mas435 / mas6370 algebraic topology 2014{15 part a

MAS435 / MAS6370 Algebraic Topology

2014–15

Part A: Semester 1

Lecturer: Miss Magdalena Kedziorek

Notes by Dr E Cheng

Weekly tests

Once a week (every Monday) at the beginning of the lecture there will be a

quick test of some definitions, theorems, examples and counterexamples from

the previous week. This will count as 20% of your final mark. The exam counts

as 80%.

Homeworks

Weekly exercises are available online.

Contents

1 Introduction and Motivation 3

1.1 Topology and algebra . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Algebraic topology . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Examples of spaces we will consider . . . . . . . . . . . . . . . . 6

1.5 Examples of groups we will be thinking about . . . . . . . . . . . 10

1.6 Questions about the fundamental group . . . . . . . . . . . . . . 10

1.7 The eventual goals of algebraic topology . . . . . . . . . . . . . . 11

2 Reminder on metric spaces 12

2.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Topological spaces 14

3.1 Spaces and continuous maps . . . . . . . . . . . . . . . . . . . . . 14

3.2 Examples of continuous maps . . . . . . . . . . . . . . . . . . . . 19

3.3 Paths and loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4 Homotopy 25

CONTENTS 2

5 The fundamental group 33

5.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5.2 Dependence on basepoint . . . . . . . . . . . . . . . . . . . . . . 35

5.3 Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.4 Homotopy invariance . . . . . . . . . . . . . . . . . . . . . . . . . 46

5.5 The circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

6 Covering spaces 52

6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

6.2 The classification of covering spaces . . . . . . . . . . . . . . . . 61

6.3 Universal covering spaces . . . . . . . . . . . . . . . . . . . . . . 68

6.4 Covering space constructions . . . . . . . . . . . . . . . . . . . . 68

7 Van Kampen’s Theorem 73

7.1 Free product of groups . . . . . . . . . . . . . . . . . . . . . . . . 74

7.2 Glueing spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

7.3 Equivalence relations on groups . . . . . . . . . . . . . . . . . . . 79

7.4 Van Kampen’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 80

7.5 Generators and relations . . . . . . . . . . . . . . . . . . . . . . . 84

7.6 Cell complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

7.7 The classification of surfaces . . . . . . . . . . . . . . . . . . . . . 93

8 Applications 96

8.1 The fundamental theorem of algebra . . . . . . . . . . . . . . . . 96

8.2 The Brouwer fixed point theorem . . . . . . . . . . . . . . . . . . 100

Plan

Week 1: Section 1

Week 2: Sections 2 and 3

Week 3: Section 4

Week 4: Section 5 (first half)

Week 5: Section 5 (second half)

Week 6: Section 6 (first half)

Week 7: Reading Week

Week 8: Section 6 (second half)

Week 9: Section 7

Week 10: Section 8

Week 11: Revision

Each week you should study the previous week’s notes and the next week’s

notes.

1 Introduction and Motivation 3

1 Introduction and Motivation

The idea of algebraic topology is to study topology using algebra. Many of the

greatest advances in mathematics have come from finding a link between two

apparently different subjects enabling us to use our understanding of one to

help our understanding of the other. Often the flow of understanding starts in

one direction and but is later discovered to go in both directions.

Topology←→ Algebra

1.1 Topology and algebra

Topology

Topology is about space, shape and form. You’ve already seen metric spaces

before, but we want to move away from the notion of distance. For example:

• How do you get from A to B? Do you need to fly? This is a separate issue

from how far it is.

• Think about things made of plasticine that we can stretch and shrink but

we don’t want to make new holes or stick things together. Stretching

changes the distance between points but we don’t care.

• We’re not worried about size, e.g.

are fundamentally “the same”.

• We’re not worried about curvature, so

is the same as

We need a better notion of “space” that doesn’t depend on having a ruler.

1.2 Algebraic topology 4

Example 1.1. A city is made up of neighbourhoods, but these are not defined

by neat circular areas around points.

Algebra

Algebra is about operations on things. Here we will mainly be thinking about

groups but later in the subject there are modules, chain complexes, groupoids,

and more.

1.2 Algebraic topology

The idea is to find a way of relating topology to algebra, and then, amazingly,

we can use algebra to help us understand space. Some examples of applications

include

• robotic arms

• computers crashing

• polymers

• ties.

Perhaps even more amazingly, we can go the other way and actually use space

to help us understand algebra. For example we will prove the Fundamental

Theorem of Algebra.

There are many ways to convert space into algebra. The main methods used in

algebraic topology are called

• homotopy,

• homology, and

• cohomology.

We will almost exclusively think about homotopy in this part of the course

(Part A).

1.3 Homotopy 5

1.3 Homotopy

Example 1.2. What if someone told you that you were living on the surface

of a torus?

Here are some questions we can ask ourselves about the space we inhabit.

1. Can we get from A to B without flying?

2. How many different ways are there of getting from A to B?

3. Are those ways really different?

4. If we all hold hands in a circle can we all walk to the same place?

That last one might sound a bit silly in this context, but it turns out that

considering “loops” like this tells us tons of stuff about the space we’re in. This

idea turns out to be the same as the idea of bending and stretching spaces

around, and this is the notion of homotopy.

The key to using these “loops” is:

The loops form a group.

This is called the “fundamental group” and helps us tell spaces apart. However,

this is only the beginning. For example, it can’t tell higher-dimensional spheres

apart—we need higher-dimensional methods to to that, and it’s very hard.

However, the fundamental group is extremely powerful. For example, it en-

ables us to classify surfaces completely. Surfaces are one of the most important

examples of spaces that we’ll look at.

Exercise 1.3. Which of the following spaces are “the same” in the sense that

a doughnut is “the same” as a teacup?

1.

2.

1.4 Examples of spaces we will consider 6

3.

4. Any of the above spaces and

(a) a pair of trousers

(b) a cardigan

(c) a sweater

(d) a skirt

(e) a woolly hat

(f) a Klein bottle

1.4 Examples of spaces we will consider

1. Discs/balls Dn or Bn: not very interesting as they’re the “same” as a

point. But they are very useful building blocks.

2. The unit interval I = [0, 1]. Also the “same” as a point, but also a crucial

building block. If you were stuck on a desert island with nothing but a unit

interval, you could (in some sense) built any topological space, provided

you knew how to make products, disjoint unions and quotients.

3. Circles/spheres Sn: these are very interesting and to date their homotopy

is still not completely understood in high dimensions.

4. The Mobius strip (also known as Mobius Band): it’s somehow the “same”

as a circle, and yet. . .

5. The torus and its many-holed cousins. These are the orientable surfaces.

6. Non-orientable surfaces such as the Klein bottle and more complicated

versions.

7. We can cut holes in a surface and glue on a Mobius Band or handle. We

get the following mysterious “equation”

handle + MB = 3MB = KB + MB.


8. Letters of the alphabet:

A D

B

C E F

9. We can “stick circles together” e.g.

10. We can also stick circles together in a “pathological” way e.g.

11. We can “stick spheres together” e.g.

12. We can mix things up, and stick anything to anything else e.g.

13. Real projective space is defined by RPn = Sn/ ∼ where ∼ is the equiv-

alence relation that sticks antipodal points together. This is hard to vi-

sualise, especially for n ≥ 3. The formal methods of studying spaces really

come into their own when we’re dealing with spaces we can’t draw/visualise,

so we can’t rely on our spatial intuitions any more.


14. We can take products e.g.

R× R = R2

I × I = I2 ∼ B2

I × I × I = I3 ∼ B3

I × R

I × S1 cylinder

S1 × S1 torus

S1 × R

15. Knots and links can be studied via their complements e.g.

• The complement of is like




16. We can take unions and intersections. This might not sound so interesting,

but will be very important later.

17. We can take a polygon and glue its edges together according to some

scheme—this is a form of quotient e.g.

18. Other quotients e.g.

• We can take I × I and identify the entire boundary with a point.

What does this give us?

• We can take D2∐D2 and identify the boundaries. What does this

give us?

• We can take MB∐D2 and identify the boundaries.

• We can quotient R by the relation

x ∼ y ⇐⇒ y − x ∈ Z.

• We can quotient R× R by the relation

(x, y) ∼ (x′, y) ⇐⇒ x′ − x ∈ Z.

1.5 Examples of groups we will be thinking about 10

1.5 Examples of groups we will be thinking about

1. Z

2. Zn = Z/nZ

3. Z× Z, Zn = Z× · · ·Z

4. Z ∗ Z, Z ∗ · · · ∗ Z

5. quotient groups

6. abelianisation.

1.6 Questions about the fundamental group

1. How do we actually calculate it?

We will build up techniques to help us understand groups in terms of simpler

ones, for example:

• covering spaces: wrapping spaces up with other spaces

• Van Kampen’s theorem: glueing spaces together to make other spaces

The crucial starting point will be to understand the fundamental group of the

circle.

2. How does the fundamental group help us?

We can’t classify all homotopy types using the fundamental group, for example

for all n ≥ 2 the fundamental group of Sn is the trivial group. However, we can

classify all surfaces using the fundamental group. That is, we can show that

any surface is entirely determined “up to homotopy” by its fundamental group.

However, for spaces in general we need something much more complicated:

homotopy equivalence⇒ same fundamental group

—but not the converse. This is why further methods are introduced such as

higher homotopy groups, and homology/cohomology in all dimensions.

1.7 The eventual goals of algebraic topology 11

1.7 The eventual goals of algebraic topology

Arguably, the eventual goal of algebraic topology is to use algebra to classify

space completely, up to homotopy. (Some would say the real goal is to classify

space up to homeomorphism but that is much harder, possibly unrealistically

so.) That is, the aim is to find some algebraic structures that correspond to

topological spaces in an extremely comprehensive way. This is a field of research

that is still wide open. In the last 15 years or so, a whole new type of algebra

called

ω-categories

has been developed to try to achieve this goal, and many arguments are going

on about what will be the most fruitful way to do this.

2 Reminder on metric spaces 12

2 Reminder on metric spaces

2.1 Basic definitions

Exercise 2.1. What can you remember about metric spaces? What do you

think we might retain if we ditch the notion of “metric”?

Definition 2.2. A metric space is a set X equipped with a metric

d : X2 −→ R≥0

such that

1. d(x, y) = 0 ⇐⇒ x = y,

2. for all x, y ∈ X, d(y, x) = d(x, y), and

3. for all x, y, z ∈ X, d(x, z) ≤ d(x, y) + d(y, z) —“triangle inequality”.

One thing we definitely want to retain even without metrics is the notion of

“continuous map”. Can you remember how to define continuous maps

• by convergent sequences:

• by ε and δ:

In fact we have the following three equivalent ways of characterising continuity.

Definition 2.3. Let f : X −→ Y be a function between metric spaces. Then

f is continuous if any of the following equivalent conditions holds:

1. Given any sequence {xn} such that xn −→ x, we have fxn −→ fx.

2. ∀a ∈ X ∀ε > 0 ∃δ > 0 s.t. ∀x ∈ X, d(a, x) < δ ⇒ d(f(a), f(x)) < ε

3. Given any open set U ⊆ Y , f−1U is open in X.

Before we go any further, we’d better remind ourselves of what the various parts

of these definitions mean. For definition (1) we need to remember convergent

sequences.

2.1 Basic definitions 13

Definition 2.4. Given a sequence {xn} in a metric space X, we say {xn}converges to x, written xn −→ x if and only if

∀ε > 0 ∃N ∈ N s.t. ∀n ≥ N d(xn, x) < ε.

Definition (2) speaks for itself, but for definition (3) we need to remind ourselves

about images and pre-images, and open sets.

Definition 2.5. Let f : X −→ Y be any function, A ⊆ X, B ⊆ X.

• The image of A under f is f(A) = {f(a) | a ∈ A} ⊆ Y .

• The pre-image of B under f is f−1B = {x ∈ X | f(x) ∈ B} ⊆ X

Definition 2.6. Let X be a metric space, a ∈ X.

• The open ball of radius r centred at a is defined by

◦Br (a) = {x ∈ X | d(a, x) < r}.

• The closed ball of radius r centred at a is defined by

Br(a) = {x ∈ X | d(a, x) ≤ r}.

Definition 2.7. Let X be a metric space and A ∈ X. Then A is open iff

∀a ∈ A ∃δ > 0 s.t.◦Bδ (a) ⊆ A.

The only definition of continuity that has a hope of working without a metric is

the last one, but to do that we have to somehow think about open sets without

using open balls. This brings us to the definition of topological space.

3 Topological spaces 14

3 Topological spaces

We are going to concentrate on the idea of open sets. But we need a notion

that doesn’t use distance. How is it possible?

We’re going to do something mind-boggling: we’re just going to say what the

open sets have to satisfy, and we’re not actually going to say what the open sets

have to be. We will define a topological space to be “a set equipped with some

subsets called open” and then specify some conditions that ensure that we have

made a sensible choice of “open” set.

Question 3.1. What do open sets look like in R2?

Note: we’re not actually going to worry about the definition of topological

space very much. We will hardly even use it. We’ll probably never prove that

something actually is a topological space using this definition. Likewise the

definition of continuity. You might wonder how and why we avoid the definition

of “topological space” in the study of topological spaces. One possible answer

is that this definition is sort of disgusting to work with, and somehow doesn’t

capture what is “really” going on with topological spaces. So we run away from

the definition as fast as possible and use algebra to study spaces instead.

The important thing we’ll need to know is:

There is some notion of space a bit like metric space and

some notion of continuous map—and some things are true

about them.

For example: we can compose maps, take products, do quotients. . .

3.1 Spaces and continuous maps

Recall that a metric space is a set equipped with a metric; similarly a topological

space is a “set equipped with a topology”, as follows.

Definition 3.2. A topology on a set X is a collection of subsets of X called

“open”, satisfying:

1. ∅ and X are open,

2. any union of open sets is open, and

3. any finite intersection of open sets is open.

3.1 Spaces and continuous maps 15

A topological space is a set X equipped with a topology.

Note that we only ask for finite intersections—it is possible to have an infinite

intersection of open sets that isn’t open, e.g.⋂n

(− 1n ,

1n ) =

Example 3.3. Any metric space is a topological space, with the metric topol-

ogy in which the open sets are the ones we defined before (Definition 2.7).

In fact many of the spaces we consider can be thought of as metric spaces, but

in a manner that might just not be very helpful. For example when we do funny

glueings like

we just don’t want to be thinking about how “far apart” points are when we’re

done.

Note that a given set X might have many different possible topologies on it.

Here are a couple of examples that sound a bit silly but are actually quite

important.

Example 3.4. Any set X can be given the following two topologies. These

may be thought of as “maximal” and “minimal”.

1. Discrete: every set is open.

2. Indiscrete: only ∅ and X are open.

We will later see (Exercises #4) that every map out of a discrete space is

continuous, and every map into an indiscrete space is continuous.

Definition 3.5. A subset A of a topological space X is called closed if X \Ais open.

For example ∅ and X are both open and closed. Note that we can state the

definition of topology using closed sets instead of open (Exercises #3).

Now that we don’t have a notion of distance any more, we don’t have a notion

of convergent sequence. The important thing is that we do have a notion of


continuity. Which definition of continuity do we use? There is clearly only one

of our three that is still feasible.

Definition 3.6. A continuous map of topological spaces f : X −→ Y is a

function such that whenever U is open in Y , f−1U is open in X.

Note that this is very different from saying that f maps open sets to open sets.

This condition on pre-images tells us nothing about images. If f is continuous

then:

• the image of an open set does not have to be open e.g.

R 0−→ R

(0, 1) −→ [0, 0]

• the image of a closed set does not have to be closed e.g. under the map

R −→ R

x 7→ e−x

the image of the subset N ⊂ R is a sequence whose limit is 0 as n −→∞.

Since this limit is not actually attained the image f(N) is not closed, but

N is closed.

The following result is important, and is also a good exercise in manipulating

pre-images.

Proposition 3.7. If f : X −→ Y and g : Y −→ Z are continuous then

gf : X −→ Z is continuous.

Proof. We need to check that if U is open in Z, (gf)−1(U) is open in X. Now

(gf)−1(U) = {x ∈ X |g(f(x)) ∈ U}

= {x ∈ X |f(x) ∈ g−1(U)}

= f−1(g−1(U)).

Since g is continuous, g−1(U) is open in Y , and thus since f is continuous,

f−1(g−1(U) is open in X. 2

We now come to two crucial constructions that we will use heavily, for building

up spaces: products and quotients. Products go pretty much as you would

expect, but perhaps quotients take a bit more thought.


Definition 3.8. If X and Y are topological spaces then X ×Y is a topological

space with the product topology whose open sets are:

• U × V where U is open in X and V open in Y , and

• all unions of these.

Exercise 3.9. Why don’t we need to specify that all intersections of these are

open as well?

Note that given continuous maps X1f−→ X2 and Y1

g−→ Y2 we get a continuous

map

X1 × Y1f×g−→ X2 × Y2.

Definition 3.10. If X is a topological space and ∼ is an equivalence relation

on the set X then X/ ∼ is a topological space with the quotient topology, in

which a subset of equivalence classes is open in X/ ∼ if their union is open in

X. The resulting space is called a quotient space.

For example a subset of equivalence classes might look like

{[x], [y], [z]}

and to check if this is open in X/ ∼ we look at the subset

[x] ∪ [y] ∪ [z] ⊂ X

that is, the subset of X consisting of all the points equivalent to x, y or z under

∼.

Example 3.11. We define an equivalence relation ∼ on R by

x ∼ y ⇐⇒ y − x ∈ Z

and form the quotient. We can depict this as:


Note that the quotient topology is defined precisely to make the quotient map

continuous. That is, the map:

X −→ X/ ∼

x 7→ [x].

Quotients are dealt with much better by topological spaces than by metric

spaces. Quotients of metric spaces are basically disgusting. We can think of

this as one of the major limitations of metric spaces, as glueing things together

is one of our most important tools for understanding complicated spaces.1

Exercise 3.12. Think about the way we glue the ends of the interval I together

to make a circle. How would that work if we thought of them as metric spaces?

We can now state the most straightforward notion of “sameness” of topological

spaces.

Definition 3.13. A homeomorphism is a continuous map with a continuous

inverse.

It is very important to stress the fact that the inverse must be continuous. A

map can certainly be continuous and bijective but not have a continuous inverse.

That is, it would have an inverse as a map between sets, but it does not have

an inverse as a continuous map between topological spaces. (This is different

from, say, groups—if a group homomorphism is bijective then its inverse is

automatically a group homomorphism. However the same problem arises if we

1This fact is expressed categorically by the fact that the category of metric spaces does

not have coequalisers.

3.2 Examples of continuous maps 19

think about smooth maps—a smooth map can certainly have an inverse that

isn’t smooth.)

Examples 3.14.

1. Consider the map

[0, 1)f−→ S1

t 7→ (cos2πt, sin2πt).

This is bijective, so it has an inverse, but its inverse is not continuous.

For example, [0, 14 ) is open in [0, 1) but its pre-image under the inverse of

f is not open in S1 (which you should prove properly in Exercises #4).

2. However, if we consider the quotient space I/ ∼ under the relation 0 ∼ 1,

then we have a well-defined map

I/ ∼ f−→ S1

t 7→ (cos2πt, sin2πt)

and it is a homeomorphism. The above problem has gone, because the

subset like [0, 14 ) is not open in I/ ∼. To see this, we have to remember

how the quotient topology goes, and [0, 14 ) ∪ 1 is not open in I.

3. The map

D2 −→ {0}

is quite obviously not a homeomorphism, but we want to think of these

spaces as being somehow “the same”. So we need more subtle ways of

telling spaces apart.

Note that from now on whenever we say “space” we will mean topological space.

3.2 Examples of continuous maps

1. For any space X we have the identity map 1X : X −→ X.


2. There are things that are a bit like identities—inclusions/embeddings e.g.

(a) R ↪→ R2 embeds in many ways.

(b) I2 ↪→ I3

(c) Sn ↪→ Sn+1

(d) S1 ↪→ I × S1

x 7→ (t, x)

—we have one of these for each t ∈ I. We can do the same to embed

X ↪→ I ×X.

(e) S1 ↪→MB by inclusion into the boundary.

(f) Bn ↪→ Sn on the north or south hemisphere.

Write Bn = {(x, y) ∈ R2 |√x2 + y2 ≤ 1}

Sn = {(x, y, z) ∈ R3 |√x2 + y2 + z2 = 1}

.

Then the map is given by

(x, y) 7→ (x, y,√

1− x2 − y2).

3. “Glueing” e.g.

(a) I2 = −→MB.

(b) I2 = −→ torus.

(c) All our other polygon glueing examples.

(d) We can glue the ends of an interval to make a circle:

Iα−→ S1

t 7→ (cos2πt, sin2πt)


(e) The above map induces a map for glueing the edges of a square to

make a cylinder:

1×α−→

I2 −→ S1 × I

(t, z) 7→ (cos2πt, sin2πt, z)

(f) D2∐D2 −→ S2

N.B. we naturally get two maps D2 ↪→ D2∐D2 −→ S2 for the upper

and lower hemisphere.

We have a diagram of continuous maps:

S1 D2

D2 S2

boundary//

//��

This diagram commutes, which means that the different ways of

following the arrows round the diagram produce the same composite.

(g) Paths and loops—we will be studying these a lot.

• A path in X from a to b is a continuous map

γ : I −→ X

such that γ(0) = a and γ(1) = b.

• A loop at a in X is a path that starts and ends at a, i.e. a

continuous map

γ : I −→ X

such that γ(0) = γ(1) = a.

3.3 Paths and loops 22

(h) There is a type of continuous map called a “retraction” e.g.

R2 \ {0} −→ S1

—see Exercises #3.

Exercise 3.15. How many “different” maps S1 −→ S1 can you think of?

3.3 Paths and loops

Definition 3.16.

• A path from a to b (written a −→ b) in a space X is a continuous map

γ : I −→ X

such that γ(0) = a and γ(1) = b.

• A loop at a ∈ X is a continuous map

γ : I −→ X

such that γ(0) = γ(1) = a.

Paths tells us some things immediately e.g. in the following space

—is there a path from a to b?

Definition 3.17. Given a space X we can put an equivalence relation on it by

x ∼ y ⇐⇒ there is a path in X from x to y.

The equivalence classes are called the path components of X, written π0X.

X is called path-connected if there is only one path componenet; equivalently

∀x, y ∈ X ∃ a path in X from x to y.

Exercise 3.18. Is that really an equivalence relation?


Definition 3.19. Let X be a space.

1. Given a point a ∈ X we define the constant path at a by

csta : I −→ X

t 7→ a.

2. Given a path γ : a −→ b in X we define the reverse path γ∗ : b −→ a by

γ∗ : I −→ X

t 7→ γ(1− t).

We can check that γ ∗ (0) = γ(1) and γ ∗ (1) = γ(0).

3. Given paths aγ1−→ b

γ2−→ c in X we define the concatenation γ2 ∗ γ1 :

a −→ c by

(γ2 ∗ γ1)(t) =

γ1(2t) 0 ≤ t ≤ 1

2

γ2(2t− 1) 12 ≤ t ≤ 1

So

1. gives reflexivity a ∼ a,

2. gives symmetry a ∼ b ⇒ b ∼ a, and

3. gives transitivity a ∼ b and b ∼ c ⇒ a ∼ c.

—so it really did make sense for us to take “path-components” π0X.

π0 is a rather crude way of classifying spaces, for example it makes these three

spaces

all the same. We want a better test—instead of just saying “Is there a path

from a to b?” we want to say something like:


How many different paths are there from a to b?

For example these paths are “different”:

whereas these should count as “the same”:

So we need a way of saying when paths are “the same really”. Note that, as you

saw in Exercises #2, just having the same image is not subtle enough—going

round a circle once or twice gives the same image but should definitely count as

different paths. So in the next section we introduce the notion of homotopy.

4 Homotopy 25

4 Homotopy

Example 4.1. Imagine policemen searching an area that happens to include a

pond.

Each individual can get to the other side, but they’ll have to break their line in

the middle somewhere.

Consider two paths f, g : I −→ Y . We want to say what it means to “deform”

f into g . At a first guess we might say

∀x ∈ I we have a path f(x)px−→ g(x)

but that isn’t quite enough. We also need to say something like

“For each t ∈ I we have a path consisting of all the px(t)’s.”

4 Homotopy 26

The easiest way to do this is to think of mapping an entire square:

which is sort of what our diagram looked like anyway. If we demand that this

map is continuous it ensures that we didn’t try and deform anything past a hole.

More generally we can do this for any continuous maps f, g : X −→ Y . This is

the notion of homotopy.

Note that from now on whenever we say “map” we will mean continuous map.

Definition 4.2. Let X, Y be spaces and f, g : X −→ Y continuous maps.

Then a homotopy α : f ⇒ g or

X Y

f

��

g

CCα

��

is a continuous map

α : I ×X −→ Y

such that for all x ∈ X

α(0, x) = f(x) also written α(0, ) = f

α(1, x) = g(x) α(1, ) = g

Note in particular this means:

• for all x ∈ X we have a path px : f(x) −→ g(x) in Y given by

I −→ Y

t 7→ α(t, x)

i.e. px = α( , x), and

4 Homotopy 27

• for all t ∈ I we have a continuous map

αt : X −→ Y

x 7→ α(t, x).

We can think of this as continuously deforming f into g via all these intermediate

maps αt.

Definition 4.3. If there is a homotopy f ⇒ g we write f ' g and say f is

homotopic to g. We will show that this is an equivalence relation.

Example 4.4. This example makes precise the idea that the map from the

circle to itself doing nothing is “more or less the same as” the map that rotates

it by an angle φ. This is simplest to write down in polar coordinates, with the

points of the circle given as (1, θ).

We have two maps f, g : S1 −→ S1 where f is the identity and

g(1, θ) = (1, θ + φ).

We can now define a homotopy α : f ⇒ g by

α : I × S1 −→ S1

(t, (1, θ)) 7→ (1, θ + tφ)

Example 4.5. This example makes precise the idea that if we map the circle

into R2 it’s “more or less the same as” mapping it to a point, because we can

shrink the circle down to a point in the plane. Again, this is simplest in polar

coordinates.

We have two maps f, g : S1 −→ R2 where f is the embedding and g sends

everything to the origin, i.e. for all θ

f(1, θ) = (1, θ)

g(1, θ) = (0, 0)


α : I × S1 −→ R2

(t, (1, θ)) 7→ (1− t, θ)

Example 4.6. The “boring but important” example. For every continuous

map f : X −→ Y there’s an identity homotopy given by

I ×X −→ Y

(t, x) 7→ f(x) ∀t

4 Homotopy 28

Example 4.7. This example is about changing the “speed” we travel along a

path. Given a path f from a to b in X there’s a new one which we can think of

as “doing f twice as fast and then just sitting still at b for the rest of the time”.

Formally, this is given by

g(t) =

f(2t) 0 ≤ t ≤ 1

2

b 12 ≤ t ≤ 1


α : I × I −→ X

(t, s) 7→

f( 2s

2−t ) 0 ≤ s ≤ 2−t2

b 2−t2 ≤ s ≤ 1

Example 4.8. This example constructs a “reverse homotopy” very much like a

“reverse path”. Informally, we know that a homotopy f ⇒ g gives for each x a

path f(x) −→ g(x). We we might wonder if we can reverse all these paths and

form a homotopy g ⇒ f . Formally we do it like this. Given maps f, g : X −→ Y

and a homotopy α : f ⇒ g we define a homotopy α∗ : g ⇒ f by

α∗ : I ×X −→ Y

(t, x) 7→ α(1− t)

Example 4.9. This example constructs a “concatenation” of two homotopies

very much like a concatenation of paths. Suppose we have homotopies

X Y

f

��

h

CCg //

α��

β��

As above, we know informally that a homotopy α : f ⇒ g gives for each x a

path

α( , x) : f(x) −→ g(x).

Now here we also have for each x a path

β( , x) : g(x) −→ h(x).

The idea is to concatenate these paths to get a path

f(x) −→ g(x) −→ h(x)

4 Homotopy 29

for each x. Formally we define the homotopy by:

β ◦ α : I ×X −→ Y

(t, x) 7→

α(2t, x) 0 ≤ t ≤ 12

β(2t− 1, x) 12 ≤ t ≤ 1

and then we must check that it is continuous.

Proposition 4.10. For any maps f, g : X −→ Y write f ' g if f is homotopic

to g. Then ' is an equivalence relation.

Proof.

• Reflexivity is Example .

• Symmetry is Example .

• Transitivity is Example .2

Example 4.11. This example is a slightly more complicated way of sticking

homotopies together. Suppose this time we have two homotopies

X Y Z

f

��

g

CCα��

h

��

k

CCβ��

Informally we know that α gives us, for each t ∈ I a continuous map

α(t, ) : X −→ Y,

and β gives us

β(t, ) : Y −→ Z.

The idea is to compose these so that for each t we have a map

β(t, ) ◦ α(t, ) : X −→ Z,

and make this into a homotopy. Formally:

β ∗ α : I ×X −→ Z

(t, x) 7→ β(t, α(t, x)

)Now that we have defined “homotopy” we are finally ready to make precise

the idea of spaces being “more or less the same” given some squashing and/or

stretching. The idea is that we want continuous maps between our two spaces

that are not exactly inverse to one another, but only up to homotopy.

4 Homotopy 30

Definition 4.12. A continuous map f : X −→ Y is called a homotopy equiv-

alence i there is a continuous map g : Y −→ X such that

g ◦ f ' 1X and

f ◦ g ' 1Y

Then the spaces X and Y are called homotopy equivalent, written X ' Y .

Equivalently, X and Y are said to have the same homotopy type.

Example 4.13. The following spaces are homotopy equivalent:

So are the following spaces:

Proposition 4.14. Homotopy equivalence is an equivalence relation.

Proof. This is mostly left as an exercise. The only part that isn’t straightfor-

ward is transitivity. 2

We’ll see some more examples of homotopy equivalence in a minute, but one

particularly important kind of homotopy equivalence is when something is ho-

motopy equivalent to a point:

Definition 4.15. A space is called contractible if it has the homotopy type

of a point, that is, it is homotopy equivalent to a point.

Lemma 4.16. A space X is contractible if and only if there is a homotopy

csta ' idX for some point a ∈ X. Here csta is the map X −→ X that sends

everything to a.

Proof.

X is contractible if and only if we have maps f : X −→ ∗ and g : ∗ −→ X

together with homotopies gf ' idX and fg ' id∗.

4 Homotopy 31

Now fg = 1∗ so the second homotopy can be taken to be the identity. For the

first homotopy, first observe that a map g : ∗ −→ X simply picks out a point in

X, so the composite gf sends everything to this point g(∗). So it is csta, where

we write a = g(∗). Thus a homotopy gf ' idX is just a homotopy csta ' idX ,

hence the result. 2

Example 4.17. R2 is contractible. We take a to be the origin and construct a

homotopy idR2 ⇒ csta by

I × R2 −→ R2

(t, (x, y)) 7→ ((1− t)x, (1− t)y)

Exercise 4.18. Show that the following two spaces are homotopy equivalent:

Example 4.19. Suppose we have maps f, g : X −→ Y where Y is a convex

subset of a vector space. Then f ' g via

α : I ×X −→ Y

(t, x) 7→ (1− t).f(x) + t.g(x)

We can check that α(0, x) = f(x) and α(1, x) = g(x). This is called a straight

line homotopy. For example, any two paths in Rn are homotopic via a straight

line homotopy. Note that if we have two paths with the same endpoints, i.e.

f, g : a −→ b ∈ Rn then the straight line homotopy keeps the endpoints fixed :

α(t, 0) = (1− t).a+ t.a = a

α(t, 1) = (1− t).b+ t.b = b

This last example leads us into a special notion of homotopy for paths.

Definition 4.20. Let f, g : I −→ X be paths from a to b in X. Then f and g

are homotopic or equivalent if they are homotopic through paths from a to

b. That is, we have a map α : I × I −→ X such that

α(0, t) = f(t)

α(1, t) = g(t)

4 Homotopy 32

and for all s ∈ Iα(s, 0) = a

α(s, 1) = b

This is called a path homotopy. Similarly we have the notion of loop homo-

topy.

Definition 4.21. As before, path homotopy and loop homotopy are equivalence

relations. Given a path (or loop) f we write [f ] for the homotopy class of f .

Note that being path homotopic is a stronger condition than being just homo-

topic. From now on we will assume that if we are dealing with paths we are

using path homotopy. We are now ready for the next section, in which we will

make the homotopy classes of loops into a group: the fundamental group.

5 The fundamental group 33

5 The fundamental group

So far we have defined a way of going from the world of spaces to the world of

sets, using path components.

π0−→

This is useful, but rather crude, and ignores the differences between many very

different spaces. So now we’re going to be a bit more subtle about it, and map

into groups instead:

π1−→

We are going to define, for each space X, a group π1X called the “fundamental

group” of X, which captures much more information than π0X did. We have

already informally seen some fundamental groups:

π1(S1) = Z

π1(∗) = 0

π1(R2) = 0

π1( ) = free group on two generators (see Exercises #4)

First we’ll show how to construct the fundamental group formally, and then

we’ll ask ourselves some questions about this construction:

1. Is π1(X × Y ) related to π1(X) and π1(Y )?

2. What about for X ∪ Y ? What about for X/ ∼?

3. What happens to maps X −→ Y ?

4. What kind of “map” is π1 itself?

5. Can we go backwards: given a group can we make a space? Does every

group arise as the fundamental group of some space?

5.1 Definition 34

6. If two spaces have the same fundamental group, what does that tell us?

7. What does π1(X) = 0 tell us?

8. What do subgroups of π1(X) tell us? What about cosets? Homomor-

phisms?

5.1 Definition

Theorem 5.1 (Fundamental group construction). Write π1(X, a) for the set

of equivalence classes of loops based at a. This can be given the structure of a

group as follows:

• multplication: [f ].[g] = [f ∗ g]

• inverses: [f ]−1 = [f∗]

• identity: e = [csta]

Proof. We need to check that:

• the operation is well-defined,

• the operation is associative,

• e really acts as a unit, and

• every element has an inverse.

This is an exercise.

2

5.2 Dependence on basepoint 35

Definition 5.2. The group π1(X, a) is the fundamental group of X based

at a.

5.2 Dependence on basepoint

When we defined the fundamental group, we had to choose a basepoint. What

would have happened if we had chosen a different one? Obviously if we chose

one in a different connected component then we’d be a bit doomed e.g.

However if there is a path aγ−→ b then it will turn out okay:

—every loop at a corresponds to a loop at b via γ:

Proposition 5.3. Let γ be a path a −→ b in X. Then there is a group isomor-

phism given by

γ : π1(X, a) −→ π1(X, b)

[f ] 7→ [γ.f.γ∗]

Its inverse is given by γ∗.

Proof.

First we need to show that γ is a group homomorphism, i.e.

γ([f ].[g]) = γ[f ].γ[g]


that is

γfgγ∗ ' γfγ∗γgγ∗.

We can construct such a homotopy as indicated by the following diagram:

We now need to show that γ∗ is really inverse to γ, i.e.

f ' γ∗γfγ∗γ

and

f ' γγ∗fγγ∗.

This follows from the fact that γγ∗ ' cstb and similarly γ∗γ ' csta. 2

So now we know that “if X is path-connected then π1 does not depend on

the choice of basepoint”. Note however that π1(X, a) and π1(X, b) are not

canonically isomorphic—we had to choose a path a −→ b which gave us a group

isomorphism. A non-equivalent path might give us a different isomorphism.

The next thing we might ask is: what happens to maps? If we have a map

φ : X −→ Y do we get a group homomorphism π1X −→ π1Y ? And the answer

is yes. Any path γ : I −→ X becomes a path in Y by

Iγ−→ X

φ−→ Y.

If γ is a path from a to b then the new path will go from φ(a) to φ(b), so in

particular if γ is a loop then φ ◦ γ is also a loop:


We would like this to give us a homomorphism of fundamental groups as well;

we will have to be a bit careful about homotopy classes.

Proposition 5.4. Given a continuous map φ : X −→ Y , we get a group

homomorphism

φ∗ : π1(X, a) −→ π1(Y, φ(a))

[γ] 7→ [φ ◦ γ]

Proof.

We need to show that this map is well-defined, and that

φ∗([γ1].[γ2]) = φ∗[γ1].φ∗[γ2].

This is an exercise.

2

We would now like to show that this process satisfies all sorts of good properties.

This is best summed up using the notions of category and functor.

5.3 Categories 38

5.3 Categories

Definition 5.5. A category C consists of:

• a collection obC of objects, and

• for every pair X,Y ∈ obC, a collection C(X,Y ) of morphisms f : X −→ Y ,

equipped with

• for each X ∈ obC, an identity morphism 1X ∈ C(X,X), and

• for all X,Y, Z ∈ obC, a composition map

mXY Z : C(Y,Z)× C(X,Y ) −→ C(X,Z)

(g, f) 7→ g ◦ f

satisfying the following axioms:

• unit laws – given f : X −→ Y we have 1Y ◦ f = f = f ◦ 1X , and

• associtiavity – given Xf−→ Y

g−→ Zh−→W we have h◦(g◦f) = (h◦g)◦f .

A category is said to be small if obC and all of the C(X,Y ) are not just collections

but actually sets, and locally small if each C(X,Y ) is a set.

Here are some examples of categories.

Examples 5.6. Large categories of mathematical structures

You may or may not have met these mathematical structures before, but it

doesn’t particularly matter for now.

1. Set of sets and functions. This is in many ways the “prototype category”.

Set has many, many wonderful features that make it a good place to start

doing mathematics, and we would like to know what other categories have

these features.

2. Categories derived from or related to Set:

5.3 Categories 39

• Rel of sets and relations; a morphism A −→ B is a subset of A×Bgiving the pairs (a, b) such that a ∼ b. Given a relation on A and B,

and a relation on B and C we produce the “composite” relation on

A and C which has a ∼ c ⇐⇒ ∃b ∈ B such that a ∼ b and b ∼ c.

• Set∗ of pointed sets and basepoint-preserving functions. A pointed

set is just a set with a chosen element as its “basepoint”; the mor-

phisms are functions which send basepoints to basepoints.

3. Algebraic structures and structure-preserving maps:

• Grp of groups and group homomorphisms;

• Ab of abelian groups and group homomorphisms;

• Ring of rings and ring homomorphisms;

• Vectk of vector spaces over a chosen field k;

• R-Mod of R-modules and their homomorphisms, for some fixed ring

R.

• Field of fields and field homomorphisms. This actually turns out not

to be a very well-behaved category, and it is often better to restrict

to fields of a fixed characteristic.

4. Categories of topological spaces: these were among the earliest motivating

examples of categories. Using this framework gives very concise statements

of results in algebraic topology.

• Top of topological spaces and continuous maps; actually unlike Set

and Vect, Top is not a well-behaved category. For this reason we

often end up working in some related category of “better” spaces,

to eliminate various pathologies that can otherwise arise and ruin

everything.

• Haus of Hausdorff spaces and continuous maps;

• CHaus of compact Hausdorff spaces and continuous maps; this is a

much better category than Top.

• Top∗ of “based spaces”, that is topological spaces equipped with

a chosen point called the basepoint; morphisms are the continuous

maps that preserve the basepoint. We have to work with this cate-

gory when we do the fundamental group, because our loops have to

be based somewhere.

• Met of metric spaces and uniformly continuous maps;

• Htpy of topological spaces and homotopy classes of maps; this cate-

gory is sometimes preferable to Top because we are often interested

in things “up to homotopy”.

5.3 Categories 40

Examples 5.7. “Very” small categories

When categories are very small indeed we can quite simply and vividly draw

them on the page using points for objects and arrows for morphisms.

1. There is a category with one object and one morphism, which has to be

the identity. We can draw a picture of this category:

��x

1x

This category is often referred to as 1.

2. There is an empty category. It has no objects and no morphisms, so must

surely be the “tiniest” and most stupid category. Still, it’s not completely

worthless and is often called 0, and after all, the number 0 is very far from

useless.

3. There is a category that we might draw like this:

· ·//

which means there are two objects and one morphism between them, as

shown. Really this category has three morphisms and could be drawn like

this:

//��x

1x

��y

1y

4. Here’s a category with some non-trivial composition:

· · ·// //

5. There is a category defined by the commutative square:

a b

c d

f1 //

f2

��

g1

��g2// .

Saying the square commutes means that “going round the square in dif-

ferent ways gives the same answer”, that is, that the two composites are

equal:

f2 ◦ f1 = g2 ◦ g2.

5.3 Categories 41

Examples 5.8. Curious categories

The following examples are a bit curious and might surprise you a bit. But

they’re a central part of category theory and yield all sorts of interesting con-

structions.

1. Every set “is” a category with only identity morphisms.

2. Every group “is” a category with only one object. That is to say we can

express a group G as a category:

• it has only one object, which we might call ∗;

• the morphisms ∗ −→ ∗ are precisely all the elements of G;

• composition is group multiplication, and the identity 1x in the cate-

gory is the unit element of the group G.

In the resulting category every morphism is invertible; the next example

is more general.

3. A category with only one object is precisely a monoid.

4. A category in which every morphism is invertible is a groupoid.

5. We can generate a category by the diagram

��x

f

where f : x −→ x is not supposed to be the identity. The result the

additive monoid N.

6. We could do something similar but impose the relation f ◦ f = 1. This

would turn our category into the one-object category corresponding to the

cyclic group of order 2.

7. Every poset is a category. Given a poset P with ordering ≤, we make

a category with objects the elements of P and precisely one morphism

x −→ y when x ≤ y, and none otherwise.

8. There is a curious category Mat of matrices. The objects are the natural

numbers; the morphisms n −→ m are all the n × m matrices (in some

fixed field, k, say). Composition is given by matrix multiplication.

5.3 Categories 42

One of the basic principles of Category Theory is that we should study struc-

tures together with their structure-preserving maps. We should look at the “to-

tality” of these structures and their maps, rather than just looking at individual

structures in isolation. The structure-preserving maps of categories are called

functors.

Definition 5.9. Let C and D be categories. A functor F : C −→ D associates:

• to every object X ∈ C an object FX ∈ D, and

• to every morphism f ∈ C(X,Y ) a morphism Ff ∈ D(FX,FY )

such that

• for all morphisms Xf−→ Y

g−→ Z, F (g ◦ f) = Fg ◦ Ff , and

• for all X ∈ C, F (1X) = 1FX .

The conditions in this definition are generally referred to as “functoriality”

conditions, and should distinctly remind you of the axioms for a group homo-

morphism.

Examples 5.10. Functors on other mathematical structures expressed

as categories

1. If C and D are groups (expressed as one-object categories) then a functor

C −→ D is precisely a group homomorphism.

2. If C and D are monoids, that is, one-object categories, then a functor

C −→ D is precisely a monoid morphism.

3. If C and D are posets (expressed as categories) then a functor C −→ D is

precisely an order-preserving map.

Examples 5.11. Forgetful functors

Forgetful functors forget all or part of the structure in question.

1. There is a forgetful functor

Gp −→ Set

which sends a group to its underlying set; the action on morphisms is

induced in the obvious way.

5.3 Categories 43

2. Similarly there are forgetful functors

Ring −→ Set

Vect −→ Set

Ab −→ Set

Top −→ Set

Mnd −→ Set

Poset −→ Set

Set∗ −→ Set

and many more.

3. Here are some forgetful functors that only forget part of the structure:

Ring −→ Ab

Ring −→ Mnd

Vect −→ Ab

Top∗ −→ Top

4. Here are some examples of functors that just forget some property :

Ab −→ Gp

Haus −→ Top

Examples 5.12. Free functors

Free functors go in the opposite direction of forgetful functors: they take some-

thing with little or no structure, and create something with more structure. The

new extra structure is added in “freely”.

1. There is a free functor

Set −→Mnd

that takes a set X and produces the free monoid on it. The result is a set

X∗ which consists of “words in X”, that is, finite lists of elements of X,

possibly with repetitions.

2. We can do something similar but harder to get a free functor

Set −→ Gp.

The reason it’s harder is that we have to make sure everything has an

inverse and this complicates matters rather.

5.3 Categories 44

3. Similarly we have a free functor

Set −→ Ring

but we also have a more subtle free functor

Mnd −→ Ring.

Examples 5.13. Mathematical constructions as functors

1. If G is a group expressed as a one-object category, a functor G −→ Set is

a set with a G-action, also known as a G-set.

2. If G is a group expressed as a one-object category, a functor G −→ Vect

is a representation of G.

3. We also get functors for the constructions of fundamental groups, higher

homotopy groups, homology, cohomology, geometric realisation, sheaves,

compactification, parallel transport, topological quantum field theory. . .

The fundamental group functor

This brings us to the functor that launched us into this whole discussion about

categories in the first place. That is, I launched into it because I wanted to

be able to say that the fundamental group construction gives us a functor as

follows.

Theorem 5.14. There is a functor π1 : Top∗ −→ Gp given by

• on objects: π1(X,x) is the fundamental group,

• on morphisms: given a morphism (X,x)φ−→ (Y, y) we put π1(φ) = φ∗.

Proof.

We need to check that this satisfies the axioms for a functor. First, given maps

(X,x)φ−→ (Y, y)

ψ−→ (Z, z)

we need to check that

π1(ψ ◦ φ) = π1(ψ) ◦ π1(φ)

i.e.

(ψ ◦ φ)∗ = ψ∗ ◦ φ∗.

5.3 Categories 45

We also need to check that for all (X,x) we have

π1(1(X,x)) = 1π1(X,x).

This is an exercise

2

Remark 5.15. We have seen something about the fundamental group being

independent of the choice of basepoint. Does this mean we have a functor

Top −→ Gp?

We would now like to know that “the fundamental group is homeomorphism

invariant” that is, if two spaces are homeomorphic then they have the same

fundamental group (or rather, isomorphic ones). This now turns out to be

immediate from the fact that π1 is a functor, because all functors preserve

isomorphisms:

Lemma 5.16. Let F : C −→ D be a functor and f : X −→ Y ∈ C an isomor-

phism with inverse g : Y −→ X, so fg = 1Y and gf = 1X . Then Ff is an

isomorphism.

Proof. See Exercises #6. 2

Corollary 5.17. π1 is homeomorphism invariant. More precisely, if

φ : (X,x) −→ (Y, y)

is a homeomorphism then

φ∗ : π1(X,x) −→ π1(Y, y)

is a group isomorphism.

5.4 Homotopy invariance 46

It is crucial that π1 be homeomorphism invariant, otherwise it would be a rather

stupid invariant of topological spaces. However, we would also like it to be

homotopy invariant and that is the subject of the next section.

5.4 Homotopy invariance

A property or calculation is said to be “homotopy invariant” if any homotopy

equivalent spaces give the same answer. Of course, we’ll need to say this more

precisely every time we come to it. We start with π0.

Proposition 5.18. π0 is homotopy invariant. That is, if X and Y are homo-

topy equivalent then we have

π0X ∼= π0Y.

In fact we prove something stronger—that a homotopy equivalence f induces

an isomorphism on path components.

Proof.

We have maps

X Y

f

!!

g

aa

and homotopies

α : gf∼

=⇒ 1X

β : fg∼

=⇒ 1Y

In particular we have

∀ x ∈ X a path gfxαx−→ x, and

∀ y ∈ Y a path fgyβy−→ y.

We now show that f induces an isomorphism on path components.

• injective: we need to show

[fx1] = [fx2] =⇒ [x1] = [x2]

i.e.

fx1 ∼ fx2 =⇒ x1 ∼ x2

5.4 Homotopy invariance 47

i.e. if there is a path fx1γ−→ fx2 then there was already a path x1 −→ x2.

We construct this path by:

x1α∗x−→−→ gfx1 −→ gfx2

αx−→ x2

• surjective: we need to show

∀y ∈ Y ∃x ∈ X s.t.fx ∼ y

that is, there is a path fx −→ y. We do this by:

fgy −→ y

with x = gy.

2

We now prove the analogous result for π1.

Proposition 5.19. π1 is homotopy invariant. That is, if X and Y are homo-

topy equivalent then π1(X) ∼= π1(Y ).

More precisely, given a homotopy equivalence f : X −→ Y , the map

π1f : π1(X,x) −→ π1(Y, f(x))

is a group isomorphism.

Proof. (Sketch)

We have homotopies α and β as in the previous proof. We now construct a left

inverse to π1f as

π1(Y, f(x))π1g−→ π1(X, gf(x))

αx−→ π1(X,x).

and similarly we construct a right inverse to π1g. This is enough as ... 2

Definition 5.20. We call a space X simply-connected if it is path-connected

and π1(X) = 0.

Intuitively, this means that the space has “no holes”—its fundamental group is

trivial so every loop is homotopic to the constant loop at a point. We really

need to start calculating some fundamental groups that aren’t trivial! The best

“building block” to start with is the circle, but actually it’s going to be easier

for us to look at that using the theory of covering spaces. However, before

we move onto that section it’s worth getting a rough idea how we want to think

about the circle, so that when we go onto covering spaces properly we have some

pictures in our head.

5.5 The circle 48

5.5 The circle

The fundamental group of the circle is Z, but we haven’t proved it yet. This

result is intuitively clear, but the proof involves a lot of technicalities. Here are

some things that follow from knowing this result.

The fundamental theorem of algebra

Every non-constant polynomial with coefficients in C has a root in C.

The Brouwer fixed point theorem

Every continuous map D2 −→ D2 has a fixed point.

The Borsuk-Ulam theorem in dimension 2

For every continuous map f : S2 −→ R2 there is a pair of antipodal points

x,−x ∈ S2 such that f(x) = f(−x).

Corollary

If S2 is expressed as the union of three closed sets then at least one must contain

a pair of antipodal points.

Question 5.21. When you’re walking up the stairs in the Hicks Building, is it

easier to count how many times you’ve gone round, or to look at the signs to

see what floor you’ve reached?

5.5 The circle 49

Sketch proof that π1(S1) = Z

We get a bit “confused” about going round our circle multiple times, so let’s

“unwind” our circle:

R

S1

p

��

Now, instead of counting how many times we went round, we count count how

many “floors” up (or down) we’ve gone. Moreover we can choose the map p so

that the preimages of the basepoint x are precisely the integers. So the number

of times we’ve gone round the circle is precisely counted by what number we

land on in the spiral version. This is very convenient! Now the question is, can

we somehow:

1. turn this into a proof that π1(S1) = Z, and

2. generalise this idea for other spaces?

We are looking for an isomorphism

[loops in S1]∼−→ p−1(x)

and we’ll do this in steps. The idea is that loops downstairs correspond to paths

upstairs, so we can fix the starting point of the path and “see where the other

end of is”. So we start by choosing a “lift” of the basepoint x i.e. x ∈ p−1(x).

1. Every loop γ “downstairs” lifts to a unique path γ “upstairs” starting at

x i.e. such that

p ◦ γ = γ

and γ(0) = x.

5.5 The circle 50

2. So we get a function

loops in S1 −→ p−1(x)

γ 7→ γ(1)

which is a good start, but we want a function from homotopy classes of

loops to p−1(x). So we need to check that this function is homotopy

invariant.

3. Now, any loop homotopy downstairs lifts to a path homotopy upstairs, so

we do indeed get a function

[loops in S1] −→ p−1(x).

4. We check that it is surjective, that is, that every pre-image of x upstairs

gets “hit”—this is true because R is path-connected. So given any a ∈p−1(x) we just taken a path x

f−→ a in R, and applying p must give a

path p(x) −→ p(a) in S1, i.e. a loop γ at x such that γ(1) = b.

5. We check that it is injective, that is

γ1(1) = γ2(1) =⇒ γ1 ' γ2.

This is true because R is simply-connected, so all paths with the same

endpoints are homotopic. Thus we have a homotopy γ1 ' γ2 and this

must map via p to a homotopy γ1 ' γ2.

This sketch proof has highlighted several of the important properties of covering

spaces that we are going to look at:

1. path lifting,

2. the lifting correspondence, and

3. homotopy lifting.

The construction we did where we “unwound” all the loops around the circle

produced was is called the “universal covering space”. We could have just

unwound some of the loops and not others e.g.

5.5 The circle 51

In this case in our proof we would still have surjectivity (4) but not injectiv-

ity (5)—some different loops downstairs would lift to the same path upstairs.

Which? Another important question is: what would the induced map on fun-

damental groups be?

These are all important questions we will consider in the next section, on cov-

ering spaces.


6 Covering spaces

So far we don’t have many ways of calculating fundamental groups. Covering

spaces give us some ways of doing this, but also much more—it’s a very specific

manifestation of the correspondence between

algebra and topology

fundamental group covering spaces

Recall that we tried to calculate the fundamental group of S1 by thinking about

a helix mapping onto the circle:

This is a good prototype example of a covering space. We are “unwinding” the

loops in the space.

Here are some more covering spaces of S1:


And some other covering spaces—can you see how the first two cover the space

?

Here are some of the main ideas about covering spaces.

• A covering space wraps itself evenly around the base space.

• The fibre over any point has n points in it, for some fixed n. (It might be

infinite.) We would then call this an “n-sheeted” cover.

• We get neighbourhoods whose pre-image splits into a disjoint union of

homeomorphic copies.

• There are various symmetries interchanging the “sheets”.

The above are all geometric features, and we will see how these correspond to

algebraic features of the fundamental group:

• subgroups, and

• group actions on sets or spaces.

We will ask ourselves questions like: can we find all the covering spaces of a

space? How do the geometric and algebraic features correspond?

The main answer is a correspondence

connected ←→ subgroups of the

covering spaces fundamental group

6.1 Definitions 54

This might remind you of Galois theory, in which you have a correspondence

between

field ←→ subgroups of the

extensions Galois group

A covering space sort of “unloops” some of the loops in your space. The cor-

responding subgroup of the fundamental group consists of the loops that you

haven’t unlooped yet.

Example 6.1. You walk around a tree and you think you’re back to where you

started but you have a layer of mud on your shoes.

Example 6.2. Pooh, Piglet and the Heffalump.

So as your covering space gets bigger and bigger, your corresponding subgroup

gets smaller and smaller, and finally you get the universal cover correspond-

ing to the trivial subgroup. This is where everything has been maximally

unwound—you no longer have any loops left.

Exercise 6.3. Look at page 58 of Hatcher, where he gives a page of lovely

diagrams of some covering spaces of (he refers to this space as S1∨S1,

a piece of notation we’ll get to later).

6.1 Definitions

Recall that a neighbourhood of a point x ∈ X is an open set containing x.

Definition 6.4. A covering space of a space X is a space X equipped with

a map

X

X

p

��

with the following property: every x ∈ X has a neighbourhood U for which

p−1(U) is a disjoint union of open sets, each of which is mapped by p homeo-

morphically to U .

6.1 Definitions 55

Example 6.5.

We can also say this in terms of open covers of the space X.

Definition 6.6. An open cover of a space X is a collection of open sets {Ui}such that ⋃

i

Ui = X.

Definition 6.7. A covering space of a space X is a space X together with a

map

X

X

p

��

where p evenly covers X, that is, there is an open cover {Ui} of X such that

for each i, p−1(Ui) is a disjoint union of open sets of X, each mapped by p

homeomorphically to Ui.

6.1 Definitions 56

Example 6.8.

Example 6.9. We can cover the torus with R2.

Example 6.10. We define a subspace of R3 called a helicoid surface:

{(rcos2πt, rsin2πt, t) | r ∈ (0,∞), t ∈ R}

This covers R2 \ {0}, with the covering map p given by:

(x, y, z) 7→ (x, y)

Example 6.11. The identity always gives us a covering space, trivially. This

amounts to “unwinding no loops”, so should correspond to the biggest possible

subgroup of the fundamental group—the whole thing.

Exercise 6.12. Can you think of a (non-trivial) cover of RP 2? S2?

6.1 Definitions 57

Proposition 6.13. “Crucial properties of covering spaces”

Let

X

X

p��

be a covering space.

1. Path lifting

Given any path f in X starting at x

and any lift x of x, i.e. p(x) = x

there is a unique lift of f to a path f in X starting at x

i.e. a unique path If−→ X such that f(0) = x

and the following diagram commutes:

I X

Xf

99

p

��

f//

2. Homotopy lifting

Given any homotopy α : I ×A −→ X starting at φ : A −→ X

and any lift φ of φ i.e.

A X

Xφ

99

p

��

φ//

there is a unique lift of α to a homotopy α starting at φ

i.e. a unique homotopy I ×A α−→ X such that α(0, ) = φ

and the following diagram commutes:

I ×A X

Xα

99

p

��α

//

3. Lifting correspondence

Given any lift x of x we have a map

π1(X,x) −→ p−1(x)

[γ] 7→ γ(1)

6.1 Definitions 58

and furthermore

• if X is path-connected then this map is surjective, and

• if X is simply connected then this map is also injective.

4. Sheets

If X is path-connected then the cardinality of p−1(x) is constant.

This is called the number of sheets of the cover.

5. Important consequence

The group π1(X, x) embeds as a subgroup of π1(X,x) via the homomor-

phism

π1(X, x)

π1(X,x)

π1(p)

��

i.e. π1(p) is injective.

Idea: π1(X, x) is “smaller” because we have “unwound some loops”.

Proof.

1. Path lifting

We start with a path f in X and construct a lift as required.

Idea:

If p were a homeomorphism it would be easy, but p is only “locally a

homeomorphism”—on the Uα’s exhibiting the covering space—so we edge our

way along, staying in one Uα at a time

We use

• the definition of covering space, and

• compactness of I—“every cover has a finite subcover”.

6.1 Definitions 59

Step 1

By the definition of covering space we have a cover {Uα} of X such that for all

α, p−1(Uα) is a disjoint union of open sets in X, each of which maps homeo-

morphically to Uα under p.

Now f : I −→ X is continuous so {f−1(Uα)} is an open cover of I.

So for all t ∈ I, there is some α with t ∈ f−1(Uα)

and there is an open ball◦Br (t) ⊆ f−1(Uα).

This gives a cover of I, so by compactness of I there is a finite subcover

[a0, b0) , (a1, b1) , (a2, b2) , . . . , (am, bm]

where a0 = 0, bm = 1 and without loss of generality

a0 < a1 < b0 < a2 < b1 < · · · < bm

i.e.

Now put

ti =ai + bi−1

2∀ 1 ≤ i ≤ m

So we have

0 = t0 < t1 < · · · < tm < tm+1 = 1

such that for each i, f [ti, ti+1] ⊂ Ui, say, where Ui ∈ {Uα}.

Step 2

We now lift the path f on each of these sub-intervals one at a time.

Consider the first “patch” [0, t1], with f [0, t1] ⊂ U0.

We have f(0) = x and we fix f(0) = x.

Now x ∈ p−1(U0) so must lie in one homeomorphic copy of U0

—we call it U0.

We can now define f on [0, t1] by

[0, t1]f−→ U0

p−1

−→ U0 ⊂ X.

6.1 Definitions 60

We can then proceed by induction—we repeat this process for each patch

[t1, t2] , [t2, t3] , . . . , [tm, 1].

This constructs a lift f of f .

Step 3

Finally we must check that this lift is unique, by essentially the same argument.

Suppose we have lifts f , f ′ : I −→ X, so

pf = pf ′ = f , and

f(0) = f ′(0) = x.

As above, we choose a partition

0 = t0 < t1 < · · · < tm+1 = 1

such that for each i, f [ti, ti+1] ⊂ some Ui.

Now we “edge our way along”:

f is continuous and [0, t1] is connected, so f [0, t1] is connected in X

so f [0, t1] lies in one homeomorphic copy of U0, say U0.

Furthermore f ′[0, t1] must lie in the same copy as f(0) = f ′(0).

But p gives a bijection U0 −→ U0

and we also know pf = pf ′

so we must have f = f ′ on [0, t1].

We can then proceed by induction and “patch” all the way along.

2. Homotopy lifting

This is basically similar to path lifting but slightly harder.

3. Lifting correspondence

We need to show that this is well-defined, i.e.

γ1 ∼ γ2 =⇒ γ1 = γ2.

This follows from the homotopy lifting property:

6.2 The classification of covering spaces 61

Given a homotopy α : γ1 ' γ2 fixing the endpoints, we get a lift

α : γ1 ' γ2

which must also fix the endpoints.

The last part is on Exercises #8.

4. Sheets

Idea: the cardinality of the pre-image must be constant on each Uα and we can

then “patch” since it must agree on intersections.

5. Important consequence

We want to show that π1(p) is injective, i.e.

[pγ1] = [pγ2] =⇒ [γ1] = [γ2]

i.e.

pγ1 ' pγ2 =⇒ γ1 ' γ2.

This follows from the homotopy lifting property. 2

6.2 The classification of covering spaces

Slogan: “Connected covering spaces of X correspond to subgroups of

π1(X).”

To make this more precise:

• we must consider covering spaces together with the covering map

(X, x)

(X,x)

p

��

• we must think of two covering spaces as “the same” if there is a homeo-


morphism f making the following diagram commute

(X, x) (X ′, x′)

(X,x)

f //

p

��p′

��

• we must restrict to “nice” spaces: X must be

– path-connected,

– locally path-connected, and

– semi-locally simply-connected.

We won’t really go into what this last part means, but for example we want to

exclude the Hawaiian earring. Here are the definitions, for completeness:

Definition 6.14. A space X is called locally path-connected if for any

point x ∈ X and any open set U containing x, there is a path-connected open

set V ⊆ U also containing x.

Definition 6.15. A space X is called semi-locally simply-connected if for

any point x ∈ X there is an open set U containing x in which every loop is

nullhomotopic.

Examples 6.16.

1. The “quasi-circle” is path-connected but not locally path-connected.

2. Let X ⊂ R2 be the space formed by taking each point (0, 0) and ( 1n , 0) for

n ∈ N and joining it to the point (0, 1). Then X is path-connected but

not locally path-connected.


3. The Hawaiian earring is locally path-connected but not semi-locally sim-

ply connected.

Theorem 6.17. Let X be path-connected, locally path connected and semi-

locally simply connected. Then there is a bijection between

• isomorphism classes of connected covering spaces

(X, x)

(X,x)

p

��

and

• subgroups of π1(X,x).

The correspondence is given by

(X, x)

(X,x)

p

��←→ Im π1p

Remark 6.18. Note that π1p is a group homomorphism

π1(X, x) −→ π1(X,x).

We know it is injective (by part 5 of Proposition 6.13), so by the First Isomor-

phism Theorem we have

However π1(X, x) could embed as a subgroup of π1(X,x) in many different

ways. For example when we consider covering spaces of the circle, we consider


all the different embeddings Z −→ Z. So in the correspondence above, we say

“Im π1p” and not just π1(X, x).

Example 6.19. Connected covering spaces of S1.

We have seen the following covering connected covering spaces of S1:

• For each n ∈ N

S1 pn−→ S1

(1, θ) 7→ (1, nθ)

and

•R p−→ S1

θ 7→ (1, θ)

We know π1(S1) = Z so we should get one connected covering space for each

subgroup of Z, i.e. for each nZ for n ∈ N including 0.

Now, if we consider the covering space

S1

S1

pn

��we get

and Im π1pn is nZ ⊆ Z.

N.B. This shows why we can’t just get a correspondence using π1 of the covering


space—in this case we would just keep on getting Z.

Finally for R

S1

p

��

we get

giving the subgroup:

This gives all subgroups of Z so we must have found all connected covering

spaces of S1.

Example 6.20. Connected covering spaces of RP 2.

We can cover RP 2 by S2 by identifying antipodal points:

S2

RP 2

1

��

x

[x] = {x,−x}

_

1

��

We will later see that π1(RP 2) = Z2.

So we get

giving the subgroup:

Z2 has no non-trivial subgroups, so this must be the only connected covering

space of RP 2. (Of course, we also have the covering of RP 2 by itself, the trivial

cover, giving the subgroup Z2.)

Note that there are many disconnected covering spaces.


Example 6.21. Connected covering spaces of S2.

We know that π1(S2) = 0, which has no non-trivial subgroups.

So the only connected covering space of S2 is the trivial one

S2

S2

1

��

Example 6.22. Conected covering spaces of RP 2 × RP 2.

We will see that π1 preserves products, so

π(RP 2 × RP 2) ∼= π1(RP 2)× π1(RP 2)

∼= Z2 × Z2

which has subgroups

• 0 and Z2 × Z2, and

• subgroups of order 2: 〈(0, 1)〉, 〈(1, 0)〉, 〈(1, 1)〉(each isomorphic to Z2)

We have a covering space

S2 × S2

RP 2 × RP 2��

(α, β)

([α], [β])

_

��

and

π1(S2 × S2) ∼= π1(S2)× π1(S2) ∼= 0

so this corresponds to the trivial subgroup.

We also have

RP 2 × S2

RP 2 × RP 2��

([α], β)

([α], [β])

_

��

and

S2 × RP 2

RP 2 × RP 2��

(α, [β])

([α], [β])

_

��


corresponding to 〈(1, 0)〉 and 〈(0, 1)〉.

Finally we seek the covering space corresponding to 〈(1, 1)〉.

It may help at this point to think of RP 2 × S2 as a quotient of S2 × S2 by the

relation

(α, β) ∼ (−α, β)

and similarly S2 × RP 2 is a quotient of S2 × S2 by the relation

(α, β) ∼ (α,−β).

It is now clearer what the last covering space should be

X = (S2 × S2)/ ∼

where ∼ is the equivalence relation defined by

(α, β) ∼ (−α,−β).

Note that this is not the same as

RP 2 × RP 2 ∼= S2/ ∼ ×S2/ ∼

∼= (S2 × S2)/ ∼

with the equivalence relation

(α, β) ∼ (±α,±β).

We have covering map

X

RP 2 × RP 2��

[(α, β)]

([α], [β])

_

��

and this covering space corresponds to 〈(1, 1)〉.

This gives us all the connected covering spaces. Note that we have a diagram

of quotients:

S2 × S2

S2 × RP 2 X RP 2 × S2

RP 2 × RP 2

��

��

4 sheets

2 sheets

1 sheet

Question 6.23. Are the upper level maps covering maps as well?

6.3 Universal covering spaces 68

6.3 Universal covering spaces

A consequence of the classification theorem is that we must have a connected

covering space corresponding to the trivial subgroup 0 (provided X is a “nice”

space). Thus we get a simply-connected covering space, that is, a connected

covering space (X, x) with π1(X, x) = 0. This is called the universal covering

space.

Remarks 6.24.

1. This is the covering space in which “all loops have been unwound”.

Or: we are stupid forever, and never notice that we’ve come back to where

we started!

2. This has a “universal property”—it is a covering space of every other

covering space.

3. Actually if we really proved the classification theorem we would start by

constructing the universal cover, and then quotient it to make all the other

covers, rather like in the RP 2 × RP 2 example.

6.4 Covering space constructions

1. Universal covering spaces

We want to “unwind” all loops.

We start with (X,x), and we make a new space by taking the points of X

together with the path we took to get there.

More precisely, we can put a topology on

{[γ] | γ is a path in X starting at x}.

For example on S1 if we go round the circle once we will no longer consider

ourselves “back where we started” because that path is not homotopic to the

constant path at x.

2. Non-universal covering spaces

Given a covering space

(X, x)

(X,x)

p

��

6.4 Covering space constructions 69

this corresponds to the subgroup of π1(X,x) of “those loops that are still loops

upstairs”.

—Every loop upstairs maps to a loop downstairs, but some loops downstairs

lift to paths upstairs that are not loops.

We make non-universal covering spaces by quotienting the universal covering

space just to make those loops we want still to be loops in X.

Comments on the classification of covering spaces

1. In fact the most rigorous statement of the classification theorem involves

a category of covering spaces and a category of subgroups of π1(X,x). We

then have an equivalence of these two categories.

2. If (X, x) corresponds to the subgroup H ⊆ G = π1(X,x) then the number

of sheets of the cover is the index of H in G i.e. |G||H| .

This can be seen in Example 6.22, the covering spaces of RP 2 × RP 2.

Wedge sums

Definition 6.25. Given based spaces (X,x) and (Y, y), their wedge sum is

defined by “glueing X and Y at the basepoint”.

(X,x) ∨ (Y, y) = (X∐

Y )/x ∼ y.

When it doesn’t matter what the basepoints are we just write X ∨ Y and un-

derstand that we have glued at one point.

For example

• S1 ∨ S1 is the familiar .

• S1 ∨ S2 is .

Here is an informal “recipe” for making the universal covering space of (X,x) ∨ (Y, y).


1. Make the universal covers X of X and Y of Y .

2. Find all the preimages of the basepoint x in X, and glue on a copy of Y

at each one.

3. Find all the preimages of the basepoint y in each copy of Y , and glue on

a copy of X at each one.

4. Iterate. . .

Examples 6.26.

• S1 ∨ S2

• RP 2 ∨ RP 2

• S1 ∨ S1


Non-example 6.27. Let X be the space given by glueing two copies of R at

the origin.

We have a map to S1 ∨ S1

but this is not a covering space.

We can see this by looking at neighbourhoods of the basepoint

—this shows that in general to make a universal cover of X ∨ Y we can’t just

stick X to Y , but rather we have to stick a copy of Y to every preimage of the

basepoint in X, and then repeat for Y and then iterate.

Non-example 6.28.

This is not a covering space; again we can see this by looking at neighbourhoods

of the basepoint.


Non-example 6.29. We have seen how covers by

“looping the middle circle around twice” so that the outer two circles land on

top of each other. But why can’t we do it by just folding the whole thing in half,

so that the middle circle folds in half, and then glueing the two points where we

folded?

This is an interesting non-example because it does “cover the space evenly” but

still isn’t a covering space. Why?

Non-example 6.30.

[0, 1)

S1��

t

(cos2πt, sin2πt)

_

��

This does have the property that the cardinality of p−1(x) is constant, but this

is not a covering space; we can see this by looking at neighbourhoods of the

point (1, 0).

7 Van Kampen’s Theorem 73

7 Van Kampen’s Theorem

This theorem is about sticking spaces together and finding the fundamental

group of the resulting space. For example

• S1 ∨ S1

• S1 ∨ · · · ∨ S1

• −→

• MB glue MB −→ KB

• MB glue −→ RP 2

Key “building block” examples

1. Take any space X, cut out a disc and glue “handle”.

2. Take any space X, cut out a disc and glue a Mobius Band.

3. Take a wedge of n circles labelled a1, . . . , an

and a “word” in the letters a1, . . . , an e.g. a23a5a41,

then glue a disc onto the loops following this word.

Key consequences

• (1) and (2) enable us to classify all 2-dimensional surfaces.

• (3) enables us to construct a space with any given fundamental group we

like.

7.1 Free product of groups 74

How we use the theorem

If we have glued X and Y , we want π1 of the result in terms of π1(X) and

π1(Y ).

Simplest example

Glueing at a point:

π1(X ∨ Y ) ∼= π1(X) ∗ π1(Y )

the “free product of groups”.

More generally: this also works if we glue along something simply-connected.

Idea: any loop in X ∨ Y can be decomposed as a concatenation of loops in X

and loops in Y (alternating).

Next simplest example

Glueing at something with non-trivial loops

e.g. A = X ∪ Y with X ∩ Y path-connected but with some non-trivial loops.

Then we have counted the loops in X ∩ Y twice, and we will need to quotient

out by something. So we get

π1(A) ∼= π1(X) ∗ π1(Y )�something

7.1 Free product of groups

Given groups A and B (not necessarily abelian)

we form the free product A ∗B.

Note that this is very different from the Cartesian product A × B in which

elements of A commute with elements of B. In A ∗ B this does not happen

(except for identities).

Idea

We try to form a disjoint union, but then we are lacking a way to multiply

elements of A with elements of B—so we throw those in.

7.1 Free product of groups 75

Analogy

Consider loops in S1 ∨ S1 .

A loop consists of a finite concatenation of a’s and b’s and their inverses. Also

if we have several a’s in a row, we can just write them as ak.

—This is exactly the idea of the free product.

Note for the categorically minded

This is a coproduct in the category Gp.

Definition 7.1. [see also Hatcher p.41]

Let A1 and A2 be groups.

The free product A1 ∗A2 is defined as follows.

• Its elements are words x1x2 · · ·xm of finite length m ≥ 0 where

– each xi is in A1 or A2 but is not the identity, and

– adjacent letters are in different groups.

• Multiplication is by juxtaposition:

(x1 · · ·xm)(y1 · · · yn) = x1 · · ·xmy1 · · · yn

but if xm and y1 are in the same group they are multiplied to produce

a single letter, and if this is the identity it is cancelled out. In this case

xm−1 and y2 must be combined or cancelled if they are in the same group,

and so on until we have a valid word.

• The identity is the “empty word” of length 0.

• The inverse of x1x2 · · ·xm is x−1m · · ·x−12 x−11 .

Remarks

1. Actually checking this is a group is quite hard (especially associativity).

2. There is an n-fold version A1 ∗A2 ∗ · · · ∗An.

7.2 Glueing spaces 76

Example 7.2. Z ∗Z is the free group of two generators 〈a, b〉 and is the funda-

mental group of S1 ∨ S1.

Note that Z is the free group on one generator 〈a〉, but Z is a group under

addition and 〈a〉 is a group under multiplication. We have

{Z,+} ∼= {〈a〉,×}

n 7→ an

Z ∗ Z should not be confused with Z × Z, which is the free abelian group on

two generators, and the fundamental group of S1 × S1, the torus.

7.2 Glueing spaces

First consider an ordinary union situation A = X ∪ Y . We have the following

commutative diagram of inclusion maps.

X ∩ Y

X

Y

X ∪ Y

f::

g $$

$$::

This is not the same as the disjoint union X∐Y because we have glued along

X ∪ Y . We can express this as

X∐

Y�∼

where ∼ is the equivalence relation generated by

∀ a ∈ X ∩ Y, f(a) ∼ g(a)

or for based spaces

X ∨ Y�∼as long as the basepoint is in the intersection.

Example 7.3. Mobius Band with disc glued on at boundary

The resulting space can be expressed as MB ∪D2 via

S1

MB

D2

MB ∪D2 ∼=

f ::

g $$

$$

::


where f and g are the inclusion maps of S1 into the boundary of MB and D2

respectively.

Example 7.4. Two Mobius Bands glued along their boundary.

S1

MB

MB

MB ∪MB ∼=

f ::

g $$

$$::

where f and g are both the inclusion map of S1 into the boundary of MB.

Example 7.5. Glueing two discs to make a sphere

S1

D2

D2

D2 ∪D2 ∼= S2

f ::

g $$

$$

::

Example 7.6. Glueing two intervals to make a circle

S0

I = D1

I = D1

S2

f ::

g $$

$$::

It is still true that every loop in X ∪Y can be decomposed as a concatenation

of loops in X and loops in Y e.g.

—provided X ∩ Y is path connected.

We have to go home and ask permission every time we want to cross

the road (into the other space).


But we have now overcounted because there can be non-trivial loops in X ∪ Y ,

for example:

If we blindly counted loops in X and loops in Y we would count loops in the

intersection twice.

So we will have to take a quotient of π1(X) ∗ π1(Y ) so that

• if γ1 is a loop in the intersection considered as being in X, and

• γ2 is the same loop but considered as being in Y

we force γ1 ∼ γ2.

This is just like the disjoint-union-and-quotient construction of X ∪ Y , and is

the idea of the theorem: Van Kampen’s theorem roughly says

If we mimic this disjoint-union-and-quotient construction on the

fundamental group, we will get the fundamental group of the union

(as long as X and Y are nice enough).

Note for the categorically inclined

These constructions are all pushouts—in Top∗ first and then Gp. The theorem

can be restated as “π1 preserves pushouts” (if the situation is nice enough).

What does this construction look like for groups?

A ∩B

A

B

“A ∪B′′

i::

j $$

$$::

7.3 Equivalence relations on groups 79

We can’t take the union of two groups, as it instead necessarily a group, so

instead we mimic

X∐

Y�∼and we form

A ∗B�∼where ∼ is the equivalence relation generated by

∀ α ∈ A ∩B, i(α) ∼ j(α).

How do we make such an equivalence relation?

7.3 Equivalence relations on groups

We’re used to equivalence relations on sets and (maybe) spaces.

When we quotient out by an equivalence relation, what do we get?

• On a set we get a set of equivalence classes.

• On a space we get a quotient space

• On a group we get a quotient group

The key

Quotienting a group G by

the equivalence relation generated by g ∼ h

is the same as

quotienting by

the equivalence relation generated by gh−1 ∼ 1

which is the same as

quotienting by

the normal subgroup generated by gh−1.

So our construction becomes

A ∗B�N

7.4 Van Kampen’s Theorem 80

where N is the normal subgroup of A ∗ B generated by all the elements of the

form

i(α).j(α)−1 for α ∈ A ∩B.

Applying this to unions of spaces

The idea is we want to find π1(X ∪ Y ).

We have this diagram

X ∩ Y

X

Y

X ∪ Y ∼= X ∨ Y�∼

f::

g $$

p""

q99

We apply π1 to the whole diagram:

π1(X ∩ Y )

π1(X)

π1(Y )

π1(X ∪ Y )

π1f ::

π1g $$

$$

::

and we get

π1(X ∪ Y ) ∼= π1(X) ∗ π1(Y )�N

where N is the normal subgroup generated by all elements of the form

π1f(α).π1g(α)−1 ∀ α ∈ π1(X ∩ Y ).

But we must say under what circumstances this is true.

7.4 Van Kampen’s Theorem

Theorem 7.7 (Van Kampen). Suppose a space Z is decomposed as a union

X ∪ Y where

• X,Y are open in Z

• X,Y, and X ∩ Y are all path connected, and

• X and Y each contain the basepoint z0 of Z

(hence X ∩ Y contains the basepoint).


Write

i : π1(X ∩ Y ) −→ π1(X)

j : π1(X ∩ Y ) −→ π1(Y )

for the homomorphisms induced by the inclusions of spaces, with all fundamental

groups taken with common basepoint z0.

Let N be the normal subgroup of π1(X) ∗π1(Y ) generated by all elements of the

form

i(α).j(α)−1 ∀ α ∈ π1(X ∩ Y ).

Then

π1(Z) ∼= π1(X) ∗ π1(Y )�N .

Remark 7.8. I think it’s easiest to think about the diagram

X ∩ Y

X

Y

X ∪ Y

f::

g $$

p

$$

q

::

and apply π1 to the whole thing.

We can now work out our examples from earlier.

Example 7.3: MB with disc glued on −→ RP 2

S1

MB

D2

RP 2

f ::

g $$

$$

::

—we can finally actually calculate π1(RP 2) !


π1(S1)

π1(MB)

π1(D2)

π1(RP 2)

π1f ::

π1g $$

$$

::

Now we know (Ex #7, q1) that π1f is “multiplication by 2”.


Also, π1g must send everything to 0.

Alert! —We are doing additive groups, so N is going to be the normal subgroup

generated by elements

i(α)− j(α) ∀ α ∈ Z

i.e.

2α ∀ α ∈ Z.

So N ∼= 2Z, and

π1(RP 2) ∼= Z ∗ 0�2Z

= Z�2Z

= Z2 (Phew)

NB G ∗ 0 is always G because the trivial group 0 only contains the identity

element.

Example 7.4: two MB glued to make KB

S1

MB

MB

MB ∪MB ∼= KB

f ::

g $$

$$::


π1(S1)

π1(MB)

π1(MB)

π1(KB)

π1f ::

π1g $$

$$

::

Alert! —in this case it is better not to use additive notation as Z ∗ Z is not

abelian. We should think of Z as 〈a〉, the infinite cyclic group generated by a,

with elements

1, a, a2, a3, . . .

Then Z ∗ Z is 〈a, b〉

and “×2” becomes “squaring”.


So our elements α ∈ π1(S1) are all of the form γn where γ represents the “basic”

loop.

Similarly the we can write the elements of π1MB as an for one MB and bn for

the other. Then we have

i : π1(S1) −→ π1(MB)

1 7→ 1

γ 7→ a2

γ2 7→ a4

γ3 7→ a6

...

γn 7→ a2n

...

Similarly

j : π1(S1) −→ π1(MB)

1 7→ 1

γ 7→ b2

...

γn 7→ b2n

...

So

N = 〈 i(α).j(α)−1 | α ∈ π1(S1) 〉

= 〈 i(γn).j(γn)−1 | n ∈ Z 〉

= 〈 a2nb−2n | n ∈ Z 〉

= 〈a2b−2〉

= 〈a2c2〉 where c = b−1

So

π1(KB) = 〈a, c〉�〈a2c2〉which we write as

〈a, c | a2c2〉.

7.5 Generators and relations 84

7.5 Generators and relations

We have given π1(KB) in terms of generators and relations. A “relation”

is just a word in the generators:

〈 generators | relations 〉

〈 a, c, | a2c2 〉

This is to be understood as “the group generated by a and c subject to the

relation generated by a2c2 ∼ 1.

It is an important result of group theory that this can always be done. e.g.

Cn ∼= 〈 a | an〉 ∼= Zn

Dn∼= 〈 a, b | an, b2, abab−1〉

D∞ ∼= 〈 a, b | b2, abab−1〉

Z ∼= 〈 a 〉

Z ∗ Z ∼= 〈 a, b 〉

Z2 ∗ Z2∼= 〈 a, b | a2, b2〉

Z× Z ∼= 〈 a, b | aba−1b−1〉

Note that a presentation by generators and relations is not necessarily unique,

e.g.

〈 a | an〉 ∼= 〈 a | a−n〉

〈 a, b | a2b−2〉 ∼= 〈 a, c | a2c2〉

∼= 〈 x, y | xyx−1y〉

We can use generators and relations to

• start with any group G, and

• construct a space whose fundamental group is G.


Method

1. Find a presentation of G by generators and relations.

2. Make a wedge of n circles, one for each generator.

3. Attach a disc for each relation, attaching it to the original circles according

to the word in question.

Then by Van Kampen’s Theorem we know that the fundamental group is G.

Example 7.9. Z× Z

1. Z× Z ∼= 〈 a, b, | aba−1b−1 〉

2. We have 2 generators:

3. We have 1 relation, so we attach one disc, and its boundary should be

aba−1b−1.

We could draw this as:

but this is the same as:

Now we can use Van Kampen’s Theorem to check its fundamental group.

We have built a space X using the diagram

S1

S1 ∨ S1

D2

X

f ::

g $$

$$::


Where f is the map that sends the circle to the loop aba−1b−1, and g is the

boundary inclusion.


π1(S1)

〈a, b〉

0

π1(X)

i ::

$$

$$

::

where i = π1(f) is the homomorphism

〈x〉 −→ 〈a, b〉

x 7→ aba−1b−1

So by Van Kampen’s Theorem we have

π1(X) = 〈a, b〉 ∗ 0�N

where N is the normal subgroup generated by elements of the form i(xn), i.e.

elements of the form (aba−1b−1)n. But this normal subgroup is just generated

by aba−1b−1, so we have

π1(X) = 〈a, b〉 ∗ 0�〈aba−1b−1〉

= 〈 a, b | aba−1b−1 〉

Example 7.10. Z2

1. Z2∼= 〈 a | a2 〉.

2. We have 1 generator:

3. We have 1 relation so we attach 1 disc:


So we get the space:

Example 7.11. Zn

1. Zn ∼=

2. Generators:

3. Relations:

See also Ex #8, q.1.

Example 7.12. Z

1. Z = 〈a〉

2. We have 1 generator, so 1 copy of S1.

3. We have no relations so no discs, so we just get S1.

Example 7.13. Z ∗ Z

1. Z ∗ Z ∼= 〈a, b〉.

2. We have 2 generators, so we have S1 ∨ S1.

3. We have no relations, so no discs, so we just get S1 ∨ S1.

Example 7.14. Dn

1. Dn∼= 〈 a, b | an, b2, abab〉

2. We have 2 generators, so 2 circles:

7.6 Cell complexes 88

3. We have 3 relations, so we attach 3 discs:

We have sketched the following theorem.

For any group G there is a space XG whose fundamental group is G.

Of course, this is not unique, and there might be many other non-homeomorphic

spaces that also have fundamental group G.

7.6 Cell complexes

We have actually been constructing 2-dimensional cell complexes. A 2-

dimensional cell complex is formed by attaching discs along circles. Cell com-

plexes of higher dimensions are formed by continuing this process for more

dimensions:

1. Attach copies of D2 along copies of S1.



4....

This produces a family of spaces that are well-behaved and (relatively) easy to

understand. We have already seen some good behaviour:

• We started with a group G and made a cell complex XG whose funda-

mental group is G.

• We can start with any cell complex and find its fundamental group rather

esily, by simply “reading off” the generators and relations.


Example 7.15. The Klein Bottle.

We usually construct the Klein Bottle as

giving fundamental group

〈 a, b, | aba−1b 〉

But in Example 7.4 we got fundamental group

〈 a, c | a2c2 〉

giving cell complex

—these are the same! (Exercise)

We will now sketch a proof of the theorem we mentioned at the end of the

previous section.

Theorem 7.16. For any group G there is a space (in fact a 2-dimensional cell

complex) XG such that

π1(XG) = G.

Proof.

Every group has a presentation by generators and relations (standard result).

Pick a presentation

G = 〈 gα | rβ 〉

which is the “free group generated by all the gα’s, quotiented by the normal

subgroup generated by all the words rβ”.

Now form the wedge of circles ∨α

S1


and attach one copy of D2 for every relation, according to the diagram

S1

∨α

S1

D2

.

rβ ::

boundary $$

$$::

for each β. So overall we have a diagram

∨β

S1

∨α

S1

∨β

D2

XG

{rβ} ::

{boundary}$$

$$

::

Applying π1 to the whole diagram we get

〈hβ〉

〈gα〉

0

π1(XG)

i ::

0 $$

$$

::

where the homorphism i sends each hβ to the word rβ ∈ 〈gα〉.

Now by Van Kampen’s theorem we have

π1(XG) = 〈gα〉�N

where N is the normal subgroup generated by all elements i(hβ) i.e. rβ .

So

π1(XG) = 〈gα | rβ〉

= G

as required. 2

Remark 7.17. Note that, strictly speaking, we either need a slightly stronger

version of Van Kampen’s theorem than the one we’ve done, or else we have to

put in some effort to make our space fit our version of the theorem—it’s not

quite expressed as a union yet.

• We can make it a union in a slightly devious way by thickening our circles

a bit.


• Actually our diagram is a pushout and the more general Van Kampen

theorem is about preserving pushouts; unions are one example of pushouts.

Example 7.18. G = Zn = 〈a | an〉

• For n = 1 we get D2.

• For n = 2 we get:

• For n ≥ 3 we get a cell complex that is not a surface:

Example 7.19. The orientable surfaces are the surfaces

Mg = torus with g holes (genus g), g ≥ 0

For g > 0 get these as XG starting with

G = 〈 a1, b1, a2, b2, . . . , ag, bg | a1b1a−11 b−11 a2b2a−12 a−12 · · · agbga−1g b−1g 〉

—so we start with 2g circles and glue one disc:

Example 7.20. We get the non-orientable surfaces Ng as XG where

G = 〈 a1, a2, . . . , ag | a21a22 · · · a2g 〉

—so we start with g circles and glue one disc:


For example

• N1 =

• N2 =

Example 7.21. The infinite dihedral group.

D∞ = 〈 a, b | b2, abab 〉

= 〈 b, c | b2, c2 〉 by putting c = ab

= Z ∗ Z

So for XG we get RP 2 ∨ RP 2.

We now get to learn some algebra from topology!

Corollary 7.22. Every subgroup of a free group is free.

Proof. (sketch)

If G is free then it has a presentation by generators and no relations, so XG is

a wedge of circles i.e. a 1-dimensional cell complex.

By the classification of covering spaces, subgroups of π1(XG) (= G) correspond

to connected covering spaces of XG.

But any such covering space must also be a 1-dimensional cell complex, so by

Van Kampen’s theorem we know that its fundamental group has only generators,

and no relations (since relations come from discs in the cell-complex).

So the fundamental group of any connected covering space is free, and hence

the corresponding subgroup of G is free.

Since we get all subgroups of G in this way, we have shown that every subgroup

of G is free. 2

7.7 The classification of surfaces 93

7.7 The classification of surfaces

We can use everything we know about the fundamental group to classify surfaces—

although we still can’t classify spaces in general. But this is a step in the right

direction.

Theorem 7.23. Every compact connected surface is homeomorphic to

• an Mg if it is orientable, or

• an Ng if it is non-orientable

and none of these is homeomorphic to any other.

Remarks 7.24.

1. The proof is hard, and usually involves some fiddly process of triangulating

surfaces.

2. Note that for this theorem to be precise we have to say exactly what we

mean by “surface”.

Definition 7.25. (Not for the exam.)

A surface is a Hausdorff space with a countable basis such that every point has

a neighbourhood homeomorphic to R2.

Hausdorff means “any two distinct points can be ‘housed off’ from each other.”

Orientable means “given any smooth curve we can choose normal and tangent

vectors, push them round, and get back to where we started”.

Visualising this result

The result can be proved by a process of “adding handles or Mobius Bands”.

Here “add” means “cut out a disc and then glue the other thing on”.

• The orientable surface Mg is a “sphere with g handles added”.

• The non-orientable surface Ng a “sphere with g Mobius Bands added”.

We make this precise using:


Definition 7.26. The connect sum X#Y of spaces X and Y is formed by

cutting a disc out of each space and glueing the resulting spaces together along

the boundary circle.

Note that the sphere S2 is a unit for # i.e.

X#S2 ∼= X.

• A “handle” is a torus T with a disc removed, so “adding a handle to X”

is achieved by forming

X#T.

• A Mobius band with a disc glued on is RP 2. So (going backwards), cutting

a disc out of RP 2 gives a Mobius band. So “adding a MB to X” is achieved

by forming

X#RP 2.

So we have:

Orientable Non-orientable

M0 = S2

M1 = S2#T = T N1 = S2#RP 2 = RP 2

M2 = T#T N2 = RP 2#RP 2 = KB

M3 = T#T#T N3 = RP 2#RP 2#RP 2

......

The classification theorem says that these are all possible surfaces. So what

about if we add some handles and some Mobius bands?

—They cancel out!

Theorem 7.27. “Adding m handles and n MBs is the same as adding 2m+ n

MBs”

Proof. By induction, starting with the case m = n = 1 (Ex. #10).

In the starting case we need to show that

T#RP 2 = RP 2#RP 2#RP 2


i.e.

T#RP 2 = KB#RP 2.

“Cutting a disc from a torus and adding a MB is the same as cutting a disc

from a KB and adding a MB.” 2

How to “see” that T#RP 2 = KB#RP 2.

1. Torus with disk removed is: KB with disk removed is:

2. The effect of glueing a MB around the edge is to make the top and bottom

sides of the (twice punctured) disk D the same.

Or: We can slide the cylinder onto the MB

and then slide one end of the cylinder around

since the MB only has one side.

8 Applications 96

8 Applications

8.1 The fundamental theorem of algebra

The theorem says:

Every non-constant polynomial has a root in C.

It was probably believed before it was rigorously proved.

This is an example of topology “helping” algebra.

The idea

We consider a polynomial

p(z) = zn + a1zn−1 + a2z

n−2 + · · ·+ an

We can assume the leading coefficient is 1 since if it is not, we can divide through

by it without changing the roots.

We assume this polynomial has no roots and show that n must then be 0.

We consider the values of p(z) as z goes round circles in C.

Now p is a continuous map C −→ C.

If we assume it has no roots then we are saying

∀z ∈ C, p(z) 6= 0

so in fact p is a continuous map

C −→ C \ 0 —the punctured plane

Aim: find a loop in C that gets mapped to a non-trivial loop in C \ 0.

This gives a contradiction—we can’t map a trivial loop to a non-trivial one.

8.1 The fundamental theorem of algebra 97

Given f : X −→ Y and a loop γ in X we have

γ ∼ cst =⇒ fγ ∼ cst

since if the first homotopy is given by α : I × I −→ X

we get the second by I × I α−→ Xf−→ Y

How do we do it?

If we go round a huge circle in C, then p(z) is dominated by zn. So applying

p gives us a loop in C \ 0 which looks like zn, which is definitely non-trivial in

C \ 0.

Summary of the proof

• topological content: a continuous map sends trivial loops to trivial loops

• analytical content : when |z| is large, p(z) behaves like zn

• technical content: the stuff that actually makes it a proof!

Note: We actually turn the “analytical content” into something topological

too—we construct a homotopy from p(z) to zn.

Theorem 8.1 (Fundamental theorem of algebra).

Every non-constant polynomial with coefficients in C has a root in C.

Proof. We assume the polynomial is of the form

p(z) = zn + a1zn−1 + · · ·+ an

Suppose p has no roots; we aim to show this implies n = 0.

Let γ be a loop in C given by

γ : I −→ C

s 7→ re2πis

We will need to make sure r is large enough, but first we have

to work out what that means. Large enough for what?


Applying p to this loop we get

Iγ−→ C p−→ C \ 0

and we want to show that this is homotopic to the loop obtained by applying

the map q(z) = zn:

Iγ−→ C q−→ C \ 0.

Certainly q ' p as maps C −→ C via

I × C α−→ C

(t, z) 7→ zn + t(a1zn−1 + · · ·+ an)

but we need to check that this gives us a homotopy qγ ' pγ as maps

I −→ C \ 0.

Why might it fail?

We have to check that the homotopy doesn’t “go across the puncture”

i.e. we need to check that we don’t have α(t, z) = 0 for any t.

—We make sure our radius r is big enough that we always get

|zn| > |t(a1zn−1 + · · ·+ an).

Let r > max( |a1|+ · · ·+ |an|, 1 ).

Then for |z| = r we have

|zn| = rn

= r.rn−1

> (|a1|+ · · ·+ |an|).rn−1

= |a1|rn−1 + |a2|rn−1 + · · ·+ |an|rn−1

> |a1|rn−1 + |a2|rn−2 + · · ·+ |an| since r > 1

= |a1|.|zn−1|+ |a1|.|zn−2|+ · · ·+ |an|

≥ |a1zn−1 + a2zn−2 + · · ·+ an|

> |t(a1zn−1 + · · ·+ an)| for t ∈ [0, 1]


So α(t, z) = zn + t(a1zn−1 + · · ·+ an) is never zero for |z| = r and t ∈ [0, 1].

So α does indeed give a homotopy pγ ' qγ as maps

I −→ C \ 0.

Note that qγ is the loop

qγ : I −→ C \ 0

s 7→ rne2πins

which, informally, “goes n times round the circle of radius rn”.

Now, we know C \ 0 is homotopy equivalent to the circle of radius rn (e.g. by

straight line homotopy) so by homotopy invariance

π1(C \ 0) ∼= Z

and

[qγ] = [pγ] correponds to n ∈ Z.

But also

p : C −→ C \ 0

so we have

π1(p) : π1(C) −→ π1(C \ 0)

0 −→ Z

so π1(p) must send everything to 0. But we have just seen that [pγ] corresponds

to n so we must have n = 0.

i.e. if p has no roots, p is constant. 2

Note that what we really did was apply p to a large disc. We showed that p

mapped the disc to another disc (up to homotopy) preserving the boundary,

and used the fact that if p is continuous then every value inside the boundary

of the second disc must be achieved—including 0 in particular.

This is really a 2-dimensional version of my favourite theorem:

The Intermediate Value Theorem

IVT it implies that baby carrots exist.

8.2 The Brouwer fixed point theorem 100

8.2 The Brouwer fixed point theorem

Theorem 8.2. Every continuous map f : D2 −→ D2 has a fixed point, that is,

α ∈ D2 s.t. f(α) = α.

Idea

If there were no fixed point we could use f to construct a continuous map

D2 −→ S1 fixing the boundary. Impossible!

Proof. Suppose f(α) 6= α for all α ∈ D2.

We define a map r : D2 −→ S1 by letting r(α) be the point of S1 reached by

starting at f(α), constructing a straight line to α and extending it to S1:

This is well-defined since f(α) 6= α, and continuous since small changes to α

produce small changes in f(α) and hence to the line joining them.

Furthermore if α ∈ S1 ⊂ D2 then r(α) = α, that is, r fixes the boundary.

Thus we have mapped a trivial loop to a non-trivial one in S1. # 2

In fact the topological content here is the same as before—you can’t map a

trivial loop to a non-trivial one. We could also say the content is

D2 and S1 are not homotopy equivalent

and for this it suffices to know π1(S1) 6= 0. (The map r in the proof is trying

to be a homotopy equivalence

r : D2 −→ S1

via the inclusion into the boundary S1 −→ D2.)

In fact this theorem is true for all dimensions n ≥ 2 and maps

f : Dn −→ Dn

8.2 The Brouwer fixed point theorem 101

—this follows from the fact that Dn and Sn−1 are not homotopy equivalent.

However for the versions with n > 2, π1 won’t help us since

π1(Sn) = 0 ∀n ≥ 2

so

π1(Dn) ∼= π1(Sn−1) ∀n > 2.

The higher homotopy groups will sort this out:

∀n ≥ 1, πn(Sn) = Z

but

∀n ≥ 1, πn(Dn) = 0

so we can deduce that

∀n ≥ 1, Dn 6= Sn−1.

mas435 / mas6370 algebraic topology 2014{15 part a

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