples, recall that a subspace C ⊆ Rn is said to be convex if for any a, b ∈ C, for anyt ∈ [0, 1], the point (1− t)a+ tb belongs to C. In convex subspaces, the linear homotopies are very useful aswe will see in the following examples and in Chapter 2.

Example 1.1.9. (i) The space R2\{(0, 0)} has the same homotopy type as the circle S1. That is,

R2\{(0, 0)} ' S1.

1.2. CONTRACTIBLE SPACES 11

(ii) The unit disk D2 is homotopy equivalent to the one-point space {(0, 0)}. That is,

D2 ' {(0, 0)}.

Solution.

(i) Define f : R2\{(0, 0)} −→ S1 by f(x) = x‖x‖ , and define g : S1 −→ R2\{(0, 0)} by g(x) = x. Of

course both f and g are continuous. Moreover, a simple calculation shows that fg(x) = x for allx ∈ S1. So fg = id, which implies in particular that fg ∼ id. To prove that gf ∼ id, defineH : R2\{(0, 0)}× I −→ R2\{(0, 0)} by H(x, t) = (1− t)gf(x) + tx. The map H is well defined becausefor any x 6= (0, 0), the segment [x, gf(x)] does not pass through the origin (0, 0). Clearly the map His continuous. Moreover, at the time t = 0 we have H(x, 0) = gf(x), and at the time t = 1 we haveH(x, 1) = x.

(ii) Let f : D2 −→ {(0, 0)} be the unique map from D2 to {(0, 0)}. And let g : {(0, 0)} −→ D2 defined byg(0, 0) = (0, 0). Of course we have fg = id. As before, the homotopy from gf to id is again the linearhomotopy defined as H(x, t) = (1− t)gf(x) + tx.

1.2 Contractible spaces

We will write pt for the one-point space. That is, the space that contains only one element.

Definition 1.2.1. A topological space X is said to be contractible if X ' pt.

Example 1.2.2. (i) The space R is contractible. (ii) The disk D2 is contractible as well.

Solution.

(i) Take pt = {0}, and define g : pt −→ R as g(0) = 0. Let f : R −→ pt be the unique map from R to pt.Certainly one has gf = id. Define H as H(x, t) = tx. Clearly this is a homotopy from fg to id. SoR ' pt.

(ii) The fact that D2 is contractible comes immediately from Example 1.1.9 (ii).

Example 1.2.3. The sphere S0 = {−1, 1} is not contractible. Indeed, assume that it is. Then thereexist continuous maps f : S0 −→ pt and g : pt −→ S0 such that gf ∼ idS0 and fg ∼ idpt. Since gf ishomotopic to idS0 , there exists a continuous map H : S0 × [0, 1] −→ S0 such that H(x, 0) = gf(x) andH(x, 1) = idS0(x) = x. Depending on the value of g, we need to deal with two cases (set pt = {∗}).

1. Assume g(∗) = 1. Then for every x ∈ S0, one has gf(x) = 1. Define φ : [0, 1] −→ S0 as φ(t) =

H(−1, t). Certainly φ is continuous since H is continuous. So φ([0, 1]) is connected because φ iscontinuous and [0, 1] is connected. Now let us compute φ([0, 1]). We have φ(0) = H(−1, 0) = gf(−1) =

1 and φ(1) = H(−1, 1) = −1. This implies that φ([0, 1]) = {−1, 1}, which is a contradiction since{−1, 1} is not connected.

2. The case g(∗) = −1 is handled in the same way.

To prove directly that a space X is not contractible (or more generally to prove that two topological spacesare not homotopy equivalent) is difficult in general because we then need to prove that certain maps arenot homotopic, and this is hard in most cases. A nice way to solve such problems is to assign an “algebraic

12 CHAPTER 1. HOMOTOPY TYPE

invariant”, A(X), to X. By algebraic invariant, we mean any correspondence that assigns to any topologicalspace X an algebraic object (such as a group, a ring, a vector space,...) A(X) such that A(X) is isomorphicto A(Y ) whenever X ' Y . The first algebraic invariant we are going to deal with in this course is thefundamental group in Chapter 2. Using that invariant, we will show for example that the circle S1 is notcontractible.

Chapter 2

The fundamental group

2.1 Definition of π1(X)

The goal of this section is to properly define the fundamental group, π1(X), of a topological space X. Thatdefinition involves the notion of loop, which is a special path. So we begin with paths. As in Chapter 1, wereserve the letter I for the interval

I = [0, 1].

2.1.1 Paths

Definition 2.1.1. A path in a topological space X is a continuous map α : [0, 1] −→ X.

One should think s ∈ I as the time. So any path α is running over within the time 1. The point α(0) iscalled the starting point or the initial point of α, and the point α(1) is called the ending point or the terminalpoint of α.

Definition 2.1.2. Let α, β : I −→ X be paths in X such that α(1) = β(0). One can then define a new path,denoted α·β, by

α·β(s) =

{α(2s) if 0 ≤ s ≤ 1

2

β(2s− 1) if 12 ≤ s ≤ 1.

The path α·β is called the composition of α and β.

A path is in particular a function. However, one has to notice that the composition of paths is quite differentto the composition of functions (recall that the composition gf of two functions makes sense only if thecodomain of f sit inside the domain of g). By Definition 2.1.2, to get α·β, we spend half of the time doingα (so we double the speed) and half of the time doing β.

The definition of the fundamental group involves the notion of homotopy between paths that we now define.

2.1.2 Homotopy of paths

We say that two paths α, β have the same endpoints if α(0) = β(0), and α(1) = β(1). From now on, thehomotopy between two paths makes sense only if they have the same endpoints.

13

14 CHAPTER 2. THE FUNDAMENTAL GROUP

Definition 2.1.3. Let α, β : I −→ X be paths with the same endpoints. We say that α is homotopic to β,and we denote α ∼ β, if there exists a continuous map H : I × I −→ X such that

(HP1) H(s, 0) = α(s), and H(s, 1) = β(s) for all s ∈ I, and

(HP2) for all t ∈ I, H(0, t) = α(0), and H(1, t) = α(1).

Such a map H is called a homotopy from α to β.

Sometimes we will write Ht(s) for H(s, t). That is, Ht(s) = H(s, t). This gives the map Ht : I −→ X, s 7→Ht(s), which is a path in X for every t. The axiom (HP1) is then equivalent to H0 = α and H1 = β. Andthe second axiom says that during the homotopy (or continuous deformation) the endpoints remain fixed.

Proposition 2.1.4. Let x, y ∈ X be two points in a space X. And let Pxy be the set of paths starting at xand ending at y. Then the relation ∼ we just defined is an equivalence relation in Pxy.

Proof. The proof is the same as that of Proposition 1.1.4.

Notation. The equivalence class of a path α, with respect to the equivalence relation ∼, will be denoted[α]. Thus α ∼ β if and only if [α] = [β].

Definition 2.1.5. Let x be a point in a space X. The path cx : I −→ X defined by cx(s) = x for all s iscalled the constant path at x.

Before we define π1X and prove that it is a group, we need four lemmas. Here is the first one.

Lemma 2.1.6. Let x, y ∈ X, and α : I −→ X be a path such that α(0) = x and α(1) = y. Then α ∼ cx·αand α ∼ α· cy.

Proof. Let us first prove cx·α ∼ α. By Definition 2.1.2, we have

cx·α(s) =

{x if 0 ≤ s ≤ 1

2

α(2s− 1) if 12 ≤ s ≤ 1.

(2.1.1)

We want to get a homotopy between α and cx·α. Define H : I × I −→ X by

H(s, t) =

{x if 0 ≤ s ≤ 1

2 t

α(

2s−t2−t

)if 1

2 t ≤ s ≤ 1.

The map H is continuous since at the time s = 12 t we have α

(2s−t2−t

)= α(0) = x. To check that H is the

required homotopy, we need to prove that H satisfies the axioms (HP1) and (HP2) from Definition 2.1.3.

(HP1) At the time t = 0, H(s, 0) = α( 2s−02−0 ) = α(s) and at the time t = 1, H(s, 1) agrees with (2.1.1).

(HP2) By the definition of H, we have H(0, t) = x = α(0), and H(1, t) = α(

2−t2−t

)= α(1) = y.

Similarly, one proves that α ∼ α· cy. This ends the proof.

Definition 2.1.7. Let α : I −→ X be a path in a space X. The path α : I −→ X defined by α(s) = α(1− s)is called the inverse path of α.

Lemma 2.1.8. Let x, y ∈ X, and let α : I −→ X be a path such that α(0) = x and α(1) = y. Then α·α ∼ cxand α·α ∼ cy.

2.1. DEFINITION OF π1(X) 15

Proof. Exercise.

Lemma 2.1.9. The composition of paths is associative up to homotopy. That is, given three paths α, β, γsuch that α(1) = β(0) and β(1) = γ(0), we have α· (β· γ) ∼ (α·β)· γ.

Proof. Exercise.

Because of Lemma 2.1.9, we have [α· (β· γ)] = [(α·β)· γ]. So we can remove parentheses and just write[α·β· γ]. The following lemma is the path version of Proposition 1.1.5. But their proofs are quite differentbecause the composition of paths is not defined in the same way as the composition of functions.

Lemma 2.1.10. Let α0, α1, β0, β1 : I −→ X be four paths such that α0(0) = α1(0), α0(1) = α1(1) = β0(0) =

β1(0), and β0(1) = β1(1). If α0 ∼ α1 and β0 ∼ β1, then α0·β0 ∼ α1·β1.

Proof. Let H : I × I −→ X be a homotopy from α0 to α1, and let Let K : I × I −→ X be a homotopy fromβ0 to β1. Define L : I × I −→ X by

L(s, t) =

{H(2s, t) if 0 ≤ s ≤ 1

2

K(2s− 1, t) if 12 ≤ s ≤ 1.

It is easy to check that the homotopy L does the work.

2.1.3 Loops and the definition of the fundamental group

Definition 2.1.11. Let X be a topological space. A loop in X is a path α : I −→ X such that α(0) = α(1).

The point x = α(0) = α(1) is called the basepoint of the loop α. We also say that α is a loop at x. Letπ1(X,x) denote the set of homotopy classes [α] of loops at x.

Proposition 2.1.12. The set π1(X,x) is a group with respect to the operation [α][β] = [α·β].

Proof. First of all, the group operation is well defined: if α ∼ α′ and β ∼ β′, then α·β ∼ α′·β′ byLemma 2.1.10. Therefore [α·β] = [α′·β′].

• The identity element in π1(X,x) is [cx]. Indeed, for any [α] we have [α][cx] = [α· cx] = [α] sinceα· cx ∼ α by Lemma 2.1.6. Similarly, one has [cx][α] = [α] by applying the same lemma.

• The group operation is associative since

[α] ([β][γ]) = [α] ([β· γ]) = [α· (β· γ)]

= [(α·β)· γ] by Lemma 2.1.9= ([α·β]) [γ] = ([α][β]) [γ].

• Every element [α] in π1(X,x) is invertible with [α] as its inverse. This follows from Lemma 2.1.8.

Now we are ready to define the fundamental group.

Definition 2.1.13. Let X be a topological space, and x ∈ X be a point in X. By Proposition 2.1.12, theset π1(X,x) is a group called the fundamental group of X at the basepoint x.


Recall that the trivial group, denoted 0, is the group with only one element. Two groups G1 and G2 are thesame if they are isomorphic. In that case, we will write G1 = G2.

Example 2.1.14. π1(R, 0) = 0.

Solution. This comes from the fact that any loop in R at 0 is homotopic to the constant loop c0. Indeed,let α : I −→ R be a loop at 0. Define H : I × I −→ R by H(s, t) = (1− t)α(s) + tc0(s). It is clear that H iscontinuous as a product of continuous functions. Moreover, at the time t = 0 we have H(s, 0) = α(s), and atthe time t = 1 we have H(s, 1) = c0(s) for all s. We also have H(0, t) = α(0) = 0 and H(1, t) = α(1) = 0 forevery t. So H satisfies conditions (HP1) and (HP2) from Definition 2.1.3. This implies that it is a homotopybetween α to c0. Thus [α] = [c0] for any [α] ∈ π1(R, 0), which gives the desired result.

Proposition 2.1.15. Let X and Y be spaces, and let (x, y) ∈ X × Y . Then

π1(X × Y, (x, y)) ∼= π1(X,x)× π1(Y, y).

Proof. The main observation is the following. Let α : I −→ X × Y be a loop at (x, y) defined as α(s) =

(α1(s), α2(s)). Then it is clear that α1 is a loop in X at x, and α2 is also a loop in Y at y. Conversely, giventwo loops β1 : I −→ X, and β2 : I −→ Y at x and y respectively, there exists a unique loop β : I −→ X × Yat (x, y) such that β(s) = (β1(s), β2(s)).

One can then define f : π1(X × Y, (x, y)) −→ π1(X,x)× π1(Y, y) by f [α] = ([α1], [α2]). One can also defineg : π1(X,x) × π1(Y, y) −→ π1(X × Y, (x, y)) by g([β1], [β2]) = [β]. Clearly f and g are homomorphism ofgroups, and gf = id and fg = id.

2.1.4 The dependence of π1(X, x) on the choice of the basepoint

Let x1, x2 ∈ X be two points in a space X (x1 may be equal to x2). Assume that there is a path γ : I −→ X

starting at x1 and ending at x2. Define

ϕγ : π1(X,x1) −→ π1(X,x2) by ϕγ [α] = [γ·α· γ].

The map ϕγ is well defined because if α ∼ β, then γ·α· γ ∼ γ·β· γ by Lemma 2.1.10.

Proposition 2.1.16. The map ϕγ is an isomorphism of groups.

Proof. First of all, the map ϕγ is a homomorphism of groups since

ϕγ([α][β]) = [γ· (α·β)· γ] = [γ· (α· γ· γ·β)· γ] = [(γ·α· γ)· (γ·β· γ)] = ϕγ([α])ϕγ([β]).

Now defineψγ : π1(X,x2) −→ π1(X,x1) by ψγ [α] = [γ·α· γ].

One can easily check that ψγ is also a homomorphism of groups, and that ϕγψγ = id and ψγϕγ = id.

Recall that a group G is abelian if for any g1, g2 ∈ G, we have g1g2 = g2g1.

Proposition 2.1.17. If π1(X,x1) is abelian, then ϕγ is independent of the choice of the path class [γ]. Thatis, if γ′ : I −→ X is another path starting at x1 and ending at x2 (γ′ need not be homotopic to γ), thenϕγ = ϕγ′ .

2.2. THE FUNDAMENTAL GROUP OF THE CIRCLE 17

Proof. Let γ′ as in the proposition, and let [α] ∈ π1(X,x1). We want to prove that ϕγ([α]) = ϕγ′([α]), thatis, [γ·α· γ] = [γ′·α· γ′]. Consider the loops γ′· γ and γ· γ′ at x1. It is clear that (γ′· γ)· (γ· γ′) ∼ cx1

, wherecx1

is the constant loop at x1. So [γ′· γ][γ· γ′] = [cx1], which implies that [γ′· γ][γ· γ′][α] = [α] because [cx1

]

is the identity element in π1(X,x1). Using now the fact that π1(X,x1) is abelian by hypothesis, the latterequality becomes

[γ′· γ][α][γ· γ′] = [α],

which is equivalent to γ′· (γ·α· γ)· γ′ ∼ α. This implies, by Lemma 2.1.10 and Lemma 2.1.6, that γ·α· γ ∼γ′·α· γ′. Thus [γ·α· γ] = [γ′·α· γ′], and we get the desired result.

Definition 2.1.18. A space X is said to be path-connected if for any pair of points x, y ∈ X, there existsa path α : I −→ X such that α(0) = x and α(1) = y.

For example Rn is path-connected for all n ≥ 0: for any x, y ∈ Rn, the path α(t) = (1 − t)x + ty does thework. One can also see that the n-sphere Sn is path-connected when n ≥ 1. For example, if n = 1, giventwo points x, y ∈ S1 we select one of the two arcs connecting x to y to be our path. For n ≥ 2, we can usearcs again, but in a more subtle way.

Proposition 2.1.16 implies that if X is path-connected π1(X,x) is, up to isomorphism, independent of thechoice of the basepoint. In that case, π1(X,x) will be just denoted π1(X). For instance we will often dropthe basepoint when dealing with Rn or Sm,m ≥ 1, or with a product of those spaces.

Definition 2.1.19. A space X is said to be simply connected if it is path-connected and π1(X) = 0.

For instance, for all n ≥ 0, the space Rn is simply connected. We will see that Sn is simply connected whenn ≥ 2. However the circle S1 fails to be simply connected as we shall see in the next section.

2.2 The fundamental group of the circle

The goal of this section is to prove Theorem 2.2.7, which states that π1(S1) = Z. To do this, we need acouple of lemmas.

2.2.1 Four lemmas in order to prove that π1(S1) = Z

By small connected neighborhood in S1 we mean a small closed arc in S1. Consider the map

p : R −→ S1 defined by p(s) = e2πis = (cos(2πs), sin(2πs)) .

Note that the map p wraps once around the circle when restricting to any interval of the form [n, n+1), n ∈ Z.Also note that the restriction of p to [n, n+ 1) is the same as the restriction to [m,m+ 1) for any n,m.

Lemma 2.2.1. Let a, b ∈ [0, 1] with a < b, and let α : [a, b] −→ S1 be a continuous map that carries [a, b]

to a small connected neighborhood. Let x0 ∈ R such that p(x0) = α(a). Then there exists a unique mapα : [a, b] −→ R such that α(a) = x0 and pα = α.

Proof. The key point of the proof is the following fact about the map p. Let x ∈ S1. Then there exists aunique sx ∈ [0, 1) such that x = p(sx). This is by the definition of p, which wraps once around S1 if onerestricts it to [0, 1). So the inverse image of x under p consists of points of the form sx + k, k ∈ Z. That is,

p−1(x) = {sx + k, k ∈ Z.}


Now if U is a small connected neighborhood containing x, then one can easily see that

Fact: p−1(U) is of the form p−1(U) =⋃k∈Z

Vk, (2.2.1)

where each Vk is a closed interval containing sx + k and contained in [k, k + 1). So the V ′ks are pairwisedisjoint and each of them is homeomorphic to U . The homeomorphism ϕk : Vk −→ U is nothing but therestriction of p on Vk. That is, ϕk = p|Vk. Set Vk = [yk, zk]. Without lost the generality, assume that thesmall closed arc U starts at α(a) and ends at α(b). This assumption implies that we must have p(yk) = α(a)

and p(zk) = α(b) for all k. Since by hypothesis x0 is such that p(x0) = α(a), it follows that x0 is one of y′ks.That is, there is an integer n, which is unique, such that x0 = yn.

Define now α : [a, b] −→ R by α(s) = ϕ−1n (α(s)). We have pα = α because ϕn = p|Vn. We also easily have

α(a) = ϕ−1n (α(a)) = yn = x0. Such a map α is unique by construction.

We prove the uniqueness. Suppose that there is another continuous map α′ : [a, b] −→ R such that α′(a) = x0

and pα′ = α. We want to prove that α = α′. By the construction of α, it follows that α([a, b]) ⊆ Vn. Forthe same n, we also have α′([a, b]) ⊆ Vn because of the following three things. (i) α′([a, b]) is connectedbecause α′ is continuous and [a, b] is connected. (ii) α′([a, b]) contains x0 since α′(a) = x0. (iii) pα′ = α.If α(s0) 6= α′(s0) for some s0 ∈ [a, b], then ϕnα(s0) 6= ϕnα′(s0) since ϕn is bijective (and then injective).However, since pα = pα′ = α, it follows that ϕnα(s0) = ϕnα′(s0). Therefore, having ϕnα(s0) 6= ϕnα′(s0) isa contradiction. So for all s ∈ [a, b] we have α(s) = α′(s).

Because the map p : R −→ S1 satisfies (2.2.1), we say that R is a covering space of S1. This gives rise to atheory called the theory of covering spaces that we won’t be studied in this course. If the reader is interestedin that theory, he can read [1, Pages 56–70].

Lemma 2.2.2. (lifting lemma) Let α : I −→ S1 be a path such that α(0) = 1. Then there exists a uniquepath α : I −→ R such that α(0) = 0 and pα = α.

Proof. Let s ∈ I, and let Ns be a small open arc containing α(s). Since α is continuous, there exists an openneighborhood Is of s such that α(Is) ⊆ Ns. Consider then the family {Is}s∈I . It indeed forms an open coverof I. Therefore, by the compactness of I, there exists a0, a1, · · · , am ∈ I with a0 = 0, am = 1, ai < ai+1 suchthat

I = [0, a1] ∪

(m−2⋃i=1

[ai, ai+1]

)∪ [am−1, 1].

Applying Lemma 2.2.1 to the restriction α|[0, a1], we get α|[0, a1]. Applying the same lemma to α|[a1, a2],we get α|[a1, a2], and so on. After a finite number of steps, we get the desired α, which is unique since byLemma 2.2.1 each α|[ai, ai+1] is unique.

Definition 2.2.3. Let α : I −→ S1 be a path such that α(0) = 1. The path α from Lemma 2.2.2 is calledthe lift of α. So α must satisfies two conditions: α(0) = 0 and pα = α.

Example 2.2.4. Let α : −→ S1 be the path defined as α(s) = e2πis. We want to find α. By definition, αhave to satisfy the equations α(0) = 0 and pα = α. That is, for all s ∈ I, one has

e2πi(α(s)) = e2πis and α(0) = 0.

Solving these equations, we get α(s) = s.

From now on, any lift will be along p. One can also lift the homotopies as it is said in the following result.

2.2. THE FUNDAMENTAL GROUP OF THE CIRCLE 19

Lemma 2.2.5. Let α, β : I −→ S1 be paths such that α(0) = β(0) = 1. Let α and β be the lifts of α and βrespectively. Let H : I × I −→ S1 be a homotopy from α to β. Then there exists a unique continuous mapH : I × I −→ R such that H(s, 0) = α(s) for all s ∈ I, and pH = H. Moreover H is a homotopy from α toβ, which implies in particular that

α(1) = β(1). (2.2.2)

Proof. Let t ∈ I be any point in [0, 1]. Consider the path Ht : I −→ S1 defined by Ht(s) = H(s, t). Byhypothesis we have H0 = α and H1 = β. Again by hypothesis H is a homotopy from α to β, and by theaxiom (HP2) from Definition 2.1.3 we have H(0, t) = α(0) = 1. So Ht(0) = 1. Applying Lemma 2.2.2 to thepath Ht, we obtain the existence of a unique path Ht : I −→ R such that Ht(0) = 0 and pHt = Ht. Definenow H : I × I −→ R by H(s, t) = Ht(s).

• The map H is continuous. Indeed, for (s, t) ∈ I × I pick a very small closed arc U in S1 such thatH(s, t) ∈ U . As in the proof of Lemma 2.2.1, there exists a closed interval Vn that maps homeomor-phically to U under ϕn = p|Vn. The map H is then defined by H(s, t) = ϕ−1

n (H(s, t)). Since ϕ−1n and

H are continuous, it follows that H is also continuous.

• Moreover, we have H0 = α since H0 is the lift of H0 = α. We also have pH = H since by constructionHt is the lift of Ht for every t ∈ I.

Now we prove that H is a homotopy from α to β. Clearly H satisfies the axiom (HP1) from Definition 2.1.3since H0 is the lift of H0 = α, and H1 is the lift of H1 = β. For the axiom (HP2), we easily haveH(0, t) = Ht(0) = 0 = α(0),∀t ∈ I, by the first property of the lifting. To see that H(1, t) = α(1) for allt, consider the path λ : I −→ S1 defined by λ(t) = H(1, t) = Ht(1). Clearly λ is the constant path at α(1)

since H is a homotopy from α to β (and thus satisfies (HP2)). It is also clear that the lift of λ is the pathλ : I −→ R defined by λ(t) = H(1, t) = Ht(1). So pλ(t) = α(1) for all t. This implies that λ is the constantpath as well. In particular we have λ(0) = λ(1), which is equivalent to α(1) = β(1).

Lemma 2.2.6. Let α, β : I −→ S1 be two loops at 1. Then the lifts α·β, α, and β of α·β, α, and β respectivelysatisfy the equation

α·β(1) = α(1) + β(1).

Proof. Exercise.

2.2.2 Proving that π1(S1) = Z

Theorem 2.2.7. The fundamental group of the circle S1 is isomorphic to Z.

Proof. Since S1 is path-connected, the choice of the basepoint does not matter. To avoid some complications,take 1 = e2πi0 to be our basepoint. The plan of the proof is to define two homomorphisms f : Z −→ π1(S1)

and g : π1(S1) −→ Z and prove that g is the inverse of f .

Definition of f . For every n ∈ Z define a loop αn : I −→ S1 by αn(s) = e2πins. That loop wraps n timesaround the circle. Define now f by

f(n) = [αn]

The map f is a homomorphism of groups since f(n+m) = [αn+m] = [αn·αm] = [αn][αm]. Indeed, one canshow that the composition αn·αm 1 is homotopic to αn+m.

1Here αn·αm is the composition of paths and not the standard product of two exponentials!


Definition of g. Define g by the formulag([α]) = α(1),

where α is the lift of α as in Lemma 2.2.2. The map g is well defined because of the following two things.The first one is the fact that α(1) is an integer. This is because p(α(1)) = α(1) = 1. The second thing is thefact that α(1) = β(1) when α ∼ β. This immediately comes from (2.2.2) in Lemma 2.2.5. The map g is ahomomorphism of groups since αβ(1) = α(1) + β(1) by Lemma 2.2.6.

The inverse of f is g. There is an explicit lift of αn: the path αn : I −→ R defined as αn(s) = ns. Of coursewe have αn(0) = 0 and pαn = αn. We also have gf(n) = g[αn] = αn(1) = n for all n ∈ Z. So gf = id. Toend the proof, it suffices to show that g is injective 2. To do this, let [α], [β] ∈ π1(S1) such that g[α] = g[β].Then α(1) = β(1), which implies that α· β is a loop at 0 in R. Since π1(R) = 0 by Example 2.1.14, it followsthat [α· β] = [c0], which gives α ∼ β. Appying now Corollary 1.1.6 –(i) we get pα ∼ pβ, that is, α ∼ β.Hence [α] = [β].

The following is a quick application of Theorem 2.2.7.

Example 2.2.8. The fundamental group of the torus T 2 is Z× Z. That is,

π1(T 2) = Z× Z.

Solution. Recall that the torus, T 2 = S1 × S1, is the product of S1 with S1. So π1(T 2) = π1(S1 × S1) ∼=π1(S1)× π1(S1) by Proposition 2.1.15. The desired result follows now from Theorem 2.2.7.

2.3 Induced homomorphism

The aim of this section is to prove Theorem 2.3.7, which asserts that the fundamental group of a topologicalspace X depends only on the homotopy type of X. In order words, the theorem says that if X is homotopyequivalent to Y , then π1(X) is isomorphic to π1(Y ).

We begin with the definition of the induced homomorphism. Let X and Y be two topological spaces, andlet x and y be the basepoints of X and Y respectively. A pointed-map from X to Y is a continuous mapf : X −→ Y such that f(x) = y. In order words, it is a continuous map that sends the basepoint to thebasepoint. Let f : X −→ Y be a pointed map. Define f∗ : π1(X,x) −→ π1(Y, y) by

f∗([α]) = [f ◦ α]. (2.3.1)

The map f∗ is well defined because, by Corollary 1.1.6 –(i), if α ∼ β then fα ∼ fβ. Furthermore f∗ isa homomorphism of groups. Indeed, if α, β : I −→ X are two loops at x, by Definition 2.1.2, we have

α·β(s) =

{α(2s) if 0 ≤ s ≤ 1

2

β(2s− 1) if 12 ≤ s ≤ 1,

which implies that

(f ◦ (α·β))(s) =

{f ◦ α(2s) if 0 ≤ s ≤ 1

2

f ◦ β(2s− 1) if 12 ≤ s ≤ 1,

= (f ◦ α)· (f ◦ β).

Definition 2.3.1. The homomorphism f∗ : π1(X,x) −→ π1(Y, y) is called the induced homomorphism by f .

Proposition 2.3.2. (i) If X is a space with x as its basepoint, then (idX)∗ = idπ1(X,x).

2In general, if f : X −→ Y and g : Y −→ X are two maps such that gf = idX and g is injective, then f is bijective. Indeed,f is injective. This is because f(x) = f(x′) implies that g(f(x)) = g(f(x′)), which is equivalent to x = x′ since gf = idX . Themap f is also surjective since for every y ∈ Y , we have f(g(y)) = y. In fact, if f(g(y)) 6= y, then gfg(y) 6= g(y) because g isinjective. We would then have g(y) 6= g(y), which is a contradiction.

2.3. INDUCED HOMOMORPHISM 21

(ii) If Xf // Y

g // Z are two pointed maps, then (gf)∗ = g∗f∗.

Proof. (i) By (2.3.1) we have (idX)∗[α] = [id ◦ α] = [α] for every [α] ∈ π1(X,x).

(ii) For every [α] ∈ π1(X,x) we have again by (2.3.1)

(gf)∗[α] = [(gf) ◦ α] = [g(f ◦ α)] = g∗([f ◦ α]) = g∗f∗([α]).

The following is an immediate consequence of Proposition 2.3.2.

Corollary 2.3.3. Let f : X −→ Y be a pointed map. If f is a homeomorphism then f∗ is an isomorphism.

Before we state the next proposition, we need to define the pointed version of the homotopy type. Let Xand Y be two spaces, and let x0 ∈ X, y0 ∈ Y be the basepoints of X and Y respectively. Let f, g : X −→ Y

be two pointed map. A continuous map H : X × I −→ Y is said to be a basepoint-preserving homotopy fromf to g if H is a homotopy in the sense of Definition 1.1.1, and H(x0, t) = y0 for all t. In that case we writef ∼0 g. Note that by definition f ∼0 g implies f ∼ g. If there is another pointed map f ′ : Y −→ X suchthat f ′f ∼0 idX and ff ′ ∼0 idY , we say that f is a base-point preserving homotopy equivalence. In thatcase we write X '0 Y meaning that X and Y have the same homotopy type in the pointed context.

Proposition 2.3.4. (i) If f ∼0 g then f∗ = g∗.

(ii) If X '0 Y then π1(X,x0) ∼= π1(Y, y0).

Proof. (i) Assume that f ∼0 g, and let [α] ∈ π1(X,x0). Then by Corollary 1.1.6 –(ii), we have fα ∼ gα,which is equivalent to [fα] = [gα]. So f∗([α]) = g∗([α]).

(ii) Suppose that X '0 Y . Then there exists two pointed maps f : X −→ Y and f ′ : Y −→ X such thatf ′f ∼0 idX and ff ′ ∼0 idY . By the first part of this proposition, we deduce that (f ′f)∗ = (idX)∗ andthat (ff ′)∗ = (idY )∗. Applying now Proposition 2.3.2, we deduce that f∗ : π1(X,x0) −→ π1(Y, y0) isan isomorphism (its inverse is indeed f ′∗). So π1(X,x0) ∼= π1(Y, y0).

Having to pay so much attention to basepoints when dealing with the fundamental group is somethingnuisance. For homotopy equivalences one does not have to be quite so careful since the conditions on thebasepoints can actually be dropped, as shown Proposition 2.3.6 below.

Let f, g : X −→ Y be two continuous maps (not necessarily pointed), and let H : X×I −→ Y be a homotopyfrom f to g (that is, f ∼ g). For every x ∈ X, define a path γx : I −→ Y by γx(t) = H(x, t). Clearly thestarting point of γx is f(x), and its ending point is g(x). Define also ϕγx : π1(Y, f(x)) −→ π1(Y, g(x)) by theformula

ϕγx([α]) = [γx·α· γx].

By Proposition 2.1.16, ϕγx is an isomorphism of groups.


Lemma 2.3.5. For any x ∈ X, the diagram

π1(Y, g(x))

π1(X,x)

g∗55

f∗

))π1(Y, f(x))

ϕγx

OO

commutes. That is, ϕγx ◦ f∗ = g∗.

Proof. Exercise.

Proposition 2.3.6. If f : X −→ Y is a homotopy equivalence in the sense of Definition 1.1.7, then for anyx ∈ X the induced homomorphism f∗ : π1(X,x) −→ π1(Y, f(x)) is an isomorphism.

Proof. Exercise.

As an immediate and nice consequence of Proposition 2.3.6, we have the following, which is the main resultof the section. It is also a stronger version of Proposition 2.3.4 –(ii) because, here, we don’t care about thebasepoint.

Theorem 2.3.7. Let X and Y be two topological spaces not necessarily pointed. Assume that X and Y arehomotopy equivalent. Then their fundamental groups are isomorphic. That is,

X ' Y ⇒ π1(X) ∼= π1(Y ).

The converse is not true: one can have π1(X) ∼= π1(Y ) without having X ' Y . For instance, by Exam-ple 2.1.14, π1(R) = 0. As we will see later in Example 2.5.6, π1(S2) = 0. So π1(R) ∼= π1(S2). But, manifestly,R is not homotopy equivalent to the sphere S2. Here are some quickly applications of Theorem 2.3.7.

Example 2.3.8. The fundamental group of the disk D2 is trivial. Indeed, by Example 1.1.9 –(ii) we haveD2 ' pt where pt = {(0, 0)}. Applying now Theorem 2.3.7 we get π1(D2) ∼= π1(pt). Since the fundamentalgroup of the one point space is trivial, it follows that π1(D2) = 0. More generally, if X is contractiblethen π1(X) = 0 by Theorem 2.3.7. Another example is π1(R2\{(0, 0)}), which is isomorphic to Z by Exam-ple 1.1.9 –(i), Theorem 2.3.7, and Theorem 2.2.7.

We close this section with the notion of retraction for spaces. Recall the inclusion map i : A ↪→ X, which isdefined by i(x) = x if A is a subspace of X.

Definition 2.3.9. Let X be a topological space, and A be a subspace of X. We say that A is a retract of Xif there is a continuous map r : X −→ A such that r ◦ i = idA. If in addition i ◦ r ∼ idX , we say that A is adeformation retract of X.

For instance, by Example 1.1.9, the circle S1 is a deformation retract of R2\{(0, 0)}. Again by the sameexample, the one-point space {(0, 0)} is a deformation retract of D2. However, as we will see in Section 2.4,S1 is not a retract of D2.

The following result immediately follows from Proposition 2.3.6.

Proposition 2.3.10. Let i : A ↪→ X be the inclusion map. If A is a deformation retract of X then theinduced homomorphism i∗ is an isomorphism.

2.4. APPLICATIONS OF THE FUNDAMENTAL GROUP 23

2.4 Applications of the fundamental group

In this section, we will see two applications (among many others) of the fundamental group. The first oneis topological, and regards the Brouwer fixed point theorem, which says that any continuous map from D2

to itself has a fixed point. The second application is more algebraic, and regards the fundamental theoremof algebra, which says that any non constant polynomial with complex coefficients has at least one root.Another interesting application is the Borsuk-Ulam theorem (see [1, Theorem 1.10]), which asserts that anycontinuous map f from the sphere S2 to the plane R2 admits a pair of antipodal points x and −x in S2 suchthat f(x) = f(−x). So at any time, there are two antipodal points on the earth with the same temperatureand the same barometric pressure!

2.4.1 The Brouwer fixed point theorem

The goal here is to prove Theorem 2.4.2 below. Recall the unit disk D2 = {(x, y) ∈ R2 : x2 + y2 ≤ 1}. Itsboundary is the circle S1.

Proposition 2.4.1. Let i : S1 ↪→ D2 be the inclusion map. There is no continuous map r : D2 −→ S1 suchthat r ◦ i = idS1 .

Proof. Assume that such a map r exists, and consider the following diagram of homomorphisms of groups

π1(S1)i∗ // π1(D2)

r∗ // π1(S1) .

The assumption r ◦ i = idS1 and Proposition 2.3.2 imply that r∗ ◦ i∗ = idπ1(S1). But this is not possiblesince π1(D2) = 0 by Example 2.3.8, and π1(S1) = Z by Theorem 2.2.7.

Before we state the theorem, recall that a map f : X −→ X admits a fixed point if there is some x0 suchthat f(x0) = x0.

Theorem 2.4.2 (Brouwer fixed point theorem). Any continuous map f : D2 −→ D2 has a fixed point.

Proof. Let f : D2 −→ D2 be a continuous map. Assume that for every x ∈ D2, f(x) 6= x. This allowsus to define r : D2 −→ S1 by r(x) = Rx ∩ S1, where Rx is the ray that starts at f(x) and passes throughx. Obviously r ◦ i(x) = x for all x ∈ S1. Moreover r is continuous because so is f . This contradictsProposition 2.4.1.

2.4.2 The fundamental theorem of algebra

Let us begin with the notion of degree of a map. Roughly speaking, the degree of a continuous mapf : S1 −→ S1 is a number that represents the number of times that f(S1) wraps around the codomain S1.Explicitly, it is defined as follows. First consider the map q : I −→ S1 defined by q(s) = e2πis. Consider nowthe composition fq = f ◦ q : I −→ S1. Clearly fq is a loop at f(1) (note that f(1) is not necessarily equal to1, the basepoint we used in the previous sections). So [fq] ∈ π1(S1, f(1)). To carry f(1) to 1, let γ : I −→ S1

be a path starting at f(1) and ending at 1. By Proposition 2.1.17, the element [γ· fq· γ] ∈ π1(S1, 1) isindependent of the choice of [γ]. Set

αf = γ· fq· γ,

and consider the lift αf : I −→ R of αf , which is unique by Lemma 2.2.2. Notice that αf (1) is an integerbecause p ◦ αf (1) = αf (1) by Lemma 2.2.2, and αf (1) = 1 (recall that p : R −→ S1 was defined byp(s) = e2πis).


Definition 2.4.3. The degree of a map f : S1 −→ S1 is an integer, denoted deg(f), and defined as theimage of 1 ∈ [0, 1] under αf . That is,

deg(f) = αf (1).

Example 2.4.4. (i) If f : S1 −→ S1 is constant, then deg(f) = 0.

(ii) Let n ∈ Z. The degree of the map fn : S1 −→ S1 defined by fn(x) = xn is n.

Solution.

(i) Assume that f : S1 −→ S1 is a constant map. Then fq : I −→ S1 is the constant path at f(1). Thatis, fq = cf(1). Therefore, for any path γ as before, we have γ· fq· γ ∼ γγ ∼ c1. This implies thatαf ∼ c1, and that deg(f) = αf (1) = c1(1) by (2.2.2) from Lemma 2.2.5. Since the lifting is unique byLemma 2.2.2, it follows that c1 : I −→ R is the constant map at 0. Thus c1(1) = 0, which implies thatdeg(f) = 0.

(ii) We want to show that deg(fn) = n. Notice first that fn(1) = 1 by the definition of fn. So we cantake γ to be the constant path at 1. This implies that αf = γ· fnq· γ ∼ fnq, and therefore we havedeg(fn) = αf (1) = fnq(1) by (2.2.2) from Lemma 2.2.5. Since fnq(s) = e2πins for all s, by theuniqueness of the lifting, it follows that fnq(s) = ns (we have used the lift s 7→ ns in the proof ofTheorem 2.2.7). So fnq(1) = n, which implies that deg(fn) = n.

Proposition 2.4.5. If two maps f, g : S1 −→ S1 are homotopic then deg(f) = deg(g).

Proof. Assume that f ∼ g. Then, by Corollary 1.1.6 –(ii), it follows that fq ∼ gq. This implies that for anypath γ in S1 such that γ(0) = f(1) and γ(1) = 1, we have γ· fq· γ ∼ γ· gq· γ. That is, αf ∼ αg. Applyingnow (2.2.2) from Lemma 2.2.5, we get αf (1) = αg(1), meaning that deg(f) = deg(g).

We are now ready to state and prove the main result of this subsection.

Theorem 2.4.6 (The fundamental theorem of algebra). Let f(x) = xn + a1xn−1 + · · · + an−1x + an be a

polynomial with complex coefficients ai, where n > 0. Then there is a complex number x such that f(x) = 0.Therefore there are n such complex numbers (counted with multiplicities).

Proof. If there is x ∈ S1 such that f(x) = 0 then the proof is over. Assume that f(x) 6= 0 for all x ∈ S1.This allows us to define f : S1 −→ S1 by f(x) = f(x)

|f(x)| where | z | stands for the modulus of z. We proceed to

calculate deg(f). Suppose that f(x) 6= 0 inside the unit disk D2. This allows us to define H : S1 × I −→ S1

by H(x, t) = f(tx)|f(tx)| . Then H is a homotopy from the constant map at f(0)

|f(0)| to f , and we conclude that

deg(f) = 0 by Proposition 2.4.5 and Example 2.4.4 –(i). Suppose next that f(x) 6= 0 for all x such that| x |≥ 1. This allows us to define K : S1 × I −→ S1 by

K(x, t) =L(x, t)

| L(x, t) |, where L(x, t) = tnf(

x

t) = xn + t(a1x

n−1 + ta2xn−2 + · · ·+ tn−1an).

Then K is a homotopy from fn to f , and we conclude that deg(f) = n by Proposition 2.4.5 and Exam-ple 2.4.4 –(ii). Since by hypothesis n > 0, it is not possible to get at the same time deg(f) = 0 inside thedisk D2, and deg(f) = n outside. So we must then have f(x) = 0 for some x either inside or outside D2.

2.5. THE VAN KAMPEN THEOREM 25

2.5 The Van Kampen theorem

The Van Kampen theorem gives a method for computing the fundamental groups of spaces than can bedecomposed into simpler spaces whose fundamental groups are already known. The theorem involves thenotion of free product of groups.

2.5.1 Free products of groups

(a) Disjoint union of sets. The disjoint union of two sets X and Y , denoted X∐Y , is a set defined by

X∐

Y = (X × {0}) ∪ (Y × {1}) .

For example, if X = {2, 3} and Y = {2, 4}, then X∐Y = {(2, 0), (3, 0), (2, 1), (4, 1)}. One can see in this

example the difference between the disjoint union and the classical union of X and Y (X ∪ Y = {1, 2, 3}).

(b) Words on an alphabet. Let A be a set, viewed as an alphabet. A word on A is a finite sequencea1a2 · · · ak, k ≥ 0, where each letter ai belongs to A. The integer k is called the length of the word. A wordof length k = 0 is called the empty word.

(c) Free product of two groups. Let G1 and G2 be two groups. Let A = G1

∐G2. Define G1 ? G2 to

be the set of words a1a2 · · · ak, k ≥ 0, on the alphabet A such that adjacent letters ai and ai+1 belong todifferent groups. Words satisfying these conditions are called reduced. One can endow G1 ?G2 with a naturalgroup structure:

• The group operation or the multiplication is the juxtaposition of words. For instance,

(a1a2 · · · ak)(b1b2 · · · bm) = a1a2 · · · akb1b2 · · · bm.

Note that the word a1a2 · · · akb1b2 · · · bm may not be reduced! The point is we can always carry anyword to a unique reduced one (see Example 2.5.2 below) by using the following two rules. Rule1: if twoadjacent letters, say a and a′, belong to the same group G1 (or G2), we replace the word aa′, which isa word with two letters, by the word a·G1 a

′ (which is a word with only one letter. Here “·G1” standsfor the group operation in G1). Rule2: if the identity element appears somewhere in a word, we takeit out. For example, if a ∈ G1 and e2 is the identity element of G2, then we replace the word ae2 bythe word a.

• The identity element is the empty word.

• The inverse of a1a2 · · · ak is the word a−1k · · · a

−12 a−1

1 .

Remark 2.5.1. The free product of two groups G1 and G2 is not abelian in most of cases. This is becausethe word ab, a ∈ G1, b ∈ G2, is not the same as the word ba. For instance Z ? Z is not abelian.

Example 2.5.2. Take G1 = G2 = Z. In this example we just want to figure out the group multiplication inZ?Z. Since the alphabet is the disjoint union A = Z

∐Z, we need to distinguish the generators. Let a be the

generator of the first copie of Z, and b be the one of the second copie. That is, G1 =< a > and G2 =< b >.

(i) (a2bab−3)(a4b) = a2bab−3a4b.

(ii) (a2bab−3)(b5a) = a2bab−3b5a = a2bab2a.

(iii) (a2bab−3)(b3a−1b−2) = a2bab−3b3a−1b−2 = a2baa−1b−2 = a2bb−2 = a2b−1.


(d) A natural fact about G1 ? G2. Let G be another group, and let i1 : G −→ G1 and i2 : G −→ G2 begroup homomorphisms. Consider the following diagram.

Gi2 //

i1

��

G2

G1

(2.5.1)

LetNG be the normal subgroup of G1?G2 generated by words of the form i1(g) (i2(g))−1, g ∈ G. Consider the

quotient group G1?G2/NG. There are canonical homomorphisms j1 : G1 −→ G1?G2/NG, x 7→ j1(x) = xNG,and j2 : G2 −→ G1 ? G2/NG, y 7→ j2(y) = yNG. So the original diagram becomes the commutative square

Gi2 //

i1

��

G2

j2

��G1

j1 // G1 ? G2/NG.

(2.5.2)

Here is the fact.

Proposition 2.5.3. Given another group A, and homomorphisms f1 : G1 −→ A and f2 : G2 −→ A such thatf1i1 = f2i2, there exists a unique homomorphism Φ: G1 ? G2/NG −→ A such that Φj1 = f1 and Φj2 = f2.

Proof. For an element a1a2 · · · akNG of G1 ? G2/NG, we will assume without lost the generality that a1 ∈G1, a2 ∈ G2, · · · , ak ∈ G2. Define Φ by

Φ(a1a2 · · · akNG) = f1(a1)f2(a2) · · · f2(ak).

We need to check four things.

• Φ is well defined. Indeed, let b1b2 · · · bm ∈ G1 ?G2 such that a1a2 · · · ak(b1b2 · · · bm)−1 = i1(g) (i2(g))−1

for some g ∈ G. A simple computation gives

f1(i1(g))f2

((i2(g))

−1)

= f1(i1(g)) (f2(i2(g)))−1

= f2(i2(g)) (f2(i2(g)))−1.

The last equality comes from the hypothesis f1i1 = f2i2. So f1(i1(g))f2

((i2(g))

−1)

= e where e is theidentity element of A. Therefore, f1(a1)f2(a2) · · · f2(ak) = f1(b1)f2(b2) · · · f2(bm), which implies thatΦ(a1a2 · · · akNG) = Φ(b1b2 · · · bmNG).

• Φ is a homomorphism of groups since

Φ((a1 · · · akNG)(b1 · · · bmNG)) = Φ(a1 · · · akb1 · · · bmNG) = (f1(a1) · · · f2(ak))(f1(b1) · · · f2(bm))

= Φ(a1 · · · akNG)Φ(b1 · · · bmNG).

• For any g1 ∈ G1, we have Φj1(g1) = Φ(g1NG) = f1(g1) by the definition of Φ. So Φj1 = f1. Similarly,we have Φj2 = f2.

• Uniqueness of Φ. Let Φ′ : G1 ? G2/NG −→ A be another homomorphism such that Φ′j1 = f1 andΦ′j2 = f2. We want to prove that Φ = Φ′. To do this, let a1a2 · · · akNG ∈ G1 ? G2/NG. By thedefinition of the group operation in G1 ? G2, and by the definition of j1 and j2, we have

a1a2 · · · akNG = (a1NG)(a2NG) · · · (akNG) = j1(a1)j2(a2) · · · j2(ak).


Therefore, because Φ′ is a homomorphism of groups, we have

Φ′(a1a2 · · · akNG) = Φ′j1(a1)Φ′j2(a2) · · ·Φ′j2(ak) = f1(a1)f2(a2) · · · f2(ak) = Φ(a1a2 · · · akNG).

The second equality comes from the assumption Φ′j1 = f1 and Φ′j2 = f2, and the last equality fromthe definition of Φ.

In a more sophisticated language, the group G1 ?G2/NG is called the pushout or the colimit of the diagram(2.5.1).

Remark 2.5.4. (i) If G = 0 then NG is trivial, and therefore G1 ? G2/NG = G1 ? G2. In that case, wehave Φ: G1 ? G2 −→ A.

(ii) If G1 = G2 = 0 then G1 ? G2 = 0, and therefore Φ: 0 −→ A is the trivial homomorphism.

We have defined the free product of two groups. One can easily extend this definition to the free product ofa finite number of groups G1, G2, · · · , Gn, and get a generalization of Φ (see for example [1, Pages 40–42]).

2.5.2 Van Kampen’s theorem

There are many versions of the Van Kampen theorem. We will use only the simplest one in this course (fora more general statement, we refer the reader to [1, Theorem 1.20]).

Let X be a topological space. Assume that X = U1 ∪ U2 where U1 and U2 are path-connected open subsetsof X such that U1 ∩ U2 is also path-connected. Choose the basepoint x0 inside U1 ∩ U2. Consider the inclu-sions maps i1 : U1 ∩ U2 ↪→ U1, i2 : U1 ∩ U2 ↪→ U2, f1 : U1 ↪→ X, and f2 : U2 ↪→ X. Applying the fundamental

group to U1 U1 ∩ U2_?

i1oo � �i2 // U2 we get the diagram

π1(U1 ∩ U2)i2∗ //

i1∗

��

π1(U2)

π1(U1),

which has the same form as (2.5.1). Set G = π1(U1 ∩ U2), A = π1(X), and consider the induced homomor-phisms f1∗ : π1(U1) −→ π1(X) and f2∗ : π1(U2) −→ π1(X). Clearly we have f1∗i1∗ = f2∗i2∗. Applying nowProposition 2.5.3, there exists a unique homomorphism

Φ: π1(U1) ? π1(U2)/NG −→ π1(X) (2.5.3)

such that Φj1 = f1∗ and Φj2 = f2∗ where j1, j2 are as in (2.5.2).

We are now ready to state the main theorem of the section.

Theorem 2.5.5 (The Van Kampen theorem). Assume that we have the underlined above things. Then thehomomorphism Φ from (2.5.3) is an isomorphism.

Proof. See [1, Pages 44–46].

Example 2.5.6. π1(Sn) = 0 for all n ≥ 2.


Solution. Write Sn = U1 ∪ U2 where U1 = Sn\{(0, · · · , 0, 1)} and U2 = Sn\{(0, · · · , 0,−1)}. It is clearthat U1, U2, and U1 ∩ U2 are path-connected. Moreover π1(U1) = π1(U2) = 0 since U1

∼= Rn ∼= U2 by thestereographic projection, and since π1(Rn) = 0. So by Theorem 2.5.5, we have

π1(Sn) ∼= π1(U1) ? π1(U2)/NG = 0 ? 0/NG = 0.

Proposition 2.5.7. For all n 6= 2, R2 is not homeomorphic to Rn.

Proof. If n = 0 then the result immediately follows since R0 is the one point space. For n = 1, the resultfollows because removing one point in R provides two connected components, while removing one point inR2 does not change the number of connected components of R2 (which is actually one). Now assume thatn ≥ 3 and suppose that there is a homeomorphism f : R2 −→ Rn. Then the map g : R2\{0} −→ Rn\{f(0)}defined by g(x) = f(x) is still a homeomorphism. Therefore, by Corollary 2.3.3, the induced homomorphismg∗ : π1(R2\{0}) −→ π1 (Rn\{f(0)}) is an isomorphism of groups. But this is not possible since

π1(R2\{0}) = π1(S1) = Z and π1(Rn\{f(0)}) = π1(Sn−1) = 0.

For the first π1 we have used Example 1.1.9 − (i), Theorem 2.3.7 and Theorem 2.2.7. And for the secondone, we have used Exercise2 − (a) (from Assignment 1), Theorem 2.3.7 and Example 2.5.6.

Remark 2.5.8. The fundamental group is not abelian in general as we can see in Example 2.5.9 below.

Let X and Y be two spaces, and let x0 ∈ X, y0 ∈ Y . The wedge of X and Y is the space denoted X ∨ Y ,and defined by

X ∨ Y = X∐

Y/(x0 ∼ y0).

That is, X ∨ Y is obtained from the disjoint union X∐Y by identifying x0 with y0. An illustration of that

is given by Figure 2.1.

Figure 2.1: An example of the wedge

Example 2.5.9. Let S1 be the circle endowed with its usual basepoint x0 = y0 = 1. Then the fundamentalgroup of S1 ∨ S1 is the free product Z ? Z. That is,

π1(S1 ∨ S1) = Z ? Z.

Solution. The space we are dealing with is shown in Figure 2.2.

Figure 2.2: The wedge of two copies of the circle

Write S1 ∨ S1 = U1 ∪ U2 where U1 and U2 are as in Figure 2.3.

Clearly U1, U2, and U1 ∩U2 are path-connected. Moreover we have U1 ' S1, U2 ' S1, and U1 ∩U2 ' pt. Soπ1(U1) = π1(U2) = Z by Theorem 2.3.7 and Theorem 2.2.7, and π1(U1∩U2) = 0 since U1∩U2 is contractible.Thus, by the Van Kampen theorem, and by Remark 2.5.4 –(i), it follows that π1(S1 ∨ S1) ∼= Z ? Z.


Figure 2.3: A decomposition of S1 ∨ S1

Chapter 3

Homology

3.1 Chain complexes

Definition 3.1.1. A chain complex is a collection {An}n∈Z, denoted A∗, of abelian groups equipped withgroup homomorphisms ∂n : An −→ An−1 such that

∂n ◦ ∂n+1 = 0 for all n ∈ Z. (3.1.1)

Example 3.1.2. Consider the collection {An}n∈Z defined as

An =

0 if n < 0

Z if n ∈ {0, 1, 2}0 if n > 2.

Define ∂2 : Z −→ Z and ∂1 : Z −→ Z as ∂2(n) = 2n and ∂1(n) = 0. The other maps ∂n are defined to be 0.Clearly one has ∂n+1∂n = 0 for all n. So {An}n∈Z is a chain complex.

A chain complex A∗ will be writing in the following way:

· · · −→ An+1

∂n+1 // An∂n // An−1 −→ · · · (3.1.2)

The elements of An are called n-chains and the homomorphisms ∂n are called the boundary maps or dif-ferentials. The groups An may be endowed with extra structure; for example, they may be vector spacesor modules over a fixed ring R. The differentials must preserve the extra structure if it exists; for example,they must be linear maps or homomorphisms of R-modules.

Notation 3.1.3. If An happens to be 0 for all n < 0, the sequence (3.1.2) will be denoted

· · ·∂n+1 // An

∂n // An−1

∂n−1 // · · · ∂2 // A1∂1 // A0.

In that case A∗ will be just denoted {An}n≥0.

Definition 3.1.4. Let A∗ be a chain complex. An element a of An is called

(i) an n-cycle (or just a cycle) if ∂n(a) = 0;

(ii) an n-boundary (or just a boundary) if there exists b ∈ An+1 such that ∂n+1(b) = a.

31

32 CHAPTER 3. HOMOLOGY

Example 3.1.5. Consider the chain complex A∗ from Example 3.1.2.

• Since ∂0 : Z −→ 0 is the zero map, for every n ∈ Z, we have ∂0(n) = 0. So every n ∈ Z = A0 is a0-cycle. For a similar reason, every element in A1 = Z is a 1-cycle. But in A2 = Z, the only 2-cycleis 0 since the kernel of ∂2 is the trivial group {0}.

• For m ∈ A2,m 6= 0, we have ∂2(m) = 2m 6= 0, and this implies that m is not an 2-cycle.

• One can see that every element of A1 of the form 2n is a 1-boundary since 2n = ∂2(n) by the definitionof ∂2. One can also see that every odd integer in A1 is not a boundary.

The kernel of ∂n is exactly the collection of all n-cycles, while the image of ∂n+1 is the collection of alln-boundaries.

Equation (3.1.1) implies that Im∂n+1 is a subgroup of Ker∂n. Indeed, if a ∈ Im∂n+1 then there existsb ∈ An+1 such that a = ∂n+1(b). Composing this latter equality by ∂n, we get ∂n(a) = ∂n ◦ ∂n+1(b). Since∂n ◦ ∂n+1 = 0 by (3.1.1), it follows that ∂n(a) = 0, which implies that a ∈ Ker∂n. So Im∂n+1 ⊆ Ker∂n.Furthermore, Im∂n+1 is a normal subgroup of Ker∂n since An is abelian and therefore Ker∂n is abelian aswell 1 . One can then consider the quotient group Ker∂n/Im∂n+1, which is denoted Hn(A). That is,

Hn(A) := Ker∂n/Im∂n+1. (3.1.3)

By the definition of a quotient group, an element of Hn(A) is the class [a] of an n-cycle a with respect tothe equivalence relation

a ∼ a′ if and only if there exists b ∈ An+1 such that a− a′ = ∂n+1(b).

In other words, a lies in the same class as a′ if and only if the difference a− a′ is a boundary. In particular,if a is an n-boundary, there exists b ∈ An+1 such that a = ∂n+1(b). This is equivalent to a − 0 = ∂n+1(b),which implies that [a] = [0]. So any boundary element in An is zero in Hn(A). The non-zero elements ofHn(A) are precisely the n-cycles which are not the boundary of something.

Definition 3.1.6. The group Hn(A) is called the nth homology group of A∗, and the collection {Hn(A)}n∈Zis called the homology of A∗, and it is denoted H∗(A).

As in the preceding chapter, the trivial group {0} will be just denoted 0. If two groups G and G′ areisomorphic, we will sometimes just write G = G′ instead of G ∼= G′. Given a group G, one can easily seethat the quotient G/{0} of G by the trivial group is isomorphic to G, that is, G/{0} = G.

Example 3.1.7. Consider the chain complex A∗ from Example 3.1.2. We want to compute the homologygroups Hn(A), n ∈ Z.

• If n < 0, then An = An−1 = 0 by the definition of An. This implies that Ker(∂n) = 0 and Im(∂n+1) = 0.Therefore Hn(A) = 0.

• If n > 2, a similar reasoning shows that Hn(A) = 0.

• Assume n = 0 and consider the sequence A1∂1−→ A0

∂0−→ A−1. Since A1 = A0 = Z, A−1 = 0 and∂1 = 0 = ∂0 by definition, it follows that Im∂1 = 0 and Ker∂0 = Z. This implies that H0(A) = Ker∂0/

Im∂1 = Z.1A classical fact in group theory says that any subgroup of an abelian group is normal.

3.2. SIMPLICIAL HOMOLOGY 33


∂1−→ A0. Since ∂1 = 0, it follows that Ker∂1 = A1 =

Z. Since ∂2 is the multiplication by 2, it follows that Im∂2 = {∂2(n) : n ∈ Z} = {2n : n ∈ Z} = 2Z.So

H1(A) = Ker∂1/Im∂2 = Z/2Z = Z2.


∂2−→ A1. Clearly Ker∂2 = 0 and Im∂3 = 0. Thisimplies H2(A) = 0.

Conclusion:

Hn(A) =

0 if n < 0

Z if n = 0

Z2 if n = 1

0 if n ≥ 2.

There is a notion of “map” between chain complexes.

Definition 3.1.8. Let A∗ and B∗ be chain complexes. A chain map from f : A∗ −→ B∗ is a collection{fn : An −→ Bn}n∈Z of group homomorphisms such that

∂nfn = fn−1∂n for all n. (3.1.4)

This amounts to saying that the following diagram commutes.

· · · // An+1

∂n+1 //

fn+1

��

An∂n //

fn

��

An−1//

fn−1

��

· · ·

· · · // Bn+1∂n+1

// Bn∂n

// Bn−1// · · ·

Proposition 3.1.9. Let f : A∗ −→ B∗ be a chain map.

(i) If a is an n-cycle, then so is fn(a).

(ii) If a is an n-boundary, then so is fn(a).

Proof. Exercise.

Proposition 3.1.9 implies that the map Hn(f) : Hn(A) −→ Hn(B) that sends [a] to [fn(a)] is well defined.

Definition 3.1.10. The collection {Hn(f)}n≥0 : H∗(A) −→ H∗(B), denoted f∗, is called the induced mapin homology.

3.2 Simplicial homology

To define the simplicial homology of a topological space, we need to define first the notion of “∆-complex”.


3.2.1 ∆-complexes

We begin with the definition of a simplex.

Definition 3.2.1. Consider a Euclidean space Rm, and let n ≤ m. Let v0, · · · , vn be a sequence of pointsof Rm such that the difference vectors v1 − v0, · · · , vn − v0 are linearly independent. Define [v0, · · · , vn] as

[v0, · · · , vn] =

{n∑i=0

tivi |n∑i=0

ti = 1 and ti ≥ 0 for all i

}. (3.2.1)

This set is called the n-simplex generated by v0, · · · , vn. The coefficients ti are called the barycentric coor-dinates of the point

∑i tivi in [v0, · · · , vn].

For purposes of homology it will be important to keep track of the order of the vertices of a simplex. Fromnow on, “n-simplex” will really mean “n-simplex with an ordering of its vertices”. Unless stated otherwise,we will always choose the natural order, that is, v0 ≤ · · · ≤ vn.

The points v0, · · · , vn are called the vertices of [v0, · · · , vn]. A 0-simplex is just a point v0, while a 1-simplexis a line segment with endpoints v0 and v1. A 2-simplex is a triangle with vertices v0, v1, and v2, while a3-simplex is a tetrahedron, and so on (see Figure 3.1).

Figure 3.1: Example of simplices

An equivalent definition of n-simplex is the following. A subset X of Rm is said to be convex if for anya, b ∈ X, the set {(1− t)a+ tb|t ∈ [0, 1]} is contained in X. One can see that the smallest convex subset inRm containing v0, · · · , vn is exactly [v0, · · · , vn] 2.

Example 3.2.2. In Rn+1, consider the points e0, · · · , en defined as ei = (ei0, · · · , eij , · · · , ein) where

eij =

{1 if j = i

0 otherwise.

For instance, if n = 2, one has e0 = (1, 0, 0), e1 = (0, 1, 0), and e2 = (0, 0, 1). The n-simplex generated bye0, · · · , en is called the standard n-simplex and it is denoted ∆n. It follows immediately from the definitionthat

∆n =

{(t0, · · · , tn) ∈ Rn+1|

n∑i=0

ti = 1 and ti ≥ 0 for all i

}. (3.2.2)

See Figure 3.2 for the standard 2-simplex.

Remark 3.2.3. There is a canonical linear homeomorphism from the standard n-simplex ∆n to any othern-simplex [v0, · · · , vn], preserving the order of vertices, namely, (t0, · · · , tn) 7→

∑i tivi. So

∆n ∼= [v0, · · · , vn].

2By “smallest” we mean the intersection of all convex subsets of Rm containing v0, · · · , vn.


Figure 3.2: The standard 2-simplex

Definition 3.2.4. Let [v0, · · · , vn] be an n-simplex, and let 0 ≤ k ≤ n. The (n − 1)-simplex generated byv0, · · · , vk−1, vk+1, · · · , vn is called a face of [v0, · · · , vn], and it is denoted [v0, · · · , vk, · · · , vn].

So [v0, · · · , vn] has exactly n + 1 faces. In other words, a face of [v0, · · · , vn] is obtained by deleting oneof the n + 1 vertices vk of [v0, · · · , vn]. We adopt the following convention: The vertices of a face, or anysubsimplex generated by a subset of the vertices, will always be ordered according to their order in the largersimplex.

The face of ∆n generated by e0, · · · , ek−1, ek+1, · · · , en is

[e0, · · · , ek, · · · , en] =

n∑

i=0,i6=k

tiei |n∑

i=0,i6=k

ti = 1 and ti ≥ 0 for all i 6= k

.

For the sake of simplicity, [e0, · · · , ek, · · · , en] will be often just denoted ∂k∆n. As in Remark 3.2.3, there isa canonical linear homeomorphism φ : ∆n−1

∼=−→ ∂k∆n preserving the order of vertices, namely,

φ(t0, · · · , tn−1) = t0e0 + · · · tk−1ek−1 + tkek+1 + · · · tn−1en. (3.2.3)

Definition 3.2.5. (i) The union of all faces of ∆n is called the boundary of ∆n, and it is denoted ∂∆n.So

∂∆n =

n⋃k=0

∂k∆n.

(ii) The interior of ∆n, denoted ∆n, is defined to be ∆n\∂∆n. That is,

∆n = ∆n\∂∆n.

Now we are ready to define a ∆-complex.

Definition 3.2.6. A ∆-complex is a pair (X,∆) where X is a topological space, and ∆ is a collection ofcontinuous maps σα : ∆n −→ X, with n depending on the index α, satisfying the following three conditions.

(i) For every σα ∈ ∆, the restriction σα|∆n is injective, and each point of X is in the image of exactlyone such restriction σα|∆n.

(ii) For any element σα : ∆n −→ X of ∆, each restriction σα|∂k∆n is such that the composition

σα|∂k∆n ◦ φ : ∆n−1 −→ X

is one of the maps σβ from ∆. Here φ is the canonical linear homeomorphism defined above.


(iii) A set A ⊂ X is open if and only if σ−1α (A) is open in ∆n for each σα ∈∆.

Among other things, this last condition rules out trivialities like regarding all the points of X as individualvertices. If the structure is understood, we will just write X for (X,∆).

Example 3.2.7. Let X be the unit circle S1. Consider the following collection ∆ = {σ0, σ1} where σ0 : ∆0 =

{1} −→ S1 is defined as σ0(1) = (1, 0), and σ1 : ∆1 −→ S1 is defined as σ1(t0, t1) = e2πit0 . We want to showthat the pair (S1,∆) is a ∆-complex. First of all, the conditions t0 + t1 = 1 and t0, t1 ≥ 0 in the definitionof ∆1 imply that t0, t1 ∈ [0, 1]. So the map σ1 wraps the circle exactly one time. Now let us check conditions(i), (ii), and (iii).

(i) We need to check (i) for every element of ∆. Let us begin with σ0. Since the boundary of ∆0 isempty, it follows that ∆0 = ∆0. Therefore the restriction σ0|∆0 is clearly injective. For σ1, a simplecomputation gives ∆1 = {(t0, t1) ∈ R2|t0 + t1 = 1 and t0, t1 ∈ (0, 1)}. This implies that the restrictionσ1|∆1 is injective (one can check that from the definition of σ1). Now let x = e2πis ∈ S1. If s = 0 thenx = (1, 0), which belongs to the image of σ0|∆0. If s ∈ (0, 1), then x definitely belongs to the image ofσ1|∆1. So (i) is satisfied.

(ii) This easily follows from the definition of σ1, ∂k∆1, and φ.

(iii) Exercise.

The idea behind a ∆-complex is as follows. Roughly speaking, a ∆-complex is a space X constructed bygluing together simplices. However, there are rules for what kinds of “gluing ” you are allowed to use.

(a) You have to start with all the 0-simplices, then glue in all the 1-simplices, then glue in all the 2-simplices,and so on. You are not allowed to drop this ordering condition.

(b) If Y is the ∆-complex you have built so far and f : ∂∆n −→ Y is any continuous map, you can build a∆-complex X = Y

∐∆n/ ∼. , where “∼” is the equivalence relation that identifies any x ∈ ∂∆n with

f(x) = y. The highly restriction when adding a new simplex is the following. For each (n− 1)-simplexF = ∂k∆n which is a face of ∂∆n, the restriction of f to F must be equal to the inclusion of oneof the (n − 1)-simplices you already have. That is, f maps the n vertices of F (with their canonicalordering) to the n vertices of some (n− 1)-simplex you have already added to your complex (with thesame ordering on the vertices).

3.2.2 Definition of the simplicial homology and examples

Let (X,∆) be a ∆-complex. An element σα : ∆n −→ X of ∆ will be called an n-simplex of X.

Definition 3.2.8. Define ∆n(X) to be the free abelian group generated by all n-simplices of X.

An element of ∆n(X) is a formal sum∑α nασα, where nα ∈ Z. We want to turn out the collection

{∆n(X)}n≥0 to a chain complex (see Definition 3.1.1). So we need to define maps ∂n : ∆n(X) −→ ∆n−1(X)

such that ∂n∂n+1 = 0 for all n. Define ∂n as

∂n(σα) =

n∑k=0

(−1)k (σα|[e0, · · · , ek, · · · , en]) ◦ φ,

where φ is the homeomorphism from (3.2.3).

Proposition 3.2.9. For all n, one has ∂n∂n+1 = 0.


Proof. Exercise.

Proposition 3.2.9 tells us that ∂n is a differential. So the collection ∆∗(X) = {∆n(X)}n≥0 is a chain complex(see Definition 3.1.1 and Notation 3.1.3).

Definition 3.2.10. The simplicial homology of X, denoted H∆∗ (X), is defined to be the homology of the

chain complex ∆∗(X). That is,H∆n (X) := Ker∂n/Im∂n+1.

The group H∆n (X) is called the nth simplicial homology group of X.

Example 3.2.11. In this example, we would like to compute the simplicial homology H∆∗ (S1) of the circle.

Consider the ∆-complex structure, ∆ = {σ0, σ1}, from Example 3.2.7. In that structure we have one 0-simplex, σ0, and one 1-simplex, σ1. So ∆0(X) is the free abelian group generated by σ0, that is, ∆0(X) = Z.Similarly ∆1(X) = Z. Since there is no n-simplex in ∆ with n /∈ {0, 1}, it follows that ∆n(X) = 0 for alln /∈ {0, 1}. So the chain complex ∆∗(X) looks like

· · · ∂3 // 0∂2 // Z ∂1 // Z ∂0 // 0

∂−1 // · · ·

• Let us compute H∆0 (X). First of all, we have Ker(∂0) = Z. A simple computation shows that ∂1(σ1) =

0. So Im∂1 = 0, and therefore H∆0 (S1) = Z.

• Now we compute H∆1 (X). Since ∂1(σ1) = 0, it follows that Ker∂1 = Z. Certainly Im∂2 = 0. So

H∆1 (S1) = Z.

• For the other values of n, we have H∆n (S1) = 0.

Hence,

H∆n (S1) =

{Z if n = 0, 1

0 otherwise.

Remark 3.2.12. For the sake of simplicity, we will often identify an n-simplex σα : ∆n −→ X with itsimage σα(∆n) (or with σα(∆n)). So instead of looking at σα as a map, we will look at it as a subspace ofX. That point of view has the advantage that it makes computations easier. And the inconvenience that itmakes much harder the verification of (i), (ii), and (iii) from Definition 3.2.6. Since we are interested incomputations, the ∆-complex structure should be always understood from the picture.

For example, let us consider again the circle S1. Define v ⊆ S1 as v = {(1, 0)} and e = S1. One shouldthink e as a 1-simplex, staring at v and ending at v, that wraps the circle one time (see Figure 3.3).

Figure 3.3: A ∆-complex structure on S1

Assume that e is oriented in the counterclockwise. One can then replace the collection {σ0, σ1} before by{v, e}. So our ∆-complex structure on S1 consists of one 0-simplex, v, and one 1-simplex, e. This impliesthat ∆0(S1) is the free abelian group generated by v that we denote 〈v〉, and ∆1(S1) = 〈e〉. Since the thefree abelian group generated by one element is Z, it follows that ∆0(S1) = Z and ∆1(S1) = Z. Since thereis no n-simplex with n /∈ {0, 1}, it follows that ∆n(S1) = 0 for all n /∈ {0, 1}. Compare to what we did inExample 3.2.11, the differential is much easier to compute here. For instance, ∂1(e) = v − v = 0.


Example 3.2.13. Here we compute the simplicial homology of the torus T , as shown Figure 3.4.

Figure 3.4: Torus T

• How many 0-simplices do we have? Only one: v. So ∆0(T ) = 〈v〉 = Z. Since ∂0 : ∆0(T ) −→ 0 is thezero morphism, it follows that Ker∂0 = Z.

• How many 1-simplices do we have? Exactly 3: a, b, and c. So ∆1(T ) = 〈a, b, c〉 = Z3. The boundary ofa is ∂1(a) = v− v = 0. Similarly, ∂1(b) = ∂1(c) = 0. This implies that Im∂1 = 0 and Ker∂1 = 〈a, b, c〉.

• How many 2-simplices do we have? Exactly 2: U and L. So ∆2(T ) = 〈U,L〉 = Z2. The boundary ofU is ∂2(U) = a+ b− c = ∂2(L). This implies that Im∂2 = 〈a+ b− c〉 and Ker∂2 = 〈U − L〉 = Z.

• How many n-simplices, n ≥ 3, do we have? Zero. So ∆n(T ) = 0 for n ≥ 3.

Now we compute H∆∗ (T ). For n = 0, H∆

0 (T ) = Ker∂0Im∂1

= Z since Ker∂0 = Z and Im∂1 = 0. For n = 1, wehave

H∆1 (T ) =

Ker∂1

Im∂2=〈a, b, c〉〈a+ b− c〉

=〈a, b, a+ b− c〉〈a+ b− c〉

∼= 〈a, b〉 ∼= Z× Z.

For n = 2, we have H∆2 (T ) = Ker∂2

Im∂3∼= Z since Ker∂2 = Z and Im∂3 = 0. Certainly, H∆

n (T ) = 0 for n ≥ 3.Hence the homology of the torus is

H∆n (T ) =

Z if n ∈ {0, 2}

Z× Z if n = 1

0 if n ≥ 3.

Notice that the orientation of simplices is very important in the calculations, especially when we want tocompute the boundary or the differential of something.

Example 3.2.14. We would like to compute the simplicial homology of the real projective plane RP2, asshown Figure 3.5

• Since there are two 0-simplices, v and w, it follows that ∆0(X) = 〈v, w〉 = Z2. Since ∂0 is the zeromorphism, Ker∂0 = 〈v, w〉.

• ∆1(X) = 〈a, b, c〉 = Z3, ∂1(a) = w − v = ∂1(b), and ∂1(c) = 0. This implies that Im∂1 = 〈w − v〉 andKer∂1 = 〈a− b, c〉.

3.3. SINGULAR HOMOLOGY 39

Figure 3.5: Real projective plane RP2

• ∆2(X) = 〈U,L〉 = Z2, ∂2(U) = a − b − c, and ∂2(L) = a − b + c. This implies that Ker∂2 = 0 andIm∂2 = 〈a− b− c, a− b+ c〉.

• Since there is no n-simplices when n ≥ 3, it follows that ∆n(X) = 0.

Now we compute H∆n (X). For n = 0, one has

H∆0 (X) =

Ker∂0

Im∂1=〈v, w〉〈w − v〉

=〈v, w − v〉〈w − v〉

∼= 〈v〉 = Z.

For n = 1,

H∆1 (X) =

Ker∂1

Im∂2=

〈a− b, c〉〈a− b− c, a− b+ c〉

=〈a− b− c, c〉〈a− b− c, 2c〉

∼=〈c〉2〈c〉

=Z2Z∼= Z2.

For n = 2, H∆2 (X) = Ker∂2

Im∂3= 0 since Ker∂2 = 0. Clearly, H∆

n (X) = 0 for n ≥ 3. So

H∆n (RP2) =

Z if n = 0

Z2 if n = 1

0 if n ≥ 2.

3.3 Singular homology

Work in progress...

3.3.1 Definition

3.3.2 Exact sequences

3.3.3 Relative homology groups

3.4 Applications

Bibliography

[1] Allen Hatcher, Algebraic Topology, (2002) Cambridge University Press, xii+544 pp.

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