II Algebraic Topology (Full)

Part II — Algebraic Topology

Based on lectures by H. Wilton

Notes taken by Dexter Chua

Michaelmas 2015

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Part IB Analysis II is essential, and Metric and Topological Spaces is highly desirable

The fundamental group

Homotopy of continuous functions and homotopy equivalence between topological

spaces. The fundamental group of a space, homomorphisms induced by maps of spaces,

change of base point, invariance under homotopy equivalence. [3]

Covering spaces

Covering spaces and covering maps. Path-lifting and homotopy-lifting properties, and

their application to the calculation of fundamental groups. The fundamental group

of the circle; topological proof of the fundamental theorem of algebra. *Construction

of the universal covering of a path-connected, locally simply connected space*. The

correspondence between connected coverings of

and conjugacy classes of subgroups

of the fundamental group of X. [5]

The Seifert-Van Kampen theorem

Free groups, generators and relations for groups, free products with amalgamation.

Statement *and proof* of the Seifert-Van Kampen theorem. Applications to the

calculation of fundamental groups. [4]

Simplicial complexes

Finite simplicial complexes and subdivisions; the simplicial approximation theorem. [3]

Homology

Simplicial homology, the homology groups of a simplex and its boundary. Functorial

properties for simplicial maps. *Proof of functoriality for continuous maps, and of

homotopy invariance*. [4]

Homology calculations

The homology groups of

, applications including Brouwer’s fixed-point theorem.

The Mayer-Vietoris theorem. *Sketch of the classification of closed combinatorical

surfaces*; determination of their homology groups. Rational homology groups; the

Euler-Poincar´e characteristic and the Lefschetz fixed-point theorem. [5]

Contents

0 Introduction

1 Definitions

1.1 Some recollections and conventions

1.2 Cell complexes

2 Homotopy and the fundamental group

2.0 Motivation

2.1 Homotopy

2.2 Paths

2.3 The fundamental group

3 Covering spaces

3.1 Covering space

3.2 The fundamental group of the circle and its applications

3.3 Universal covers

3.4 The Galois correspondence

4 Some group theory

4.1 Free groups and presentations

4.2 Another view of free groups

4.3 Free products with amalgamation

5 Seifert-van Kampen theorem

5.1 Seifert-van Kampen theorem

5.2 The effect on π

of attaching cells

5.3 A refinement of the Seifert-van Kampen theorem

5.4 The fundamental group of all surfaces

6 Simplicial complexes

6.1 Simplicial complexes

6.2 Simplicial approximation

7 Simplicial homology

7.1 Simplicial homology

7.2 Some homological algebra

7.3 Homology calculations

7.4 Mayer-Vietoris sequence

7.5 Continuous maps and homotopy invariance

7.6 Homology of spheres and applications

7.7 Homology of surfaces

7.8 Rational homology, Euler and Lefschetz numbers

0 Introduction

In topology, a typical problem is that we have two spaces

and

, and we want

to know if

∼

, i.e. if

and

are homeomorphic. If they are homeomorphic,

we can easily show this by writing down a homeomorphism. But what if they

are not? How can we prove that two spaces are not homeomorphic?

For example, are

and

homeomorphic (for

m 6

)? Intuitively, they

should not be, since they have different dimensions, and in fact they are not.

But how can we actually prove this?

The idea of algebraic topology is to translate these non-existence problems in

topology to non-existence problems in algebra. It turns out we are much better

at algebra than topology. It is much easier to show that two groups are not

isomorphic. For example, we will be able to reduce the problem of whether

and

are homeomorphic (for

m 6

) to the question of whether

and

{e}

are isomorphic, which is very easy.

While the statement that

∼

for

n 6

is intuitively obvious, algebraic

topology can be used to prove some less obvious results.

Let

be the

dimensional unit disk, and

n−1

be the

n−

1 dimensional unit

sphere. We will be able to show that there is no continuous map

→ S

n−1

such that the composition

n−1

is the identity, where the first arrow is the inclusion map. Alternatively, this

says that we cannot continuously map the disk onto the boundary sphere such

that the boundary sphere is fixed by the map.

Using algebraic topology, we can translate this statement into an algebraic

statement: there is no homomorphism F : {0} → Z such that

Z {0} Z

is the identity. This is something we can prove in 5 seconds.

By translating a non-existence problem of a continuous map to a non-existence

problem of a homomorphism, we have made our life much easier.

In algebraic topology, we will be developing a lot of machinery to do this sort

of translation. However, this machinery is not easy. It will take some hard work,

and will be rather tedious and boring at the beginning. So keep in mind that

the point of all that hard work is to prove all these interesting theorems.

In case you are completely uninterested in topology, and don’t care if

and

are homeomorphic, further applications of algebraic topology include solving

equations. For example, we will be able to prove the fundamental theorem of

algebra (but we won’t), as well as Brouwer’s fixed point theorem (which says

that every continuous function from

→ D

has a fixed point). If you are not

interested in these either, you may as well drop this course.

1 Definitions

1.1 Some recollections and conventions

We will start with some preliminary definitions and conventions.

Definition (Map). In this course, the word map will always refer to continuous

maps. We are doing topology, and never care about non-continuous functions.

We are going to build a lot of continuous maps in algebraic topology. To do

so, we will often need to glue maps together. The gluing lemma tells us that

this works.

Lemma (Gluing lemma). If

X → Y

is a function of topological spaces,

C ∪ K

and

are both closed, then

is continuous if and only if the

restrictions f |

and f |

are continuous.

Proof. Suppose f is continuous. Then for any closed A ⊆ Y , we have

−1

(A) = f

−1

(A) ∩ C,

which is closed. So f|

is continuous. Similarly, f |

is continuous.

If f |

and f |

are continuous, then for any closed A ⊆ Y , we have

−1

(A) = f|

−1

(A) ∪ f|

−1

(A),

which is closed. So f is continuous.

This lemma is also true with “closed” replaced with “open”. The proof is

the same proof with “closed” replaced with “open”.

We will also need the following technical lemma about metric spaces.

Lemma. Let (

X, d

) be a compact metric space. Let

}

α∈A

be an open

cover of

. Then there is some

such that for each

x ∈ X

, there is some

α ∈ A

such that B

(x) ⊆ U

. We call δ a Lebesgue number of this cover.

Proof.

Suppose not. Then for each

n ∈ N

, there is some

∈ X

such that

1/n

(

) is not contained in any

. Since

is compact, the sequence (

)

has a convergent subsequence. Suppose this subsequence converges to y.

Since

is an open cover, there is some

α ∈ A

such that

y ∈ U

. Since

is open, there is some

r >

0 such that

(

)

⊆ U

. But then we can find a

sufficiently large n such that

and d(x

, y) <

. But then

1/n

) ⊆ B

(y) ⊆ U

Contradiction.

1.2 Cell complexes

In topology, we can construct some really horrible spaces. Even if we require them

to be compact, Hausdorff etc, we can often still produce really ugly topological

spaces with weird, unexpected behaviour. In algebraic topology, we will often

restrict our attention to some nice topological spaces, known as cell complexes.

To build cell complexes, we are not just gluing maps, but spaces.

Definition (Cell attachment). For a space

, and a map

n−1

→ X

, the

space obtained by attaching an n-cell to X along f is

X ∪

= (X q D

)/∼,

where the equivalence relation

∼

is the equivalence relation generated by

x ∼ f

(

)

for all x ∈ S

n−1

⊆ D

(and q is the disjoint union).

Intuitively, a map

n−1

→ X

just picks out a subset of

that looks like

the sphere. So we are just sticking a disk onto

by attaching the boundary of

the disk onto a sphere within X.

Definition (Cell complex). A (finite) cell complex is a space X obtained by

(i) Start with a discrete finite set X

(0)

(ii)

Given

(n−1)

, form

(n)

by taking a finite set of maps

n−1

→

(n−1)

} and attaching n-cells along the f

(n)

(n−1)

/{x ∼ f

(x)}.

For example, given the

(0)

above, we can attach some loops and lines to

obtain the following X

(1)

We can add surfaces to obtain the following X

(2)

(iii) Stop at some X = X

(k)

. The minimum such k is the dimension of X.

To define non-finite cell complexes, we just have to remove the words “finite” in

the definition and remove the final stopping condition.

We have just had an example of a cell complex above, and it is easy to

produce many more examples, such as the

-sphere. So instead let’s look at a

non-cell complex.

Example. The following is not a cell complex: we take

, and add a circle with

radius

and center (0

). Then we add another circle with radius

and center

), then a circle with radius

and center (0

) etc. We obtain something like

This is known as the Hawaiian Earring.

Why is this not an (infinite) cell complex? We did obtain it by attaching

lots of 1-cells to the single point (0

0). However, in the definition of a cell

complex, the cells are supposed to be completely unrelated and disjoint, apart

from intersecting at the origin. However, here the circles clump together at the

origin.

In particular, if we take the following sequence (0

)

, · · ·

, it

converges to (0

0). If this were a cell complex, then this shouldn’t happen

because the cells are unrelated, and picking a point from each cell should not

produce a convergent sequence (if you are not convinced, if we actually did

produce by attaching cells, then note that during the attaching process, we

needn’t have attached them this way. We could have made it such that the

cell has radius

. Then clearly picking the topmost point of each cell will not

produce a convergent sequence).

We will see that the Hawaiian Earring will be a counterexample to a lot of

our theorems here.

2 Homotopy and the fundamental group

This will be our first trick to translate topological spaces into groups.

2.0 Motivation

Recall that we wanted to prove that

∼

for

n 6

. Let’s first do the

simple case, where m = 1, n = 2. We want to show that R 6

∼

This is not hard. We know that

is a line, while

is a plane. Let’s try to

remove a point from each of them. If we remove a point from

, the space stops

being path connected. However, removing a point does not have this effect on

. Since being path connected is a topological property, we have now showed

that R and R

are not homeomorphic.

Unfortunately, this does not extend very far. We cannot use this to show

that R

and R

are not homeomorphic. What else can we do?

Notice that when we remove a point from

, sure it is still connected, but

something has changed.

Consider a circle containing the origin in

\ {

}

. If the origin were there,

we can keep shrinking the circle down until it becomes a point. However, we

cannot do this if the origin is missing.

The strategy now is to exploit the fact that

\ {

}

has circles which cannot be

deformed to points.

2.1 Homotopy

We have just talked about the notion of “deforming” circles to a point. We can

think of a circle in

as a map

→ X

, and we want to “deform” this map to

a point. This process of deformation is known as homotopy. Here we are going

to use the interval [0, 1] ⊆ R a lot, and we will just call it I.

Notation.

I = [0, 1] ⊆ R.

Definition (Homotopy). Let

f, g

X → Y

be maps. A homotopy from

is a map

H : X × I → Y

such that

H(x, 0) = f (x), H(x, 1) = g(x).

We think of the interval

as time. For each time

(

· , t

) defines a map

X → Y

So we want to start from f, move with time, and eventually reach g.

If such an

exists, we say

is homotopic to

, and write

f ' g

. If we want

to make it explicit that the homotopy is H, we write f '

As mentioned at the beginning, by calling

a map, we are requiring it to

be continuous.

Sometimes, we don’t want a general homotopy. We might want to make sure

that when we are deforming from a path

, the end points of the path don’t

move. In general, we have

Definition (Homotopy

rel

A). We say

is homotopic to

g rel A

, written

f ' g rel A, if for all a ∈ A ⊆ X, we have

H(a, t) = f (a) = g(a).

This notion will find itself useful later, but we don’t have to pay too much

attention to this yet.

Our notation suggests that homotopy is an equivalence relation. Indeed, we

have

Proposition. For spaces

X, Y

, and

A ⊆ X

, the “homotopic

rel A

” relation is

an equivalence relation. In particular, when

∅

, homotopy is an equivalence

relation.

Proof.

(i) Reflexivity: f ' f since H(x, t) = f(x) is a homotopy.

(ii)

Symmetry: if

(

x, t

) is a homotopy from

, then

(

− t

) is a

homotopy from g to f.

(iii)

Transitivity: Suppose

f, g, h

X → Y

and

f '

g rel A

g '

h rel A

We want to show that

f ' h rel A

. The idea is to “glue” the two maps

together.

We know how to continuously deform

, and from

. So we just

do these one after another. We define H

: X × I → Y by

(x, t) =

(

H(x, 2t) 0 ≤ t ≤

(x, 2t − 1)

≤ t ≤ 1

This is well-defined since

(

1) =

(

) =

(

0). This is also continuous

by the gluing lemma. It is easy to check that

is a homotopy

rel A

We now have a notion of equivalence of maps — two maps are equivalent if

they are homotopic. We can extend the notion of homotopy to spaces as well.

Recall that when we defined homeomorphism, we required that there be some

f, g such that f ◦ g = id, g ◦ f = id. Here, we replace equality by homotopy.

Definition (Homotopy equivalence). A map

X → Y

is a homotopy equiva-

lence if there exists a

Y → X

such that

f ◦ g ' id

and

g ◦ f ' id

. We call

g a homotopy inverse for f.

If a homotopy equivalence

X → Y

exists, we say that

and

are

homotopy equivalent and write X ' Y .

We are soon going to prove that this is indeed an equivalence relation on

spaces, but we first look at some examples of homotopy equivalent spaces. Clearly,

homeomorphic spaces are homotopy equivalent. However, we will see that we

can do much more “violent” things to a space and still be homotopy equivalent.

Example. Let

\ {

}

. We have a natural inclusion map

X → Y

. To obtain a map

Y → X

, we can project each point onto the circle.

r(y)

In particular, we define r : Y → X by

r(y) =

kyk

We immediately have

r ◦ i

. We will now see that

i ◦ r ' id

. The

composition

i ◦ r

first projects each object into

, and then includes it back

into

\ {

}

. So this is just the projection map. We can define a homotopy

H : Y × I → Y by

H(y, t) =

t + (1 − t)kyk

This is continuous, and H( · , 0) = i ◦ r, H( · , 1) = id

As we have said, homotopy equivalence can do really “violent” things to a

space. We started with a 2-dimensional

\ {

}

space, and squashed it into a

one-dimensional sphere.

Hence we see that homotopy equivalence doesn’t care about dimensions.

Dimensions seem to be a rather fundamental thing in geometry, and we are

discarding it here. So what is left? What does homotopy equivalence preserve?

While

and

}

seem rather different, they have something in common —

they both have a “hole”. We will later see that this is what homotopy equivalence

preserves.

Example. Let

{

}

∗

. Let

X → Y

be the inclusion map, and

Y → X

be the unique map that sends everything to

{

}

. Again, we have

r ◦ i = id

. We can also obtain a homotopy from i ◦ r to id

H(y, t) = ty.

Again, from the point of view of homotopy theory,

is just the same as a

point! You might think that this is something crazy to do — we have just given

up a lot of structure of topological spaces. However, by giving up these structures,

it is easier to focus on what we really want to care about — holes. For example,

it is often much easier to argue about the one-point space

∗

than the whole of

! By studying properties that are preserved by homotopy equivalence, and

not just homeomorphism, we can simplify our problems by reducing complicated

spaces to simpler ones via homotopy equivalence.

In general things homotopy equivalent to a single point are known as con-

tractible spaces.

Notation. ∗ denotes the one-point space {0}.

Definition (Contractible space). If X ' ∗, then X is contractible.

We now show that homotopy equivalence of spaces is an equivalence relation.

To do this, we first need a lemma.

Lemma. Consider the spaces and arrows

X Y Z

If f

and g

, then g

◦ f

' g

◦ f

Proof.

We will show that

◦ f

' g

◦ f

' g

◦ f

. Then we are done since

homotopy between maps is an equivalence relation. So we need to write down

two homotopies.

(i) Consider the following composition:

X × I Y Z

It is easy to check that this is the first homotopy we need to show

◦ f

(ii) The following composition is a homotopy from g

◦ f

to g

◦ f

X × I Y × I Z

×id

Proposition. Homotopy equivalence of spaces is an equivalence relation.

Proof.

Symmetry and reflexivity are trivial. To show transitivity, let

X → Y

and

Y → Z

be homotopy equivalences, and

Y → X

and

Z → Y

be their homotopy inverses. We will show that

h ◦ f

X → Z

is a homotopy

equivalence with homotopy inverse g ◦ k. We have

(h ◦ f) ◦ (g ◦ k) = h ◦ (f ◦ g) ◦ k ' h ◦ id

◦ k = h ◦ k ' id

Similarly,

(g ◦ k) ◦ (h ◦ f ) = g ◦ (k ◦ h) ◦ f ' g ◦ id

◦ f = g ◦ f ' id

So done.

Definition (Retraction). Let

A ⊆ X

be a subspace. A retraction

X → A

is a map such that

r ◦ i

, where

A → X

is the inclusion. If such an

exists, we call A a retract of X.

This map sends everything in

without moving things in

. Roughly

speaking, if such a retraction exists, then A is no more complicated than X.

Definition (Deformation retraction). The retraction

is a deformation retrac-

tion if

i ◦ r ' id

. A deformation retraction is strong if we require this homotopy

to be a homotopy rel A.

Roughly, this says that A is as complicated as X.

Example. Take

any space, and

{x} ⊆ X

. Then the constant map

X → A

is a retraction. If

is contractible, then

is a deformation retract

of X.

2.2 Paths

Definition (Path). A path in a space

is a map

I → X

. If

(0) =

and

γ(1) = x

, we say γ is a path from x

to x

, and write γ : x

If γ(0) = γ(1), then γ is called a loop (based at x

Note that this map does not have to be injective. It can be self-intersecting

or do all sorts of weird stuff.

Recall that the basic idea of algebraic topology is to assign to each space

an (algebraic) object, which is (hopefully) easier to deal with. In homotopy

theory, we do so using the idea of paths.

To do so, we need to be able to perform operations on paths.

Definition (Concatenation of paths). If we have two paths

from

;

and γ

from x

to x

, we define the concatenation to be

(γ

· γ

)(t) =

(

(2t) 0 ≤ t ≤

(2t − 1)

≤ t ≤ 1.

This is continuous by the gluing lemma.

Note that we concatenate left to right, but function composition goes from

right to left.

Definition (Inverse of path). The inverse of a path γ : I → X is defined by

−1

(t) = γ(1 − t).

This is exactly the same path but going in the opposite direction.

What else do we need? We need an identity.

Definition (Constant path). The constant path at a point

x ∈ X

is given by

(t) = x.

We haven’t actually got a good algebraic system. We have

and

−1

, but

when we compose them, we get a path from

and back, and not the

identity. Also, we are not able to combine arbitrary paths in a space.

Before we make these into proper algebraic operations, we will talk about

something slightly different. We can view this as a first attempt at associating

things to topological spaces.

Definition (Path components). We can define a relation on

∼ x

there exists a path from

. By the concatenation, inverse and constant

paths,

∼

is an equivalence relation. The equivalence classes [

] are called path

components. We denote the quotient X/∼ by π

(X).

In the above space, we have three path components.

This isn’t really a very useful definition, since most spaces we care about are

path-connected, i.e. only have one path component. However, this is a first step

at associating something to spaces. We can view this as a “toy model” for the

more useful definitions we will later have.

One important property of this

is that not only does it associate a set to

each topological space, but also associates a function between the corresponding

sets to each continuous map.

Proposition. For any map f : X → Y , there is a well-defined function

(f) : π

(X) → π

(Y ),

defined by

(f)([x]) = [f(x)].

Furthermore,

(i) If f ' g, then π

(f) = π

(g).

(ii) For any maps A B C

h k

, we have π

(k ◦ h) = π

(k) ◦ π

(h).

(iii) π

(id

) = id

(X)

Proof.

To show this is well-defined, suppose [

] = [

]. Then let

I → X

be a

path from x to y. Then f ◦ γ is a path from f (x) to f (y). So [f(x)] = [f (y)].

(i)

f ' g

, let

X × I → Y

be a homotopy from

. Let

x ∈ X

. Then

(

x, ·

) is a path from

(

) to

(

). So [

(

)] = [

(

)], i.e.

(

)([

]) =

(g)([x]). So π

(f) = π

(g).

(ii) π

(k ◦ h)([x]) = π

(k) ◦ π

(h)([x]) = [k(h(x))].

(iii) π

(id

)([x]) = [id

(x)] = [x]. So π

(id

) = id

(X)

Corollary. If f : X → Y is a homotopy equivalence, then π

(f) is a bijection.

Example. The two point space

{−

}

is not contractible, because

|π

(X)| = 2, but |π

(∗)| = 1.

This is a rather silly example, since we can easily prove it directly. However,

this is an example to show how we can use this machinery to prove topological

results.

Now let’s return to our operations on paths, and try to make them algebraic.

Definition (Homotopy of paths). Paths

γ, γ

I → X

are homotopic as paths if

they are homotopic rel

{

} ⊆ I

, i.e. the end points are fixed. We write

γ ' γ

Note that we would necessarily want to fix the two end points. Otherwise, if we

allow end points to move, we can shrink any path into a constant path, and our

definition of homotopy would be rather silly.

This homotopy works well with our previous operations on paths.

Proposition. Let

, γ

I → X

be paths,

(1) =

(0). Then if

' γ

and

' γ

, then γ

· γ

' γ

· γ

Proof. Suppose that γ

and γ

. Then we have the diagram

We can thus construct a homotopy by

H(s, t) =

(

(s, 2t) 0 ≤ t ≤

(s, 2t − 1)

≤ t ≤ 1

To solve all our previous problems about operations on paths not behaving

well, we can look at paths up to homotopy.

Proposition. Let γ

: x

, γ

: x

, γ

: x

be paths. Then

(i) (γ

· γ

) · γ

' γ

· (γ

· γ

)

(ii) γ

· c

' γ

' c

· γ

(iii) γ

· γ

−1

' c

and γ

−1

· γ

' c

Proof.

(i) Consider the following diagram:

(ii) Consider the following diagram:

(iii) Consider the following diagram:

−1

Turning these into proper proofs is left as an exercise for the reader.

2.3 The fundamental group

The idea is to take spaces and turn them into groups. We want to try to do

this using paths. We’ve seen that if we want to do this, we should not work

directly with paths, but paths up to homotopy. According to our proposition,

this operation satisfies associativity, inverses and identity. The last issue we have

to resolve is that we can’t actually put two paths together unless they have the

same start and end points.

The idea is to fix one of the points

in our space, and only think about

loops that start and end at x

. So we can always join two paths together.

This tells us that we aren’t going to just think about spaces, but spaces with

basepoints. We’ve now got everything we need.

Definition (Fundamental group). Let

be a space and

∈ X

. The fun-

damental group of

(based at

), denoted

(

X, x

), is the set of homotopy

classes of loops in

based at

(i.e.

(0) =

(1) =

). The group operations

are defined as follows:

We define an operation by [

][

] = [

· γ

]; inverses by [

]

−1

= [

−1

]; and

the identity as the constant path e = [c

Often, when we write the homotopy classes of paths [

], we just get lazy and

write γ.

Theorem. The fundamental group is a group.

Proof. Immediate from our previous lemmas.

Often in mathematics, after defining a term, we give lots of examples of

it. Unfortunately, it is rather difficult to prove that a space has a non-trivial

fundamental group, until we have developed some relevant machinery. Hence we

will have to wait for a while before we have some concrete examples. Instead,

we will look at some properties of the fundamental group first.

Definition (Based space). A based space is a pair (

X, x

) of a space

and a

point x

∈ X, the basepoint. A map of based spaces

f : (X, x

) → (Y, y

)

is a continuous map

X → Y

such that

(

) =

. A based homotopy is a

homotopy rel {x

Recall that for

, to every map

X → Y

, we can associate a function

(f) : π

(X) → π

(Y ). We can do the same for π

Proposition. To a based map

f : (X, x

) → (Y, y

there is an associated function

∗

= π

(f) : π

(X, x

) → π

(Y, y

defined by [γ] 7→ [f ◦ γ]. Moreover, it satisfies

(i) π

(f) is a homomorphism of groups.

(ii) If f ' f

, then π

(f) = π

(iii)

For any maps

(A, a) (B, b) (C, c)

h k

, we have

(

k ◦ h

) =

(k) ◦ π

(h).

(iv) π

(id

) = id

(X,x

)

Proof. Exercise.

In category-theoretic language, we say that π

is a functor.

So far so good. However, to define the fundamental group, we had to make a

compromise and pick a basepoint. But we just care about the space. We don’t

want a basepoint! Hence, we should look carefully at what happens when we

change the basepoint, and see if we can live without it.

The first observation is that

(

X, x

) only “sees” the path component of

since all loops based at

can only live inside the path component of

. Hence,

the first importance of picking a basepoint is picking a path component.

For all practical purposes, we just assume that

is path connected, since if

we weren’t, the fundamental group just describes a particular path component

of the original space.

Now we want to compare fundamental groups with different basepoints.

Suppose we have two basepoints

and

. Suppose we have a loop

How can we turn this into a loop based at

? This is easy. We first pick a path

. Then we can produce a new loop at

by going along

−1

take the path γ, and then return to x

by u, i.e. consider u

−1

· γ · u.

Proposition. A path u : x

induces a group isomorphism

: π

(X, x

) → π

(X, x

)

[γ] 7→ [u

−1

· γ · u].

This satisfies

(i) If u ' u

, then u

= u

(ii) (c

)

= id

(X,x

)

(iii) If v : x

. Then (u · v)

= v

◦ u

(iv) If f : X → Y with f(x

) = y

, f (x

) = y

, then

(f ◦ u)

◦ f

∗

= f

∗

◦ u

: π

(X, x

) → π

(Y, y

A nicer way of writing this is

(X, x

) π

(Y, y

)

(X, x

) π

(Y, y

)

∗

(f◦u)

∗

The property says that the composition is the same no matter which way

we go from

(

X, x

) to

(

Y, y

). We say that the square is a commutative

diagram. These diagrams will appear all of the time in this course.

(v)

, then

is an automorphism of

(

X, x

) given by conjugation

by u.

It is important (yet difficult) to get the order of concatenation and composition

right. Path concatenation is from left to right, while function composition is

from right to left.

Proof.

Yet another exercise. Note that (

−1

)

= (

)

−1

, which is why we have

an isomorphism.

The main takeaway is that if

and

are in the same path component,

then

(X, x

)

∼

(X, x

So the basepoint isn’t really too important. However, we have to be careful.

While the two groups are isomorphic, the actual isomorphism depends on which

path

we pick. So there is no natural isomorphism between the two

groups. In particular, we cannot say “let

α ∈ π

(

X, x

). Now let

be the

corresponding element in π

(X, x

)”.

We can also see that if (

X, x

) and (

Y, y

) are based homotopy equivalent,

then

(

X, x

)

∼

(

Y, y

). Indeed, if they are homotopy equivalent, then there

are some f : X → Y , g : Y → X such that

f ◦ g ' id

, g ◦ f ' id

∗

◦ g

∗

= id

(Y,y

)

, g

∗

◦ f

∗

= id

(X,x

)

and f

∗

and g

∗

are the isomorphisms we need.

However, can we do this with non-based homotopies? Suppose we have

a space

, and a space

, and functions

f, g

X → Y

, with a homotopy

H : X × I → Y from f to g. If this were a based homotopy, we know

∗

= g

∗

: π

(X, x

) → π

(Y, f(x

)).

Now we don’t insist that this homotopy fixes

, and we could have

(

)

(

). How can we relate

∗

(

X, x

)

→ π

(

Y, f

(

)) and

∗

(

X, x

)

→

(Y, g(x

))?

First of all, we need to relate the groups

(

Y, f

(

)) and

(

Y, g

(

)). To

do so, we need to find a path from

(

) to

(

). To produce this, we can use

the homotopy

. We let

(

)

(

) with

(

, ·

). Now we have

three maps f

∗

, g

∗

and u

. Fortunately, these do fit together well.

X Y

f(x

)

g(x

)

Lemma. The following diagram commutes:

(Y, f(x

))

(X, x

)

(Y, g(x

))

∗

In algebra, we say

∗

= u

◦ f

∗

Proof. Suppose we have a loop γ : I → X based at x

We need to check that

∗

([γ]) = u

◦ f

∗

([γ]).

In other words, we want to show that

g ◦ γ ' u

−1

· (f ◦ γ) · u.

To prove this result, we want to build a homotopy.

Consider the composition:

F : I × I X × I Y.

γ×id

Our plan is to exhibit two homotopic paths 

and 

−

in I × I such that

F ◦ 

= g ◦ γ, F ◦ 

−

= u

−1

· (f ◦ γ) · u.

This is in general a good strategy —

is a complicated and horrible space we

don’t understand. So to construct a homotopy, we make ourselves work in a

much nicer space I × I.

Our 

and 

−

are defined in a rather simple way.

I × I



−



−1

More precisely,



is the path

s 7→

(

1), and



−

is the concatenation of the

paths s 7→ (0, 1 − s), s 7→ (s, 0) and s 7→ (1, s).

Note that



and



−

are homotopic as paths. If this is not obvious, we can

manually check the homotopy

L(s, t) = t

(s) + (1 − t)

−

(s).

This works because I × I is convex. Hence F ◦ 

F ◦L

F ◦ 

−

as paths.

Now we check that the compositions

F ◦ 

are indeed what we want. We

have

F ◦ 

(s) = H(γ(s), 1) = g ◦ γ(s).

Similarly, we can show that

F ◦ 

−

(s) = u

−1

· (f ◦ γ) · u(s).

So done.

It is worth looking at this proof hard and truly understand what is going

on, since this is a really good example of how we can construct interesting

homotopies.

With this lemma, we can show that fundamental groups really respect

homotopies.

Theorem. If

X → Y

is a homotopy equivalence, and

∈ X

, then the

induced map

∗

: π

(X, x

) → π

(Y, f(x

)).

is an isomorphism.

While this seems rather obvious, it is actually non-trivial if we want to do

it from scratch. While we are given a homotopy equivalence, we are given no

guarantee that the homotopy respects our basepoints. So this proof involves

some real work.

Proof.

Let

Y → X

be a homotopy inverse. So

f ◦g '

and

g◦f '

X Y

f(x

)

g ◦ f(x

)

We have no guarantee that

g ◦ f

(

) =

, but we know that our homotopy

gives us u

= H

, · ) : x

g ◦ f(x

Applying our previous lemma with id

for “f ” and g ◦ f for “g”, we get

◦ (id

)

∗

= (g ◦ f)

∗

Using the properties of the

∗

operation, we get that

∗

◦ f

∗

= u

However, we know that

is an isomorphism. So

∗

is injective and

∗

surjective.

Doing it the other way round with

f ◦ g

instead of

g ◦ f

, we know that

∗

injective and f

∗

is surjective. So both of them are isomorphisms.

With this theorem, we can finally be sure that the fundamental group

is a property of the space, without regards to the basepoint (assuming path

connectedness), and is preserved by arbitrary homotopies.

We now use the fundamental group to define several simple notions.

Definition (Simply connected space). A space

is simply connected if it is

path connected and π

(X, x

)

∼

1 for some (any) choice of x

∈ X.

Example. Clearly, a point

∗

is simply connected since there is only one path on

∗

(the constant path). Hence, any contractible space is simply connected since

it is homotopic to ∗. For example, R

is simply connected for all n.

There is a useful characterization of simply connected spaces:

Lemma. A path-connected space

is simply connected if and only if for any

, x

∈ X, there exists a unique homotopy class of paths x

Proof.

Suppose

is simply connected, and let

u, v

be paths. Now

note that

u · v

−1

is a loop based at

, it is homotopic to the constant path, and

−1

· v is trivially homotopic to the constant path. So we have

u ' u · v

−1

· v ' v.

On the other hand, suppose there is a unique homotopy class of paths

for all

, x

∈ X

. Then in particular there is a unique homotopy class of loops

based at x

. So π

(X, x

) is trivial.

3 Covering spaces

We can ask ourselves a question — what are groups? We can write down a

definition in terms of operations and axioms, but this is not what groups are.

Groups were created to represent symmetries of objects. In particular, a group

should be acting on something. If not, something wrong is probably going on.

We have created the fundamental group. So what do they act on? Can we

find something on which these fundamental groups act on?

An answer to this would also be helpful in more practical terms. So far we

have not exhibited a non-trivial fundamental group. This would be easy if we

can make the group act on something — if the group acts non-trivially on our

thing, then clearly the group cannot be trivial.

These things we act on are covering spaces.

3.1 Covering space

Intuitively, a covering space of

is a pair (

X, p

X → X

), such that if we take

any

∈ X

, there is some neighbourhood

such that the pre-image of

the neighbourhood is “many copies” of U.

Definition (Covering space). A covering space of

is a pair (

X, p

X → X

such that each x ∈ X has a neighbourhood U which is evenly covered.

Whether we require

to be surjective is a matter of taste. However, it will

not matter if X is path-connected, which is the case we really care about.

Definition (Evenly covered). U ⊆ X is evenly covered by p :

X → X if

−1

(U)

∼

α∈Λ

where p|

: V

→ U is a homeomorphism, and each of the V

⊆

X is open.

Example. Homeomorphisms are covering maps. Duh.

Example. Consider

R → S

⊆ C

defined by

t 7→ e

2πit

. This is a covering

space.

−1

(1)

Here we have p

−1

(1) = Z.

As we said, we are going to use covering spaces to determine non-trivial

fundamental groups. Before we do that, we can guess what the fundamental

group of

is. If we have a loop on

, it could be nothing, it can loop around

the circle once, or loop around many times. So we can characterize each loop by

the number of times it loops around the circle, and it should not be difficult to

convince ourselves that two loops that loop around the same number of times

can be continuously deformed to one another. So it is not unreasonable to guess

that

(

∼

. However, we also know that

−1

(1) =

. In fact, for any

point

z ∈ S

−1

(

) is just “

many copies” of

. We will show that this is not

a coincidence.

Example. Consider

→ S

(for any

n ∈ Z \ {

}

) defined by

z 7→ z

. We

can consider this as “winding” the circle

times, or as the following covering

map:

where we join the two red dots together.

This time the pre-image of 1 would be

copies of 1, instead of

copies of 1.

Example. Consider

, which is the real projective plane. This is defined

/∼

, where we identify every

x ∼ −x

, i.e. every pair of antipodal points.

We can also think of this as the space of lines in

. This is actually really

difficult to draw (and in fact impossible to fully embed in

), so we will not

attempt. There is, however, a more convenient way of thinking about

. We

just use the definition — we imagine an

, but this time, instead of a point

being a “point” on the sphere, it is a pair of antipodal points.

We define

→ RP

to be the quotient map. Then this is a covering map,

since the pre-image of a small neighbourhood of any

is just two copies of the

neighbourhood.

As we mentioned, a covering space of

is (locally) like many “copies” of

. Hence, given any function

Y → X

, a natural question is whether we can

“lift” it to a function f : Y →

X, by mapping to one of the copies of X in

This is not always possible, but when it is, we call this a lift.

Definition (Lifting). Let

Y → X

be a map, and

X → X

a covering

space. A lift of

is a map

Y →

such that

p ◦

, i.e. the following

diagram commutes:

Y X

We can visualize this in the case where

is the unit interval

and the map

is just a path in X.

f(Y )

It feels that if we know which “copy” of

we lifted our map to, then we

already know everything about

, since we are just moving our

from

that particular copy of X. This is made precise by the following lemma:

Lemma. Let

X → X

be a covering map,

Y → X

be a map, and

be both lifts of f. Then

S = {y ∈ Y :

(y) =

(y)}

is both open and closed. In particular, if

is connected,

and

agree either

everywhere or nowhere.

This is sort of a “uniqueness statement” for a lift. If we know a point in the

lift, then we know the whole path. This is since once we’ve decided our starting

point, i.e. which “copy” of

we work in, the rest of

has to follow what

does.

Proof.

First we show it is open. Let

be such that

(

) =

(

). Then

there is an evenly covered open neighbourhood

U ⊆ X

(

). Let

such that

(

)

∈

(

) =

and

U → U

is a homeomorphism. Let

V =

−1

(

U) ∩

−1

(

U). We will show that

on V .

Indeed, by construction

◦

= p|

◦

Since p|

is a homeomorphism, it follows that

Now we show

is closed. Suppose not. Then there is some

y ∈

S \ S

. So

(

)

(

). Let

be an evenly covered neighbourhood of

(

). Let

−1

(

) =

Let

(

)

∈ U

and

(

)

∈ U

, where

β 6

. Then

−1

(

)

∩

−1

(

) is

an open neighbourhood of

, and hence intersects

by definition of closure. So

there is some

x ∈ V

such that

(

) =

(

). But

(

)

∈ U

and

(

)

∈ U

and hence

and

have a non-trivial intersection. This is a contradiction. So

S is closed.

We just had a uniqueness statement. How about existence? Given a map,

is there guarantee that we can lift it to something? Moreover, if I have fixed a

“copy” of

I like, can I also lift my map to that copy? We will later come up

with a general criterion for when lifts exist. However, it turns out homotopies

can always be lifted.

Lemma (Homotopy lifting lemma). Let

X → X

be a covering space,

Y × I → X

be a homotopy from

. Let

be a lift of

. Then there

exists a unique homotopy

H : Y × I →

X such that

(i)

H( · , 0) =

; and

(ii)

H is a lift of H, i.e. p ◦

H = H.

This lemma might be difficult to comprehend at first. We can look at the

special case where

∗

. Then a homotopy is just a path. So the lemma

specializes to

Lemma (Path lifting lemma). Let

X → X

be a covering space,

I → X

path, and

˜x

∈

such that

(

˜x

) =

(0). Then there exists a unique path

˜γ : I →

X such that

(i) ˜γ(0) = ˜x

; and

(ii) ˜γ is a lift of γ, i.e. p ◦ ˜γ = γ.

This is exactly the picture we were drawing before. We just have to start

at a point

˜x

, and then everything is determined because locally, everything

upstairs in

is just like

. Note that we have already proved uniqueness. So

we just need to prove existence.

In theory, it makes sense to prove homotopy lifting, and path lifting comes

immediately as a corollary. However, the proof of homotopy lifting is big and

scary. So instead, we will prove path lifting, which is something we can more

easily visualize and understand, and then use that to prove homotopy lifting.

Proof. Let

S = {s ∈ I : ˜γ exists on [0, s] ⊆ I}.

Observe that

(i) 0 ∈ S.

(ii) S

is open. If

s ∈ S

and

˜γ

(

)

∈ V

⊆ p

−1

(

), we can define

˜γ

on some

small neighbourhood of s by

˜γ(t) = (p|

)

−1

◦ γ(t)

(iii) S

is closed. If

s 6∈ S

, then pick an evenly covered neighbourhood

(

Suppose

((

s − ε, s

))

⊆ U

. So

s −

6∈ S

. So (

s −

∩ S

∅

. So

closed.

Since

is both open and closed, and is non-empty, we have

. So

˜γ

exists.

How can we promote this to a proof of the homotopy lifting lemma? At

every point

y ∈ Y

, we know what to do, since we have path lifting. So

(

y, ·

)

is defined. So the thing we have to do is to show that this is continuous. Steps

of the proof are

(i)

Use compactness of

to argue that the proof of path lifting works on small

neighbourhoods in Y .

(ii)

For each

, we pick an open neighbourhood

, and define a good path

lifting on U × I.

(iii)

By uniqueness of lifts, these path liftings agree when they overlap. So we

have one big continuous lifting.

With the homotopy lifting lemma in our toolkit, we can start to use it to

do stuff. So far, we have covering spaces and fundamental groups. We are now

going to build a bridge between these two, and show how covering spaces can be

used to reflect some structures of the fundamental group.

At least one payoff of this work is that we are going to exhibit some non-trivial

fundamental groups.

We have just showed that we are allowed to lift homotopies. However, what

we are really interested in is homotopy as paths. The homotopy lifting lemma

does not tell us that the lifted homotopy preserves basepoints. This is what we

are going to show.

Corollary. Suppose

γ, γ

I → X

are paths

and

˜γ, ˜γ

I →

are lifts

of γ and γ

respectively, both starting at ˜x

∈ p

−1

γ ' γ

as paths, then

˜γ

and

˜γ

are homotopic as paths. In particular,

˜γ(1) = ˜γ

(1).

Note that if we cover the words “as paths” and just talk about homotopies,

then this is just the homotopy lifting lemma. So we can view this as a stronger

form of the homotopy lifting lemma.

Proof.

The homotopy lifting lemma gives us an

, a lift of

with

(

· ,

0) =

˜γ

lift

˜γ

˜x

In this diagram, we by assumption know the bottom of the

square is

˜γ

. To

show that this is a path homotopy from

˜γ

, we need to show that the other

edges are c

˜x

, c

˜x

and ˜γ

respectively.

Now

(

· ,

1) is a lift of

(

· ,

1) =

, starting at

˜x

. Since lifts are unique,

we must have

(

· ,

1) =

˜γ

. So this is indeed a homotopy between

˜γ

and

˜γ

Now we need to check that this is a homotopy of paths.

We know that

, ·

) is a lift of

, ·

) =

. We are aware of one lift of

, namely

˜x

. By uniqueness of lifts, we must have

, ·

) =

˜x

. Similarly,

H(1, · ) = c

˜x

. So this is a homotopy of paths.

So far, our picture of covering spaces is like this:

Except. . . is it? Is it possible that we have four copies of

but just three copies

? This is obviously possible if

is not path connected — the component

containing

and the one containing

are completely unrelated. But what if

X is path connected?

Corollary. If

is a path connected space,

, x

∈ X

, then there is a bijection

−1

) → p

−1

Proof.

Let

be a path. We want to use this to construct a bijection

between each preimage of

and each preimage of

. The obvious thing to do

is to use lifts of the path γ.

Define a map

−1

(

)

→ p

−1

(

) that sends

˜x

to the end point of the

unique lift of γ at ˜x

The inverse map is obtained by replacing

with

−1

, i.e.

−1

. To show this

is an inverse, suppose we have some lift

˜γ

˜x

, so that

(

˜x

) =

˜x

. Now

notice that

˜γ

−1

is a lift of

−1

starting at

˜x

and ending at

˜x

. So

−1

(

˜x

) =

˜x

So f

−1

is an inverse to f

, and hence f

is bijective.

Definition (

-sheeted). A covering space

X → X

of a path-connected space

X is n-sheeted if |p

−1

(x)| = n for any (and hence all) x ∈ X.

Each covering space has a number associated to it, namely the number of

sheets. Is there any number we can assign to fundamental groups? Well, the

index of a subgroup might be a good candidate. We’ll later see if this is the case.

One important property of covering spaces is the following:

Lemma. If p :

X → X is a covering map and ˜x

∈

X, then

∗

: π

(

X, ˜x

) → π

(X, x

)

is injective.

Proof.

To show that a group homomorphism

∗

is injective, we have to show

that if p

∗

(x) is trivial, then x must be trivial.

Consider a based loop

˜γ

. We let

p ◦ ˜γ

. If

is trivial, i.e.

γ ' c

paths, the homotopy lifting lemma then gives us a homotopy upstairs between

˜γ

and c

˜x

. So ˜γ is trivial.

As we have originally said, our objective is to make our fundamental group

act on something. We are almost there already.

Let’s look again at the proof that there is a bijection between

−1

(

) and

−1

(

). What happens if

is a loop? For any

˜x

∈ p

−1

(

), we can look at

the end point of the lift. This end point may or may not be our original

˜x

. So

each loop γ “moves” our ˜x

to another ˜x

However, we are not really interested in paths themselves. We are interested

in equivalence classes of paths under homotopy of paths. However, this is fine.

is homotopic to

, then this homotopy can be lifted to get a homotopy

between

˜γ

and

˜γ

. In particular, these have the same end points. So each (based)

homotopy class gives a well-defined endpoint.

˜γ

˜x

Now this gives an action of

(

X, x

) on

−1

(

)! Note, however, that this will

not be the sort of actions we are familiar with. We usually work with left-actions,

where the group acts on the left, but now we will have right-actions, which

may confuse you a lot. To see this, we have to consider what happens when we

perform two operations one after another, which you shall check yourself. We

write this action as ˜x

· [γ].

When we have an action, we are interested in two things — the orbits, and

the stabilizers. This is what the next lemma tells us about.

Lemma. Suppose X is path connected and x

∈ X.

(i)

The action of

(

X, x

) on

−1

(

) is transitive if and only if

is path

connected. Alternatively, we can say that the orbits of the action correspond

to the path components.

(ii) The stabilizer of ˜x

∈ p

−1

) is p

∗

(π

(

X, ˜x

)) ⊆ π

(X, x

(iii) If

X is path connected, then there is a bijection

∗

(π

(

X, ˜x

))\π

(X, x

) → p

−1

Note that

∗

(

X, ˜x

))

\π

(

X, x

) is not a quotient, but simply the set of

cosets. We write it the “wrong way round” because we have right cosets

instead of left cosets.

Note that this is great! If we can find a covering space

and a point

such that

−1

(

) is non-trivial, then we immediately know that

(

X, x

) is

non-trivial!

Proof.

(i)

˜x

, ˜x

∈ p

−1

(

), then since

is path connected, we know that there

is some

˜γ

˜x

. Then we can project this to

p ◦ ˜γ

. Then

a path from

, i.e. a loop. Then by the definition of the action,

˜x

· [γ] = ˜γ(1) = ˜x

(ii)

Suppose [

]

∈ stab

(

˜x

). Then

˜γ

is a loop based at

˜x

. So

˜γ

defines

[˜γ] ∈ π

(

X, ˜x

) and γ = p ◦ ˜γ.

(iii) This follows directly from the orbit-stabilizer theorem.

We now want to use this to determine that the fundamental group of a space

is non-trivial. We can be more ambitious, and try to actually find

(

X, x

In the best possible scenario, we would have

(

X, ˜x

) trivial. Then we have a

bijection between

(

X, x

) and

−1

(

). In other words, we want our covering

space

X to be simply connected.

Definition (Universal cover). A covering map

X → X

is a universal cover

X is simply connected.

We will look into universal covers in depth later and see what they really are.

Corollary. If

X → X

is a universal cover, then there is a bijection



(X, x

) → p

−1

Note that the orbit-stabilizer theorem does not provide a canonical bijection

between

−1

(

) and

∗

(

X, ˜x

)

\π

(

X, x

). To obtain a bijection, we need to

pick a starting point

˜x

∈ p

−1

(

). So the above bijection



depends on a choice

of ˜x

3.2

The fundamental group of the circle and its applica-

tions

Finally, we can exhibit a non-trivial fundamental group. We are going to consider

the space S

and a universal covering R.

−1

(1)

Then our previous corollary gives

Corollary. There is a bijection π

, 1) → p

−1

(1) = Z.

What’s next? We just know that

(

1) is countably infinite, but can we

work out the group structure?

We can, in fact, prove a stronger statement:

Theorem. The map  : π

, 1) → p

−1

(1) = Z is a group isomorphism.

Proof.

We know it is a bijection. So we need to check it is a group homomorphism.

The idea is to write down representatives for what we think the elements should

be.

˜u

−2

−1

Let

˜u

I → R

be defined by

t 7→ nt

, and let

p ◦ ˜u

. Since

is simply

connected, there is a unique homotopy class between any two points. So for any

[

]

∈ π

(

1), if

˜γ

is the lift to

at 0 and

˜γ

(1) =

, then

˜γ ' ˜u

as paths. So

[γ] = [u

To show that this has the right group operation, we can easily see that

· u

= ˜u

m+n

, since we are just moving by n + m in both cases. Therefore

([u

][u

]) = ([u

· u

]) = m + n = ([u

m+n

]).

So  is a group isomorphism.

What have we done? In general, we might be given a horrible, crazy loop

. It would be rather difficult to work with it directly in

. So we pull it

up to the universal covering

. Since

is nice and simply connected, we can

easily produce a homotopy that “straightens out” the path. We then project

this homotopy down to S

, to get a homotopy from γ to u

It is indeed possible to produce a homotopy directly inside

from each loop

to some

, but that would be tedious work that involves messing with a lot of

algebra and weird, convoluted formulas.

With the fundamental group of the circle, we do many things. An immediate

application is that we can properly define the “winding number” of a closed

curve. Since

C \ {

}

is homotopy equivalent to

, its fundamental group is

as well. Any closed curve

→ C \ {

}

thus induces a group homomorphism

Z → Z

. Any such group homomorphism must be of the form

t 7→ nt

, and the

winding number is given by

. If we stare at it long enough, it is clear that this

is exactly the number of times the curve winds around the origin.

Also, we have the following classic application:

Theorem (Brouwer’s fixed point theorem). Let

{

(

x, y

)

∈ R

≤

}

be the unit disk. If

→ D

is continuous, then there is some

x ∈ D

such

that f (x) = x.

Proof. Suppose not. So x 6= f(x) for all x ∈ D

f(x)

g(x)

We define

→ S

as in the picture above. Then we know that

continuous and

is a retraction from

onto

. In other words, the following

composition is the identity:

Then this induces a homomorphism of groups whose composition is the identity:

Z {0} Z

∗

But this is clearly nonsense! So we must have had a fixed point.

But we have a problem. What about

? Can we prove a similar theorem?

Here the fundamental group is of little use, since we can show that the funda-

mental group of

for

n ≥

2 is trivial. Later in the course, we will be able to

prove this theorem for higher dimensions, when we have developed more tools

to deal with stuff.

3.3 Universal covers

We have defined universal covers mysteriously as covers that are simply connected.

We have just shown that

R → S

is a universal cover. In general, what do

universal covers look like?

Let’s consider a slightly more complicated example. What would be a

universal cover of the torus

× S

? An obvious guess would be

p × p

R × R →

× S

. How can we visualize this?

First of all, how can we visualize a torus? Often, we just picture it as the

surface of a doughnut. Alternatively, we can see it as a quotient of the square,

where we identify the following edges:

Then what does it feel like to live in the torus? If you live in a torus and look

around, you don’t see a boundary. The space just extends indefinitely for ever,

somewhat like

. The difference is that in the torus, you aren’t actually seeing

free space out there, but just seeing copies of the same space over and over again.

If you live inside the square, the universe actually looks like this:

As we said, this looks somewhat likes

, but we know that this is not

since we can see some symmetry in this space. Whenever we move one unit

horizontally or vertically, we get back to “the same place”. In fact, we can move

horizontally by

units and vertically by

units, for any

n, m ∈ Z

, and still get

back to the same place. This space has a huge translation symmetry. What is

this symmetry? It is exactly Z × Z.

We see that if we live inside the torus

× S

, it feels like we are actually

living in the universal covering space

R × R

, except that we have an additional

symmetry given by the fundamental group Z × Z.

Hopefully, you are convinced that universal covers are nice. We would like to

say that universal covers always exist. However, this is not always true.

Firstly, we should think — what would having a universal cover imply?

Suppose

has a universal cover

. Pick any point

∈ X

, and pick an

evenly covered neighbourhood

. This lifts to some

U ⊆

. If we draw a

teeny-tiny loop

around

inside

, we can lift this

˜γ

. But we know

that

is simply connected. So

˜γ

is homotopic to the constant path. Hence

is also homotopic to the constant path. So all loops (contained in

) at

are

homotopic to the constant path.

It seems like for every

∈ X

, there is some neighbourhood of

that is

simply connected. Except that’s not what we just showed above. The homotopy

from

˜γ

to the constant path is a homotopy in

, and can pass through anything

, not just

. Hence the homotopy induced in

is also a homotopy in

not a homotopy in

. So

itself need not be simply connected. What we have

is a slightly weaker notion.

Definition (Locally simply connected).

is locally simply connected if for all

∈ X, there is some neighbourhood U of x

such that U is simply connected.

As we mentioned, what we actually want is a weaker condition.

Definition (Semi-locally simply connected).

is semi-locally simply connected

if for all

∈ X

, there is some neighbourhood

such that any loop

based at x

is homotopic to c

as paths in X.

We have just argued that if a universal cover

X → X

exists, then

semi-locally simply connected. This is really not interesting, since we don’t care

if a space is semi-locally simply connected. What is important is that this is the

other direction. We still need one more additional condition:

Definition (Locally path connected). A space

is locally path connected if for

any point

and any neighbourhood

, there is some open path connected

U ⊆ V such that x ∈ U.

It is important to note that a path connected space need not be locally path

connected. It is an exercise for the reader to come up with a counterexample.

Theorem. If

is path connected, locally path connected and semi-locally

simply connected, then X has a universal covering.

Note that we can alternatively define a universal covering as a covering space

that is also a covering space of all other covers of

. If we use this definition,

then we can prove this result easily using Zorn’s lemma. However, that proof

is not too helpful since it does not tell us where the universal covering comes

from. Instead, we will provide a constructive proof (sketch) that will hopefully

be more indicative of what universal coverings are like.

Proof.

(idea) We pick a basepoint

∈ X

for ourselves. Suppose we have a

universal covering

. Then this lifts to some

˜x

. If we have any other point

˜x ∈

, since

should be path connected, there is a path

˜α

˜x

. If we

have another path, then since

is simply connected, the paths are homotopic.

Hence, we can identify each point in

X with a path from ˜x

, i.e.

{points of

X} ←→ {paths ˜α from ˜x

∈

X}/'.

This is not too helpful though, since we are defining

in terms of things in

However, by path lifting, we know that paths

˜α

from

˜x

biject with paths

from

. Also, by homotopy lifting, homotopies of paths in

can be

lifted to homotopies of paths in

X. So we have

{points of

X} ←→ {paths α from x

∈ X}/'.

So we can produce our

X by picking a basepoint x

∈ X, and defining

X = {paths α : I → X such that α(0) = x

}/'.

The covering map p :

X → X is given by [α] 7→ α(1).

One then has to work hard to define the topology, and then show this is

simply connected.

3.4 The Galois correspondence

Recall that at the beginning, we wanted to establish a correspondence between

covering spaces and fundamental groups. We have already established the result

that covering maps are injective on

. Therefore, given a (based) covering

space

: (

X, ˜x

)

→

(

X, x

), we can give a subgroup

∗

(

X, ˜x

)

≤ π

(

X, x

). It

turns out that as long as we define carefully what we mean for based covering

spaces to be “the same”, this is a one-to-one correspondence — each subgroup

corresponds to a covering space.

We can have the following table of correspondences:

Covering spaces Fundamental group

(Based) covering spaces ←→ Subgroups of π

Number of sheets ←→ Index

Universal covers ←→ Trivial subgroup

We now want to look at some of these correspondences.

Recall that we have shown that

(

X, x

) acts on

−1

(

). However, this is

not too interesting an action, since

−1

(

) is a discrete group with no structure.

Having groups acting on a cube is fun because the cube has some structure. So

we want something more “rich” for π

(X, x

) to act on.

We note that we can make

(

X, x

) “act on” the universal cover. How?

Recall that in the torus example, each item in the fundamental group corresponds

to translating the whole universal covering by some amount. In general, a point

can be thought of as a path

starting from

. Then it is easy to

make a loop

act on this: use the concatenation

γ ·α

: [

]

[

] = [

γ ·α

˜x

˜γ

We will use this idea and return to the initial issue of making subgroups corre-

spond to covering spaces. We want to show that this is surjective — every sub-

group arises from some cover. We want to say “For any subgroup

H ≤ π

(

X, x

there is a based covering map

: (

X, ˜x

)

→

(

X, x

) such that

∗

(

X, ˜x

) =

”.

Except, this cannot possibly be true, since by taking the trivial subgroup, this

would imply that there is a universal covering for every space. So we need some

additional assumptions.

Proposition. Let

be a path connected, locally path connected and semi-

locally simply connected space. For any subgroup

H ≤ π

(

X, x

), there is a

based covering map p : (

X, ˜x

) → (X, x

) such that p

∗

(

X, ˜x

) = H.

Proof.

Since

is a path connected, locally path connected and semi-locally

simply connected space, let

be a universal covering. We have an intermediate

group

such that

(

X, ˜x

) = 1

≤ H ≤ π

(

X, x

). How can we obtain a

corresponding covering space?

Note that if we have

and we want to recover

, we can quotient

by the

action of

(

X, x

). Since

(

X, x

) acts on

, so does

H ≤ π

(

X, x

). Now we

can define our covering space by taking quotients. We define

∼

to be

the orbit relation for the action of

, i.e.

˜x ∼

˜y

if there is some

h ∈ H

such

that ˜y = h˜x. We then let

X be the quotient space

X/∼

We can now do the messy algebra to show that this is the covering space we

want.

We have just showed that every subgroup comes from some covering space,

i.e. the map from the set of covering spaces to the subgroups of

is surjective.

Now we want to prove injectivity. To do so, we need a generalization of the

homotopy lifting lemma.

Suppose we have path-connected spaces (

Y, y

), (

X, x

) and (

X, ˜x

), with

: (

Y, y

)

→

(

X, x

) a continuous map,

: (

X, ˜x

)

→

(

X, x

) a covering map.

When does a lift of

: (

Y, y

)

→

(

X, ˜x

) exist? The answer is given by the

lifting criterion.

Lemma (Lifting criterion). Let

: (

X, ˜x

)

→

(

X, x

) be a covering map of path-

connected based spaces, and (

Y, y

) a path-connected, locally path connected

based space. If

: (

Y, y

)

→

(

X, x

) is a continuous map, then there is a (unique)

lift

: (

Y, y

)

→

(

X, ˜x

) such that the diagram below commutes (i.e.

p ◦

(

X, ˜x

)

(Y, y

) (X, x

)

if and only if the following condition holds:

∗

(Y, y

) ≤ p

∗

(

X, ˜x

Note that uniqueness comes from the uniqueness of lifts. So this lemma is

really about existence.

Also, note that the condition holds trivially when

is simply connected, e.g.

when it is an interval (path lifting) or a square (homotopy lifting). So paths and

homotopies can always be lifted.

Proof.

One direction is easy: if

exists, then

p ◦

. So

∗

◦

∗

. So we

know that im f

∗

⊆ im p

∗

. So done.

In the other direction, uniqueness follows from the uniqueness of lifts. So we

only need to prove existence. We define

f as follows:

Given a

y ∈ Y

, there is some path

. Then

maps this to

(

) in

. By path lifting, this path lifts uniquely to

. Then

we set

(

) =

(1). Note that if

exists, then this must be what

sends

to.

What we need to show is that this is well-defined.

Suppose we picked a different path

. Then this

would have

differed from α

by a loop γ in Y .

Our condition that

∗

(

Y, y

)

≤ p

∗

(

X, ˜x

) says

f ◦ γ

is the image of a

loop in

. So

and

also differ by a loop in

, and hence have the same

end point. So this shows that

f is well-defined.

Finally, we show that

is continuous. First, observe that any open set

U ⊆

can be written as a union of

such that

V → p

(

) is a homeomorphism.

Thus, it suffices to show that if

V → p

(

) =

is a homeomorphism, then

−1

(

V ) is open.

Let

y ∈

−1

(

), and let

(

). Since

−1

(

) is open and

is locally

path-connected, we can pick an open

W ⊆ f

−1

(

) such that

y ∈ W

and

path connected. We claim that W ⊆

−1

(

V ).

Indeed, if

z ∈ W

, then we can pick a path

from

. Then

sends

this to a path from

(

). The lift of this path to

is given by

−1

(

)),

whose end point is p|

−1

(f(z)) ∈

V . So it follows that

f(z) = p|

−1

(f(z)) ∈

V .

Now we prove that every subgroup of

comes from exactly one covering

space. What this statement properly means is made precise in the following

proposition:

Proposition. Let (

X, x

), (

, ˜x

), (

, ˜x

) be path-connected based spaced,

and p

: (

, ˜x

) → (X, x

) be covering maps. Then we have

1∗

(

, ˜x

) = p

2∗

(

, ˜x

)

if and only if there is some homeomorphism

such that the following diagram

commutes:

(

, ˜x

) (

, ˜x

)

(X, x

)

i.e. p

= p

◦ h.

Note that this is a stronger statement than just saying the two covering

spaces are homeomorphic. We are saying that we can find a nice homeomorphism

that works well with the covering map p.

Proof.

If such a homeomorphism exists, then clearly the subgroups are equal. If

the subgroups are equal, we rotate our diagram a bit:

(

, ˜x

)

(

, ˜x

) (X, x

)

h=˜p

Then

˜p

exists by the lifting criterion. By symmetry, we can get

−1

˜p

To show

˜p

is indeed the inverse of

˜p

, note that

˜p

◦ ˜p

is a lift of

◦ ˜p

Since

is also a lift, by the uniqueness of lifts, we know

˜p

◦ ˜p

is the identity

map. Similarly, ˜p

◦ ˜p

is also the identity.

(

, ˜x

)

(

, ˜x

) (

, ˜x

) (X, x

)

˜p

Now what we would like to do is to forget about the basepoints. What

happens when we change base points? Recall that the effect of changing base

points is that we will conjugate our group. This doesn’t actually change the

group itself, but if we are talking about subgroups, conjugation can send a

subgroup into a different subgroup. Hence, if we do not specify the basepoint,

we don’t get a subgroup of π

, but a conjugacy class of subgroups.

Proposition. Unbased covering spaces correspond to conjugacy classes of

subgroups.

4 Some group theory

Algebraic topology is about translating topology into group theory. Unfortu-

nately, you don’t know group theory. Well, maybe you do, but not the right

group theory. So let’s learn group theory!

4.1 Free groups and presentations

Recall that in IA Groups, we defined, say, the dihedral group to be

= hr, s | r

= s

= e, srs = r

−1

What does this expression actually mean? Can we formally assign a meaning to

this expression?

We start with a simple case — the free group. This is, in some sense, the

“freest” group we can have. It is defined in terms of an alphabet and words.

Definition (Alphabet and words). We let

α ∈

}

be our alphabet,

and we have an extra set of symbols

−1

α ∈

}

. We assume that

S ∩ S

−1

= ∅. What do we do with alphabets? We write words with them!

We define

∗

to be the set of words over

S ∪ S

−1

, i.e. it contains

-tuples

· · · x

for any 0 ≤ n < ∞, where each x

∈ S ∪ S

−1

Example. Let

{a, b}

. Then words could be the empty word

∅

, or

aba

−1

, or

−1

aaaaabbbb

, etc. We are usually lazy and write

−1

aaaaabbbb

as aa

−1

When we see things like

−1

, we would want to cancel them. This is called

elementary reduction.

Definition (Elementary reduction). An elementary reduction takes a word

−1

v and gives uv, or turns us

−1

v into uv.

Since each reduction shortens the word, and the word is finite in length, we

cannot keep reducing for ever. Eventually, we reach a reduced state.

Definition (Reduced word). A word is reduced if it does not admit an elementary

reduction.

Example. ∅, a, aba

−1

are reduced words, while aa

−1

aaaaabbbb is not.

Note that there is an inclusion map

S → S

∗

that sends the symbol

to the

word s

Definition (Free group). The free group on the set

, written

(

), is the set

of reduced words on S

∗

together with some operations:

(i)

Multiplication is given by concatenation followed by elementary reduction

to get a reduced word. For example, (

aba

−1

)

(

bab

) =

aba

−1

bab

(ii) The identity is the empty word ∅.

(iii) The inverse of x

· · · x

is x

−1

· · · x

−1

, where, of course, (s

−1

)

−1

= s

The elements of S are called the generators of F (S).

Note that we have not showed that multiplication is well-defined — we might

reduce the same word in different ways and reach two different reduced words.

We will show that this is indeed well-defined later, using topology!

Some people like to define the free group in a different way. This is a cleaner

way to define the free group without messing with alphabets and words, but is

(for most people) less intuitive. This definition also does not make it clear that

the free group

(

) of any set

exists. We will state this definition as a lemma.

Lemma. If

is a group and

S → G

is a set map, then there exists a unique

homomorphism f : F (S) → G such that the following diagram commutes:

F (S)

S G

where the arrow not labeled is the natural inclusion map that sends

(as a

symbol from the alphabet) to s

(as a word).

Proof.

Clearly if

exists, then

must send each

(

) and

−1

(

)

−1

Then the values of f on all other elements must be determined by

f(x

· · · x

) = f(x

) · · · f(x

)

since

is a homomorphism. So if

exists, it must be unique. So it suffices to

show that this f is a well-defined homomorphism.

This is well-defined if we define

(

) to be the set of all reduced words, since

each reduced word has a unique representation (since it is defined to be the

representation itself).

To show this is a homomorphism, suppose

x = x

· · · x

· · · a

, y = a

−1

· · · a

−1

· · · y

where y

6= x

−1

. Then

xy = x

· · · x

· · · y

Then we can compute

f(x)f(y) =



φ(x

) · · · φ(x

)φ(a

) · · · φ(a

)



φ(a

)

−1

· · · φ(a

)

−1

φ(y

) · · · φ(y

)



= φ(x

) · · · φ(x

) · · · φ(y

)

= f(xy).

So f is a homomorphism.

We call this a “universal property” of

(

). We can show that

(

) is the

unique group satisfying the conditions of this lemma (up to isomorphism), by

taking G = F (S) and using the uniqueness properties.

Definition (Presentation of a group). Let

be a set, and let

R ⊆ F

(

) be any

subset. We denote by

hhRii

the normal closure of

, i.e. the smallest normal

subgroup of F (S) containing R. This can be given explicitly by

hhRii =

(

i=1

−1

: n ∈ N, r

∈ R, g

∈ F (S)

)

Then we write

hS | Ri = F (S)/hhRii.

This is just the usual notation we have for groups. For example, we can write

= hr, s | r

, s

, srsri.

Again, we can define this with a universal property.

Lemma. If

is a group and

S → G

is a set map such that

(

) = 1 for all

r ∈ R

(i.e. if

±1

· · · s

±1

, then

(

) =

(

)

±1

(

)

±1

· · · φ

(

)

±1

= 1),

then there exists a unique homomorphism

hS | Ri → G

such that the

following triangle commutes:

hS | Ri

S G

Proof is similar to the previous one.

In some sense, this says that all the group

hS | Ri

does is it satisfies the

relations in R, and nothing else.

Example (The

stupid

canonical presentation). Let

be a group. We can view

the group as a set, and hence obtain a free group

(

). There is also an obvious

surjection

G → G

. Then by the universal property of the free group, there is a

surjection

(

)

→ G

. Let

ker f

hhRii

. Then

hG | Ri

is a presentation

for G, since the first isomorphism theorem says

∼

F (G)/ ker f.

This is a really stupid example. For example, even the simplest non-trivial

group

2 will be written as a quotient of a free group with two generators.

However, this tells us that every group has a presentation.

Example. ha, b | bi

∼

hai

∼

Example. With a bit of work, we can show that ha, b | ab

−3

, ba

−2

i = Z/5.

4.2 Another view of free groups

Recall that we have not yet properly defined free groups, since we did not show

that multiplication is well-defined. We are now going to do this using topology.

Again let

be a set. For the following illustration, we will just assume

S = {a, b}, but what we will do works for any set S. We define X by

We call this a “rose with 2 petals”. This is a cell complex, with one 0-cell and

|S|

1-cells. For each

s ∈ S

, we have one 1-cell,

, and we fix a path

: [0

→ e

that goes around the 1-cell once. We will call the 0-cells and 1-cells vertices and

edges, and call the whole thing a graph.

What’s the universal cover of

? Since we are just lifting a 1-complex, the

result should be a 1-complex, i.e. a graph. Moreover, this graph is connected

and simply connected, i.e. it’s a tree. We also know that every vertex in the

universal cover is a copy of the vertex in our original graph. So it must have 4

edges attached to it. So it has to look like something this:

˜x

, we know that at each vertex, there should be an edge labeled

going in;

an edge labeled

going out; an edge labeled

going in; an edge labeled

going

out. This should be the case in

X as well. So

X looks like this:

˜x

0 aa

The projection map is then obvious — we send all the vertices in

∈ X

and then the edges according to the labels they have, in a way that respects the

direction of the arrow. It is easy to show this is really a covering map.

We are now going to show that this tree “is” the free group. Notice that

every word

w ∈ S

∗

denotes a unique “edge path” in

starting at

˜x

, where an

edge path is a sequence of oriented edges

˜e

, · · · , ˜e

such that the “origin” of

˜e

i+1

is equal to the “terminus” of ˜e

For example, the following path corresponds to w = abb

−1

˜x

We can note a few things:

(i)

X is connected. So for all ˜x ∈ p

−1

), there is an edge-path ˜γ : ˜x

˜x.

(ii)

If an edge-path

˜γ

fails to be locally injective, we can simplify it. How can

an edge fail to be locally injective? It is fine if we just walk along a path,

since we are just tracing out a line. So it fails to be locally injective if two

consecutive edges in the path are the same edge with opposite orientations:

˜e

i−1

˜e

i+1

˜e

i+2

We can just remove the redundant lines and get

˜e

i−1

˜e

i+2

This reminds us of two things — homotopy of paths and elementary

reduction of words.

(iii)

Each point

˜x ∈ p

−1

(

) is joined to

˜x

by a unique locally injective

edge-path.

(iv)

For any

w ∈ S

∗

, from (ii), we know that

˜γ

is locally injective if and only if

w is reduced.

We can thus conclude that there are bijections

F (S) p

−1

) π

(X, x

)

that send

˜x

to the word

w ∈ F

(

) such that

˜γ

is a locally injective edge-path

to ˜x

˜x; and ˜x to [γ] ∈ π

(X, x

) such that ˜x

· [γ] = ˜x.

So there is a bijection between

(

) and

(

X, x

). It is easy to see that the

operations on

(

) and

(

X, x

) are the same, since they are just concatenating

words or paths. So this bijection identifies the two group structures. So this

induces an isomorphism F (S)

∼

(X, x

4.3 Free products with amalgamation

We managed to compute the fundamental group of the circle. But we want

to find the fundamental group of more things. Recall that at the beginning,

we defined cell complexes, and said these are the things we want to work with.

Cell complexes are formed by gluing things together. So we want to know what

happens when we glue things together.

Suppose a space

is constructed from two spaces

A, B

by gluing (i.e.

A ∪ B

). We would like to describe

(

) in terms of

(

) and

(

). To

understand this, we need to understand how to “glue” groups together.

Definition (Free product). Suppose we have groups

| R

i, G

i, where we assume S

∩ S

= ∅. The free product G

∗ G

is defined to be

∗ G

= hS

∪ S

| R

∪ R

This is not a really satisfactory definition. A group can have many different

presentations, and it is not clear this is well-defined. However, it is clear that

this group exists. Note that there are natural homomorphisms

→ G

∗ G

that send generators to generators. Then we can have the following universal

property of the free product:

Lemma.

∗ G

is the group such that for any group

and homomorphisms

→ K

, there exists a unique homomorphism

∗ G

→ K

such that

the following diagram commutes:

∗ G

Proof.

It is immediate from the universal property of the definition of presenta-

tions.

Corollary. The free product is well-defined.

Proof.

The conclusion of the universal property can be seen to characterize

∗ G

up to isomorphism.

Again, we have a definition in terms of a concrete construction of the group,

without making it clear this is well-defined; then we have a universal property

that makes it clear this is well-defined, but not clear that the object actually

exists. Combining the two would give everything we want.

However, this free product is not exactly what we want, since there is little

interaction between

and

. In terms of gluing spaces, this corresponds to

gluing

and

when

A ∩ B

is trivial (i.e. simply connected). What we really

need is the free product with amalgamation, as you might have guessed from

the title.

Definition (Free product with amalgamation). Suppose we have groups

and H, with the following homomorphisms:

H G

The free product with amalgamation is defined to be

∗

= G

∗ G

/hh{(j

◦ i

(h))

−1

◦ i

)(h) : h ∈ H}ii.

Here we are attempting to identify things “in H” as the same, but we need

to use the maps

and

to map the things from

∗ G

. So we want to

say “for any

◦ i

(

) =

◦ i

(

)”, or (

◦ i

(

))

−1

(

◦ i

)(

) =

. So we

quotient by the (normal closure) of these things.

We have the universal property

Lemma.

∗

is the group such that for any group

and homomorphisms

→ K

, there exists a unique homomorphism

∗

→ K

such that the

following diagram commutes:

H G

∗

This is the language we will need to compute fundamental groups. These

definitions will (hopefully) become more concrete as we see more examples.

5 Seifert-van Kampen theorem

5.1 Seifert-van Kampen theorem

The Seifert-van Kampen theorem is the theorem that tells us what happens

when we glue spaces together.

Here we let X = A ∪ B, where A, B, A ∩ B are path-connected.

A B

We pick a basepoint

∈ A ∩ B

for convenience. Since we like diagrams, we can

write this as a commutative diagram:

A ∩ B B

A X

where all arrows are inclusion (i.e. injective) maps. We can consider what

happens when we take the fundamental groups of each space. Then we have the

induced homomorphisms

(A ∩ B, x

) π

(B, x

)

(A, x

) π

(X, x

)

We might guess that π

(X, x

) is just the free product with amalgamation

(X, x

) = π

(A, x

) ∗

(A∩B,x

)

(B, x

The Seifert-van Kampen theorem says that, under mild hypotheses, this guess is

correct.

Theorem (Seifert-van Kampen theorem). Let

A, B

be open subspaces of

such

that

A ∪ B

, and

A, B, A ∩ B

are path-connected. Then for any

∈ A ∩ B

we have

(X, x

) = π

(A, x

) ∗

(A∩B,x

)

(B, x

Note that by the universal property of the free product with amalgamation,

we by definition know that there is a unique map

(

A, x

)

∗

(A∩B,x

)

(

B, x

)

→

(X, x

). The theorem asserts that this map is an isomorphism.

Proof is omitted because time is short.

Example. Consider a higher-dimensional sphere

{

∈ R

n+1

= 1

}

for

n ≥ 2. We want to find π

The idea is to write

as a union of two open sets. We let

= e

∈ S

⊆

n+1

be the North pole, and

−

be the South pole. We let

\ {n}

and

\ {s}

. By stereographic projection, we know that

A, B

∼

. The

hard part is to understand the intersection.

To do so, we can draw a cylinder

n−1

(

−

1), and project our

A ∩ B

onto the cylinder. We can similarly project the cylinder onto

A ∩ B

. So

A ∩ B

∼

n−1

× (−1, 1) ' S

n−1

, since (−1, 1) is contractible.

We can now apply the Seifert-van Kampen theorem. Note that this works only

if S

n−1

is path-connected, i.e. n ≥ 2. Then this tells us that

)

∼

) ∗

n−1

)

∼

1 ∗

n−1

)

It is easy to see this is the trivial group. We can see this directly from the

universal property of the amalgamated free product, or note that it is the quotient

of 1 ∗ 1, which is 1.

So for n ≥ 2, π

)

∼

We have found yet another simply connected space. However, this is unlike

our previous examples. Our previous spaces were simply connected because they

were contractible. However, we will later see that

is not contractible. So this

is genuinely a new, interesting example.

Why did we go though all this work to prove that

(

)

∼

1? It feels like

we can just prove this directly — pick a point that is not in the curve as the

North pole, project stereographically to

, and contract it there. However,

the problem is that space-filling curves exist. We cannot guarantee that we can

pick a point not on the curve! It is indeed possible to prove directly that given

any curve on

(with

n ≥

1), we can deform it slightly so that it is no longer

surjective. Then the above proof strategy works. However, using Seifert-van

Kampen is much neater.

Example (

). Recall that

∼

/{± id}

, and the quotient map

→

is a covering map. Now that we have proved that

is simply connected,

we know that S

is a universal cover of RP

For any x

∈ RP

, we have a bijection

(RP

, x

) ↔ p

−1

Since

−1

(

) has two elements by definition, we know that

|π

(

, x

)

= 2.

So π

(RP

, x

) = Z/2.

You will prove a generalization of this in example sheet 2.

Example (Wedge of two circles). We are going to consider the operation of

wedging. Suppose we have two topological spaces, and we want to join them

together. The natural way to join them is to take the disjoint union. What if

we have based spaces? If we have (

X, x

) and (

Y, y

), we cannot just take the

disjoint union, since we will lose our base point. What we do is take the wedge

sum, where we take the disjoint union and then identify the base points:

X Y

X ∧ Y

∼ y

Suppose we take the wedge sum of two circles

∧ S

. We would like to pick

A, B

to be each of the circles, but we cannot, since

and

have to be open.

So we take slightly more, and get the following:

A B

Each of A and B now look like this:

We see that both

and

retract to the circle. So

(

)

∼

(

)

∼

, while

A ∩ B is a cross, which retracts to a point. So π

(A ∩ B) = 1.

Hence by the Seifert-van Kampen theorem, we get

∧ S

, x

) = π

(A, x

) ∗

(A∩B,x

)

(B, x

)

∼

Z ∗

∼

Z ∗ Z

∼

where

is just

(

) for

|S|

= 2. We can see that

Z ∗ Z

∼

by showing that

they satisfy the same universal property.

Note that we had already figured this out when we studied the free group,

where we realized F

is the fundamental group of this thing.

More generally, as long as

, y

and

are “reasonable”,

(

X ∧ Y

)

∼

(X) ∗ π

(Y ).

Next, we would exhibit some nice examples of the covering spaces of

∧ S

i.e. the “rose with 2 petals”.

Recall that π

∧ S

, x

)

∼

ha, bi.

Example. Consider the map

→ Z/

3 which sends

a 7→

, b 7→

1. Note

that 1 is not the identity, since this is an abelian group, and the identity is 0.

This exists since the universal property tells us we just have to say where the

generators go, and the map exists (and is unique).

Now

ker φ

is a subgroup of

. So there is a based covering space of

∧ S

corresponding to it, say,

X. Let’s work out what it is.

First, we want to know how many sheets it has, i.e. how many copies of

we have. There are three, since we know that the number of sheets is the index

of the subgroup, and the index of

ker φ

|Z/

= 3 by the first isomorphism

theorem.

˜x

Let’s try to lift the loop

˜x

. Since

a 6∈ ker φ

(

X, ˜x

does not lift to

a loop. So it goes to another vertex.

˜x

Similarly, a

6∈ ker φ = π

(

X, ˜x

). So we get the following

˜x

Since a

∈ ker φ, we get a loop labelled a.

˜x

Note that

−1

∈ ker φ

. So

−1

gives a loop. So

goes in the same direction as

˜x

This is our covering space.

This is a fun game to play at home:

(i) Pick a group G (finite groups are recommended).

(ii) Then pick α, β ∈ G and let φ : F

→ G send a 7→ α, b 7→ β.

(iii) Compute the covering space corresponding to φ.

5.2 The effect on π

of attaching cells

We have started the course by talking about cell complexes, but largely ignored

them afterwards. Finally, we are getting back to them.

The process of attaching cells is as follows: we start with a space

, get a

function

n−1

→ X

, and attach

by gluing along the image of

to get

X ∪

Since we are attaching stuff, we can use the Seifert-van Kampen theorem to

analyse this.

Theorem. If

n ≥

3, then

(

X ∪

)

∼

(

). More precisely, the map

(

X, x

)

→ π

(

X ∪

, x

) induced by inclusion is an isomorphism, where

is a point on the image of f .

This tells us that attaching a high-dimensional disk does not do anything to

the fundamental group.

Proof.

Again, the difficulty of applying Seifert-van Kampen theorem is that we

need to work with open sets.

Let 0

∈ D

be any point in the interior of

. We let

X ∪

(

\ {

}

Note that

}

deformation retracts to the boundary

n−1

. So

deformation

retracts to X. Let B =

D, the interior of D

. Then

A ∩ B =

\ 0

∼

n−1

× (−1, 1)

We cannot use

as our basepoint, since this point is not in

A ∩ B

. Instead,

pick an arbitrary

∈ A ∩ B

. Since

is path connected, we have a path

, and we can use this to recover the fundamental groups based at

Now Seifert-van Kampen theorem says

(X ∪

, y

)

∼

(A, y

) ∗

(A∩B,y

)

(B, y

Since

is just a disk, and

A ∩ B

is simply connected (

n ≥

3 implies

n−1

simply connected), their fundamental groups are trivial. So we get

(X ∪

, y

)

∼

(A, y

We can now use γ to change base points from y

to y

. So

(X ∪

, y

)

∼

(A, y

)

∼

(X, y

The more interesting case is when we have smaller dimensions.

Theorem. If

= 2, then the natural map

(

X, x

)

→ π

(

X ∪

, x

) is

surjective, and the kernel is

[

]

. Note that this statement makes sense, since

n−1

is a circle, and f : S

n−1

→ X is a loop in X.

This is what we would expect, since if we attach a disk onto the loop given

by f, this loop just dies.

Proof. As before, we get

(X ∪

, y

)

∼

(A, y

) ∗

(A∩B,y

)

(B, y

Again,

is contractible, and

(

B, y

)

∼

1. However,

(

A ∩ B, y

)

∼

. Since

(A ∩ B, y

) is just (homotopic to) the loop induced by f, it follows that

(A, y

) ∗

(A∩B,y

)

1 = (π

(A, y

) ∗ 1)/hhπ

(A ∩ B, y

)ii

∼

(X, x

)/hhfii.

In summary, we have

(X ∪

) =

(

(X) n ≥ 3

(X)/hhfii n = 2

This is a useful result, since this is how we build up cell complexes. If we want to

compute the fundamental groups, we can just work up to the two-cells, and know

that the higher-dimensional cells do not have any effect. Moreover, whenever

is a cell complex, we should be able to write down the presentation of π

(X).

Example. Let

be the 2-torus. Possibly, our favorite picture of the torus is

(not a doughnut):

This is already a description of the torus as a cell complex!

We start with our zero complex X

(0)

We then add our 1-cells to get X

(1)

We now glue our square to the cell complex to get X = X

(2)

matching up the colors and directions of arrows.

So we have our torus as a cell complex. What is its fundamental group?

There are many ways we can do this computation, but this time we want to do

it as a cell complex.

We start with

(0)

. This is a single point. So its fundamental group is

(0)

) = 1.

When we add our two 1-cells, we get π

(1)

) = F

∼

ha, bi.

Finally, to get

(

), we have to quotient out by the boundary of our square,

which is just aba

−1

. So we have

(2)

) = F

/hhaba

−1

ii = ha, b | aba

−1

∼

We have the last congruence since we have two generators, and then we make

them commute by quotienting the commutator out.

This procedure can be reversed — given a presentation of a group, we can just

add the right edges and squares to produce a cell complex with that presentation.

Corollary. For any (finite) group presentation

hS | Ri

, there exists a (finite)

cell complex (of dimension 2) X such that π

(X)

∼

hS | Ri.

There really isn’t anything that requires that finiteness in this proof, but

finiteness makes us feel more comfortable.

Proof.

Let

, · · · , a

}

and

, · · · , r

}

. We start with a single point,

and get our

(1)

by adding a loop about the point for each

∈ S

. We then get

our 2-cells

for

= 1

, · · · , n

, and attaching them to

(1)

→ X

(1)

given by a based loop representing r

∈ F (S).

Since all groups have presentations, this tells us that all groups are funda-

mental groups of some spaces (at least those with finite presentations).

5.3 A refinement of the Seifert-van Kampen theorem

We are going to make a refinement of the theorem so that we don’t have to

worry about that openness problem. We first start with a definition.

Definition (Neighbourhood deformation retract). A subset

A ⊆ X

is a neigh-

bourhood deformation retract if there is an open set

A ⊆ U ⊆ X

such that

a strong deformation retract of

, i.e. there exists a retraction

U → A

and

r ' id

rel A.

This is something that is true most of the time, in sufficiently sane spaces.

Example. If

is a subcomplex of a cell complex, then

is a neighbourhood

deformation retract.

Theorem. Let

be a space,

A, B ⊆ X

closed subspaces. Suppose that

and

A ∩ B

are path connected, and

A ∩ B

is a neighbourhood deformation

retract of A and B. Then for any x

∈ A ∩ B.

(X, x

) = π

(A, x

) ∗

(A∩B,x

)

(B, x

A B

This is just like Seifert-van Kampen theorem, but usually easier to apply, since

we no longer have to “fatten up” our A and B to make them open.

Proof.

Pick open neighbourhoods

A ∩ B ⊆ U ⊆ A

and

A ∩ B ⊆ V ⊆ B

that strongly deformation retract to

A ∩ B

. Let

be such that

retracts

A ∩ B

. Since

retracts to

, it follows that

is path connected since

path-connectedness is preserved by homotopies.

Let

A∪V

and

B∪U

. Since

= (

X \B

)

∪V

, and

= (

X \A

)

∪U

it follows that A

and B

are open.

Since

and

retract to

A ∩ B

, we know

' A

and

' B

. Also,

∩B

= (

A∪V

)

∩

(

B ∪ U

) =

U ∪ V ' A∩ B

. In particular, it is path connected.

So by Seifert van-Kampen, we get

(A ∪ B) = π

, x

) ∗

∩B

)

, x

) = π

(A, x

) ∗

(A∩B,x

)

(B, x

This is basically what we’ve done all the time when we enlarge our

and

to become open.

Example. Let

∧ S

, the rose with two petals. Let

A, B

∼

be the

circles.

A B

Then since

}

A ∩ B

is a neighbourhood deformation retract of

and

we know that

∼

∗ π

5.4 The fundamental group of all surfaces

We have found that the torus has fundamental group

, but we already knew

this, since the torus is just

× S

, and the fundamental group of a product is

the product of the fundamental groups, as you have shown in the example sheet.

So we want to look at something more interesting. We look at all surfaces.

We start by defining what a surface is. It is surprisingly difficult to get

mathematicians to agree on how we can define a surface. Here we adopt the

following definition:

Definition (Surface). A surface is a Hausdorff topological space such that every

point has a neighbourhood U that is homeomorphic to R

Some people like

more that

, and sometimes they put

instead of

the definition, which is confusing, since that would have two complex dimensions

and hence four real dimensions. Needless to say, the actual surfaces will also

be different and have different homotopy groups. We will just focus on surfaces

with two real dimensions.

To find the fundamental group of all surfaces, we rely on the following

theorem that tells us what surfaces there are.

Theorem (Classification of compact surfaces). If

is a compact surface, then

X is homeomorphic to a space in one of the following two families:

(i)

The orientable surface of genus

, Σ

includes the following (please excuse

my drawing skills):

A more formal definition of this family is the following: we start with the

2-sphere, and remove a few discs from it to get

\ ∪

i=1

. Then we take

g tori with an open disc removed, and attach them to the circles.

(ii)

The non-orientable surface of genus

{RP

, K, · · · }

(where

the Klein bottle). This has a similar construction as above: we start with

the sphere S

, make a few holes, and then glue M¨obius strips to them.

It would be nice to be able to compute fundamental groups of these guys.

To do so, we need to write them as polygons with identification.

Example. To obtain a surface of genus two, written Σ

, we start with what we

had for a torus:

If we just wanted a torus, we are done (after closing the loop), but now we want

a surface with genus 2, so we add another torus:

To visualize how this works, imagine cutting this apart along the dashed line.

This would give two tori with a hole, where the boundary of the holes are just the

dashed line. Then gluing back the dashed lines would give back our orientable

surface with genus 2.

In general, to produce Σ

, we produce a polygon with 4g sides. Then we get

= ha

, b

, · · · , a

, b

| a

−1

· · · a

−1

We do we care? The classification theorem tells us that each surface is homeomor-

phic to some of these orientable and non-orientable surfaces, but it doesn’t tell us

there is no overlap. It might be that Σ

∼

241

, via some weird homeomorphism

that destroys some holes.

However, this result lets us know that all these orientable surfaces are

genuinely different. While it is difficult to stare at this fundamental group

and say that

∼

for

g 6

, we can perform a little trick. We can

take the abelianization of the group

, where we further quotient by all

commutators. Then the abelianized fundamental group of Σ

will simply be

. These are clearly distinct for different values of

. So all these surfaces are

distinct. Moreover, they are not even homotopy equivalent.

The fundamental groups of the non-orientable surfaces is left as an exercise

for the reader.

6 Simplicial complexes

So far, we have taken a space

, and assigned some things to it. The first was

easy —

(

). It was easy to calculate and understand. We then spent a lot of

time talking about

(

). What are they good for? A question we motivated

ourselves with was to prove

∼

implies

was pretty good for the

case when

= 1. If

∼

, then we would have

\ {

}

∼

R \ {

} ' S

. We

know that

|π

(

)

= 2, while

|π

(

\ {

}

)

= 1 for

m 6

= 1. This is just a fancy

way of saying that R \ {0} is disconnected while R

\ {0} is not for m 6= 1.

We can just add 1 to

, and add 1 to our subscript. If

∼

, then we

have

}

∼

} ' S

. We know that

(

)

∼

, while

(

}

)

∼

m−1

)

∼

1 unless m = 2.

The obvious thing to do is to create some

(

). But as you noticed,

took us quite a long time to define, and was really hard to compute. As we

would expect, this only gets harder as we get to higher dimensions. This is

indeed possible, but will only be done in Part III courses.

The problem is that π

works with groups, and groups are hard. There are

too many groups out there. We want to do some easier algebra, and a good

choice is linear algebra. Linear algebra is easy. In the rest of the course, we

will have things like

(

) and

(

), instead of

, which are more closely

related to linear algebra.

Another way to motivate the abandoning of groups is as follows: recall last

time we saw that if

is a finite cell, reasonably explicitly defined, then we can

write down a presentation

hS | Ri

for

(

). This sounds easy, except that we

don’t understand presentations. In fact there is a theorem that says there is no

algorithm that decides if a particular group presentation is actually the trivial

group.

So even though we can compute the fundamental group in terms of presenta-

tions, this is not necessarily helpful. On the other hand, linear algebra is easy.

Computers can do it. So we will move on to working with linear algebra instead.

This is where homology theory comes in. It takes a while for us to define it,

but after we finish developing the machinery, things are easy.

6.1 Simplicial complexes

There are many ways of formulating homology theory, and these are all equivalent

at least for sufficiently sane spaces (such as cell complexes). In this course, we

will be using simplicial homology, which is relatively more intuitive and can be

computed directly. The drawback is that we will have to restrict to a particular

kind of space, known as simplicial complexes. This is not a very serious restriction

per se, since many spaces like spheres are indeed simplicial complexes. However,

the definition of simplicial homology is based on exactly how we view our space

as a simplicial complex, and it will take us quite a lot of work to show that the

simplicial homology is indeed a property of the space itself, and not how we

represent it as a simplicial complex.

We now start by defining simplicial complexes, and developing some general

theory of simplicial complexes that will become useful later on.

Definition (Affine independence). A finite set of points

, · · · , a

} ⊆ R

affinely independent iff

i=1

= 0 with

i=1

= 0 ⇔ t

= 0 for all i.

Example. When n = 3, the following points are affinely independent:

The following are not:

The proper way of understanding this definition is via the following lemma:

Lemma.

, · · · , a

∈ R

are affinely independent if and only if

−a

, · · · , a

−

are linearly independent.

Alternatively,

+ 1 affinely independent points span an

-dimensional thing.

Proof. Suppose a

, · · · , a

are affinely independent. Suppose

i=1

− a

) = 0.

Then we can rewrite this as

−

i=1

+ λ

+ · · · + λ

= 0.

Now the sum of the coefficients is 0. So affine independence implies that all

coefficients are 0. So a

− a

, · · · , a

− a

are linearly independent.

On the other hand, suppose

− a

, · · · , a

− a

are linearly independent.

Now suppose

i=0

= 0,

i=0

= 0.

Then we can write

= −

i=1

Then the first equation reads

0 =

−

i=1

+ t

+ · · · + t

i=1

− a

So linear independence implies all t

= 0.

The relevance is that these can be used to define simplices (which are simple,

as opposed to complexes).

Definition (

-simplex). An

-simplex is the convex hull of (

+ 1) affinely

independent points a

, · · · , a

∈ R

, i.e. the set

σ = ha

, · · · , a

i =

(

i=0

= 1, t

≥ 0

)

The points

, · · · , a

are the vertices, and are said to span

. The (

+ 1)-tuple

, · · · , t

) is called the barycentric coordinates for the point

Example. When n = 0, then our 0-simplex is just a point:

When n = 1, then we get a line:

When n = 2, we get a triangle:

When n = 3, we get a tetrahedron:

The key motivation of this is that simplices are determined by their vertices.

Unlike arbitrary subspaces of

, they can be specified by a finite amount of

data. We can also easily extract the faces of the simplices.

Definition (Face, boundary and interior). A face of a simplex is a subset (or

subsimplex) spanned by a subset of the vertices. The boundary is the union of

the proper faces, and the interior is the complement of the boundary.

The boundary of

is usually denoted by

∂σ

, while the interior is denoted by

˚σ, and we write τ ≤ σ when τ is a face of σ.

In particular, the interior of a vertex is the vertex itself. Note that these

notions of interior and boundary are distinct from the topological notions of

interior and boundary.

Example. The standard

-simplex is spanned by the basis vectors

{

, · · · ,

}

in R

n+1

. For example, when n = 2, we get the following:

We will now glue simplices together to build complexes, or simplicial com-

plexes.

Definition. A (geometric) simplicial complex is a finite set

of simplices in

such that

(i) If σ ∈ K and τ is a face of σ, then τ ∈ K.

(ii) If σ, τ ∈ K, then σ ∩ τ is either empty or a face of both σ and τ .

Definition (Vertices). The vertices of

are the zero simplices of

, denoted

Example. This is a simplicial complex:

These are not:

Technically, a simplicial complex is defined to be a set of simplices, which

are just collections of points. It is not a subspace of

. Hence we have the

following definition:

Definition (Polyhedron). The polyhedron defined by

is the union of the

simplices in K, and denoted by |K|.

We make this distinction because distinct simplicial complexes may have the

same polyhedron, such as the following:

Just as in cell complexes, we can define things like dimensions.

Definition (Dimension and skeleton). The dimension of

is the highest

dimension of a simplex of

. The

-skeleton

(d)

is the union of the

n-simplices in K for n ≤ d.

Note that since these are finite and live inside

, we know that

|K|

is always

compact and Hausdorff.

Usually, when we are given a space, say

, it is not defined to be a simplicial

complex. We can “make” it a simplicial complex by a triangulation.

Definition (Triangulation). A triangulation of a space

is a homeomorphism

h : |K| → X, where K is some simplicial complex.

Example. Let

be the standard

-simplex. The boundary

∂σ

is homeomorphic

n−1

(e.g. the boundary of a (solid) triangle is the boundary of the triangle,

which is also a circle). This is called the simplicial (n − 1)-sphere.

We can also triangulate our S

in a different way:

Example. In

n+1

, consider the simplices

h±

, · · · , ±

for each possible

combination of signs. So we have 2

n+1

simplices in total. Then their union

defines a simplicial complex K, and

|K|

∼

The nice thing about this triangulation is that the simplicial complex is

invariant under the antipodal map. So not only can we think of this as a

triangulation of the sphere, but a triangulation of RP

as well.

As always, we don’t just look at objects themselves, but also maps between

them.

Definition (Simplicial map). A simplicial map

K → L

is a function

→ V

such that if

, · · · , a

is a simplex in

, then

(

)

, · · · , f

(

)

}

spans a simplex of L.

The nice thing about simplicial maps is that we only have to specify where

the vertices go, and there are only finitely many vertices. So we can completely

specify a simplicial map by writing down a finite amount of information.

It is important to note that we say

(

)

, · · · , f

(

)

}

as a set span a simplex

of L. In particular, they are allowed to have repeats.

Example. Suppose we have the standard 2-simplex K as follows:

The following does not define a simplicial map because

, a

is a simplex in

K, but {f(a

), f(a

)} does not span a simplex:

f(a

), f(a

) f(a

)

On the other hand, the following is a simplicial map, because now

(

)

, f

(

)

}

spans a simplex, and note that

(

)

, f

(

)

, f

(

)

}

also spans a 1-simplex

because we are treating the collection of three vertices as a set, and not a

simplex.

f(a

), f(a

) f(a

)

Finally, we can also do the following map:

f(a

), f(a

) f(a

)

The following lemma is obvious, but we will need it later on.

Lemma. If

is a simplicial complex, then every point

x ∈ |K|

lies in the

interior of a unique simplex.

As we said, simplicial maps are nice, but they are not exactly what we want.

We want to have maps between spaces.

Lemma. A simplicial map

K → L

induces a continuous map

|f|

|K| → |L|

and furthermore, we have

|f ◦ g| = |f| ◦ |g|.

There is an obvious way to define this map. We know how to map vertices,

and then just extend everything linearly.

Proof. For any point in a simplex σ = ha

, · · · , a

i, we define

|f|

i=0

f(a

The result is in

because

(

)

}

spans a simplex. It is not difficult to see this

is well-defined when the point lies on the boundary of a simplex. This is clearly

continuous on σ, and is hence continuous on |K| by the gluing lemma.

The final property is obviously true by definition.

6.2 Simplicial approximation

This is all very well, but we are really interested in continuous maps. So given a

continuous map

, we would like to find a related simplicial map

. In particular,

we want to find a simplicial map

that “approximates”

in some sense. The

definition we will write down is slightly awkward, but it turns out this is the

most useful definition.

Definition (Open star and link). Let

x ∈ |K|

. The open star of

is the union

of all the interiors of the simplices that contain x, i.e.

(x) =

[

x∈σ∈K

˚σ.

The link of

, written

(

), is the union of all those simplices that do not

contain x, but are faces of a simplex that does contain x.

Definition (Simplicial approximation). Let

|K| → |L|

be a continuous map

between the polyhedra. A function

→ V

is a simplicial approximation

to f if for all v ∈ V

f(St

(v)) ⊆ St

(g(v)). (∗)

The following lemma tells us why this is a good definition:

Lemma. If

|K| → |L|

is a map between polyhedra, and

→ V

is a simplicial approximation to

, then

is a simplicial map, and

|g| ' f

Furthermore, if

is already simplicial on some subcomplex

M ⊆ K

, then we get

= f|

, and the homotopy can be made rel M .

Proof.

First we want to check

is really a simplicial map if it satisfies (

∗

). Let

, · · · , a

be a simplex in

. We want to show that

(

)

, · · · , g

(

)

}

spans a simplex in L.

Pick an arbitrary

x ∈ ˚σ

. Since

contains each

, we know that

x ∈ St

(

)

for all i. Hence we know that

f(x) ∈

i=0

f(St

)) ⊆

i=0

(g(a

)).

Hence we know that there is one simplex, say,

that contains all

(

) whose

interior contains

(

). Since each

(

) is a vertex in

, each

(

) must be a

vertex of τ . So they span a face of τ, as required.

We now want to prove that

|g| ' f

. We let

|K| × I → |L| ⊆ R

defined by

(x, t) 7→ t|g|(x) + (1 − t)f(x).

This is clearly continuous. So we need to check that

im H ⊆ |L|

. But we

know that both

|g|

(

) and

(

) live in

and

is convex. It thus follows that

H(x × I) ⊆ τ ⊆ |L|.

To prove the last part, it suffices to show that every simplicial approximation

to a simplicial map must be the map itself. Then the homotopy is

rel M

the construction above. This is easily seen to be true — if

is a simplicial

approximation to

, then

(

)

∈ f

(

))

⊆ St

(

)). Since

(

) is a vertex

and

(

) is the only vertex in

(

)), we must have

(

) =

(

). So done.

What’s the best thing we might hope for at this point? It would be great if

every map were homotopic to a simplicial map. Is this possible? Let’s take a

nice example. Let’s consider the following K:

How many homotopy classes of continuous maps are there

K → K

? Countably

many, one for each winding number. However, there can only be at most 3

= 27

simplicial maps. The problem is that we don’t have enough vertices to realize all

those interesting maps. The idea is to refine our simplicial complexes. Suppose

we have the following simplex:

What we do is to add a point in the center of each simplex, and join them up:

This is known as the barycentric subdivision. After we subdivide it once, we can

realize more homotopy classes of maps. We will show that for any map, as long

as we are willing to barycentrically subdivide the simplex many times, we can

find a simplicial approximation to it.

Definition (Barycenter). The barycenter of σ = ha

, · · · , a

i is

ˆσ =

i=0

n + 1

Definition (Barycentric subdivision). The (first) barycentric subdivision

K is the simplicial complex:

= {hˆσ

, · · · , ˆσ

i : σ

∈ K and σ

< σ

< · · · < σ

If you stare at this long enough, you will realize this is exactly what we have

drawn above.

The

th barycentric subdivision

(r)

is defined inductively as the barycentric

subdivision of the r − 1th barycentric subdivision, i.e.

(r)

= (K

(r−1)

)

Proposition. |K| = |K

| and K

really is a simplicial complex.

Proof. Too boring to be included in lectures.

We now have a slight problem. Even though

and

|K|

are equal, the

identity map from |K

| to |K| is not a simplicial map.

To solve this problem, we can choose any function

K → V

σ 7→ v

with

∈ σ

, i.e. a function that sends any simplex to any of its vertices. Then we can

define

→ K

by sending

ˆσ 7→ v

. Then this is a simplicial map, and indeed

a simplicial approximation to the identity map

| → |K|

. We will revisit this

idea later when we discuss homotopy invariance.

The key theorem is that as long as we are willing to perform barycentric

subdivisions, then we can always find a simplicial approximation.

Theorem (Simplicial approximation theorem). Let

and

be simplicial

complexes, and

|K| → |L|

a continuous map. Then there exists an

and

a simplicial map

(r)

→ L

such that

is a simplicial approximation of

Furthermore, if

is already simplicial on

M ⊆ K

, then we can choose

such

that |g||

= f|

The first thing we have to figure out is how far we are going to subdivide.

To do this, we want to quantify how “fine” our subdivisions are.

Definition (Mesh). Let K be a simplicial complex. The mesh of K is

µ(K) = max{kv

− v

k : hv

, v

i ∈ K}.

We have the following lemma that tells us how large our mesh is:

Lemma. Let dim K = n, then

µ(K

(r)

) ≤



n + 1



µ(K).

The key point is that as

r → ∞

, the mesh goes to zero. So indeed we can

make our barycentric subdivisions finer and finer. The proof is purely technical

and omitted.

Now we can prove the simplicial approximation theorem.

Proof of simplicial approximation theorem.

Suppose we are given the map

|K| → |L|

. We have a natural cover of

|L|

, namely the open stars of all vertices.

We can use f to pull these back to |K| to obtain a cover of |K|:

−1

(St

(w)) : w ∈ V

The idea is to barycentrically subdivide our

such that each open star of

contained in one of these things.

By the Lebesgue number lemma, there exists some

, the Lebesgue number

of the cover, such that for each x ∈ |K|, B

(x) is contained in some element of

the cover. By the previous lemma, there is an r such that µ(K

(r)

) < δ.

Now since the mesh

(

(r)

) is the smallest distance between any two vertices,

the radius of every open star

(r)

(

) is at most

(

(r)

). Hence it follows that

(r)

(

)

⊆ B

(

) for all vertices

x ∈ V

(r)

. Therefore, for all

x ∈ V

(r)

, there

is some w ∈ V

such that

(r)

(x) ⊆ B

(x) ⊆ f

−1

(St

(w)).

Therefore defining g(x) = w, we get

f(St

(r)

(x)) ⊆ St

(g(x)).

So g is a simplicial approximation of f.

The last part follows from the observation that if

is a simplicial map, then

it maps vertices to vertices. So we can pick g(v) = f(v).

7 Simplicial homology

7.1 Simplicial homology

For now, we will forget about simplicial approximations and related fluff, but

just note that it is fine to assume everything can be considered to be simplicial.

Instead, we are going to use this framework to move on and define some new

invariants of simplicial complexes

, known as

(

). These are analogous

, π

, · · ·

, but only use linear algebra in the definitions, and are thus much

simpler. The drawback, however, is that the definitions are slightly less intuitive

at first sight.

Despite saying “linear algebra”, we won’t be working with vector spaces most

of the time. Instead, we will be using abelian groups, which really should be

thought of as

-modules. At the end, we will come up with an analogous theory

using

-modules, i.e.

-vector spaces, and most of the theory will carry over.

Using

makes some of our work easier, but as a result we would have lost some

information. In the most general case, we can replace

with any abelian group,

but we will not be considering any of these in this course.

Recall that we defined an

-simplex as the collection of

+ 1 vertices that

span a simplex. As a simplex, when we permute the vertices, we still get the

same simplex. What we want to do is to remember the orientation of the simplex.

In particular, we want think of the simplices (a, b) and (b, a) as different:

b b

Hence we define oriented simplices.

Definition (Oriented

-simplex). An oriented

-simplex in a simplicial complex

is an (

+ 1)-tuple (

, · · · , a

) of vertices

∈ V

such that

, · · · , a

i ∈ K

where we think of two (

+ 1)-tuples (

, · · · , a

) and (

π(0)

, · · · , a

π(n)

) as the

same oriented simplex if π ∈ S

is an even permutation.

We often denote an oriented simplex as

, and then

¯σ

denotes the same

simplex with the opposite orientation.

Example. As oriented 2-simplices, (

, v

) and (

, v

) are equal, but

they are different from (

, v

). We can imagine the two different orientations

of the simplices as follows:

One and two dimensions are the dimensions where we can easily visualize

the orientation. This is substantially harder in higher dimensions, and often we

just work with the definition instead.

Definition (Chain group

(

)). Let

be a simplicial complex. For each

n ≥ 0, we define C

(K) as follows:

Let

{σ

, · · · , σ

}

be the set of

-simplices of

. For each

, choose an

orientation on

. That is, choose an order for the vertices (up to an even

permutation). This choice is not important, but we need to make it. Now when

we say σ

, we mean the oriented simplex with this particular orientation.

Now let

(

) be the free abelian group with basis

{σ

, · · · , σ

}

, i.e.

(

)

∼

. So an element in C

(K) might look like

− 7σ

+ 52σ

− 28σ

1000000

In other words, an element of C

(K) is just a formal sum of n-simplices.

For convenience, we define

(

) = 0 for

 <

0. This will save us from

making exceptions for n = 0 cases later.

For each oriented simplex, we identify −σ

with ¯σ

, at least when n ≥ 1.

In this definition, we have to choose a particular orientation for each of our

simplices. If you don’t like making arbitrary choices, we could instead define

(K) as some quotient, but it is slightly more complicated.

Note that if there are no

-simplices (e.g. when

−

1), we can still

meaningfully talk about C

(K), but it’s just 0.

Example. We can think of elements in the chain group

(

) as “paths” in

For example, we might have the following simplex:

Then the path

around the left triangle is represented

by the member

− σ

+ σ

= 2σ

− σ

+ σ

Of course, with this setup, we can do more random things, like adding 57 copies

to it, and this is also allowed. So we could think of these as disjoint union

of paths instead.

When defining fundamental groups, we had homotopies that allowed us to

“move around”. Somehow, we would like to say that two of these paths are

“equivalent” under certain conditions. To do so, we need to define the boundary

homomorphisms.

Definition (Boundary homomorphisms). We define boundary homomorphisms

: C

(K) → C

n−1

(K)

, · · · , a

) 7→

i=0

(−1)

, · · · , ˆa

, · · · , a

where (

, · · · , ˆa

, · · · , a

) = (

, · · · , a

i−1

, a

i+1

, · · · , a

) is the simplex with

removed.

This means we remove each vertex in turn and add them up. This is clear if

we draw some pictures in low dimensions:

−v

If we take a triangle, we get

An important property of the boundary map is that the boundary of a boundary

is empty:

Lemma. d

n−1

◦ d

= 0.

In other words, im d

n+1

⊆ ker d

Proof.

This just involves expanding the definition and working through the

mess.

With this in mind, we will define the homology groups as follows:

Definition (Simplicial homology group

(

)). The

th simplicial homology

group H

(K) is defined as

(K) =

ker d

im d

n+1

This is a nice, clean definition, but what does this mean geometrically?

Somehow,

(

) describes all the “

-dimensional holes” in

|K|

. Since we

are going to draw pictures, we are going to start with the easy case of

= 1.

Our

(

) is made from the kernel of

and the image of

. First, we give

these things names.

Definition (Chains, cycles and boundaries). The elements of

(

) are called

-chains of

, those of

ker d

are called

-cycles of

, and those of

im d

k+1

are

called k-boundaries of K.

Suppose we have some

c ∈ ker d

. In other words,

= 0. If we interpret

as a “path”, if it has no boundary, then it represents some sort of loop, i.e. a

cycle. For example, if we have the following cycle:

We have

c = (e

, e

) + (e

, e

) + (e

, e

We can then compute the boundary as

dc = (e

− e

) + (e

− e

) + (e

− e

) = 0.

So this c is indeed a cycle.

Now if

c ∈ im d

, then

for some 2-chain

, i.e.

is the boundary of

some two-dimensional thing. This is why we call this a 1-boundary. For example,

suppose we have our cycle as above, but is itself a boundary of a 2-chain.

We see that a cycle that has a boundary has been “filled in”. Hence the “holes”

are, roughly, the cycles that haven’t been filled in. Hence we define the homology

group as the cycles quotiented by the boundaries, and we interpret its elements

as k-dimensional “holes”.

Example. Let

be the standard simplicial 1-sphere, i.e. we have the following

in R

Our simplices are thus

K = {he

i, he

, e

i, he

, e

i, he

, e

i}.

Our chain groups are

(K) = h(e

), (e

∼

(K) = h(e

, e

), (e

, e

), (e

, e

∼

All other chain groups are zero. Note that our notation is slightly confusing

here, since the brackets

h · i

can mean the simplex spanned by the vertices, or

the group generated by certain elements. However, you are probably clueful

enough to distinguish the two uses.

Hence, the only non-zero boundary map is

: C

(K) → C

(K).

We can write down its matrix with respect to the given basis.





−1 0 1

1 −1 0

0 1 −1





We have now everything we need to know about the homology groups, and we

just need to do some linear algebra to figure out the image and kernel, and thus

the homology groups. We have

(K) =

ker(d

: C

(K) → C

−1

(K))

im(d

: C

(K) → C

(K))

∼

(K)

im d

∼

im d

After doing some row operations with our matrix, we see that the image of

a two-dimensional subspace generated by the image of two of the edges. Hence

we have

(K) = Z.

What does this

(

) represent? We initially said that

(

) should represent

the

-dimensional holes, but when

= 0, this is simpler. As for

just

represents the path components of

. We interpret this to mean

has one

path component. In general, if

has

path components, then we expect

(

)

to be Z

Similarly, we have

(K) =

ker d

im d

∼

ker d

It is easy to see that in fact we have

ker d

= h(e

, e

) + (e

, e

) + (e

, e

∼

So we also have

(K)

∼

We see that this

(

) is generated by precisely the single loop in the triangle.

The fact that

(

) is non-trivial means that we do indeed have a hole in the

middle of the circle.

Example. Let L be the standard 2-simplex (and all its faces) in R

Now our chain groups are

(L) = C

(K)

∼

h(e

), (e

(L) = C

(K)

∼

h(e

, e

), (e

, e

), (e

, e

(L)

∼

Z = h(e

, e

)i.

Since

is the same as before, the only new interesting boundary map is

. We

compute

((e

, e

)) = (e

, e

) + (e

, e

) + (e

, e

We know that

(

) depends only on

and

, which are the same as for

(L)

∼

Again, the interpretation of this is that

is path-connected. The first homology

group is

(L) =

ker d

im d

h(e

, e

) + (e

, e

) + (e

, e

h(e

, e

) + (e

, e

) + (e

, e

∼

This precisely illustrates the fact that the “hole” is now filled in L.

Finally, we have

(L) =

ker d

im d

= ker d

∼

This is zero since there aren’t any two-dimensional holes in L.

We have hopefully gained some intuition on what these homology groups

mean. We are now going to spend a lot of time developing formalism.

7.2 Some homological algebra

We will develop some formalism to help us compute homology groups

(

) in

lots of examples. Firstly, we axiomatize the setup we had above.

Definition (Chain complex and differentials). A chain complex

is a sequence

of abelian groups

, C

, · · ·

equipped with maps

→ C

n−1

such that

n−1

◦ d

= 0 for all n. We call these maps the differentials of C

0 C

· · ·

Whenever we have some of these things, we can define homology groups in

exactly the same way as we defined them for simplicial complexes.

Definition (Cycles and boundaries). The space of n-cycles is

The space of n-boundaries is

n+1

Definition (Homology group). The

-th homology group of

is defined to be

ker d

im d

n+1

(C)

In mathematics, when we have objects, we don’t just talk about the objects

themselves, but also functions between them. Suppose we have two chain

complexes. For the sake of simplicity, we write the chain maps of both as d

In general, we want to have maps between these two sequences. This would

correspond to having a map

→ D

for each

, satisfying some commuta-

tivity relations.

Definition (Chain map). A chain map

→ D

is a sequence of homo-

morphisms f

: C

→ D

such that

n−1

◦ d

= d

◦ f

for all n. In other words, the following diagram commutes:

0 C

· · ·

0 D

· · ·

We want to have homotopies. So far, chain complexes seem to be rather rigid,

but homotopies themselves are rather floppy. How can we define homotopies for

chain complexes? It turns out we can have a completely algebraic definition for

chain homotopies.

Definition (Chain homotopy). A chain homotopy between chain maps

, g

→ D

is a sequence of homomorphisms h

: C

→ D

n+1

such that

− f

= d

n+1

◦ h

+ h

n−1

◦ d

We write f

' g

if there is a chain homotopy between f

and g

The arrows can be put in the following diagram, which is not commutative:

n−1

n+1

n−1

n+1

The intuition behind this definition is as follows: suppose

(

) and

(

) for

K, L

simplicial complexes, and

and

are “induced” by

simplicial maps f, g : K → L (if f maps an n-simplex σ to a lower-dimensional

simplex, then

•

= 0). How can we detect if

and

are homotopic via the

homotopy groups?

Suppose

|K| × I → |L|

is a homotopy from

. We further suppose

that

actually comes from a simplicial map

K × I → L

(we’ll skim over the

technical issue of how we can make

K × I

a simplicial complex. Instead, you are

supposed to figure this out yourself in example sheet 3).

Let σ be an n-simplex of K, and here is a picture of H(σ × I):

Let

(σ) = H(σ × I).

We think of this as an (

+ 1)-chain. What is its boundary? We’ve got the

vertical sides plus the top and bottom. The bottom should just be

(

), and

the top is just

(

), since

is a homotopy from

. How about the sides?

They are what we get when we pull the boundary

∂σ

up with the homotopy,

i.e.

(

∂σ × I

) =

n−1

◦ d

(

). Now note that

(

) and

(

) have opposite

orientations, so we get the result

n+1

◦ h

(σ) = h

n−1

◦ d

(σ) + g

(σ) − f

(σ).

Rearranging and dropping the σs, we get

n+1

◦ h

− h

n−1

◦ d

= g

− f

This looks remarkably like our definition for chain homotopies of maps, with the

signs a bit wrong. So in reality, we will have to fiddle with the sign of

a bit

to get it right, but you get the idea.

Lemma. A chain map f

: C

→ D

induces a homomorphism:

∗

: H

(D)

[c] 7→ [f(c)]

Furthermore, if f

and g

are chain homotopic, then f

∗

= g

∗

Proof.

Since the homology groups are defined as the cycles quotiented by the

boundaries, to show that

∗

defines a homomorphism, we need to show

sends

cycles to cycles and boundaries to boundaries. This is an easy check. If

(

) = 0,

then

(σ)) = f

(0) = 0.

So f

(σ) ∈ Z

(D).

Similarly, if σ is a boundary, say σ = d

(τ), then

(σ) = f

(τ)) = d

(τ)).

So f

(σ) is a boundary. It thus follows that f

∗

is well-defined.

Now suppose

is a chain homotopy between

and

. For any

c ∈ Z

(

we have

n+1

◦ h

n−1

◦ d

(c).

Since c ∈ Z

(C), we know that d

n+1

◦ h

(D).

Hence

(

) and

(

) differ by a boundary. So [

(

)]

−

[

(

)] = 0 in

(

i.e. f

∗

(c).

The following statements are easy to check:

Proposition.

(i) Being chain-homotopic is an equivalence relation of chain maps.

(ii) If a

: A

→ C

is a chain map and f

' g

, then f

◦ a

' g

◦ a

(iii) If f : C

→ D

and g : D

→ A

are chain maps, then

∗

◦ f

∗

= (g

◦ f

)

∗

(iv) (id

)

∗

= id

∗

(C)

The last two statements can be summarized in fancy language by saying that

is a functor.

Definition (Chain homotopy equivalence). Chain complexes

and

are

chain homotopy equivalent if there exist

→ D

and

→ C

such

that

◦ g

' id

, g

◦ f

' id

We should think of this in exactly the same way as we think of homotopy

equivalences of spaces. The chain complexes themselves are not necessarily the

same, but the induced homology groups will be.

Lemma. Let

→ D

be a chain homotopy equivalence, then

∗

(

)

→

(D) is an isomorphism for all n.

Proof.

Let

be the homotopy inverse. Since

◦ g

' id

, we know

∗

◦ g

∗

(D)

. Similarly,

∗

◦ f

∗

(C)

. So we get isomorphisms between

(

)

and H

(D).

7.3 Homology calculations

We’ll get back to topology and compute some homologies.

Here, K is always a simplicial complex, and C

= C

(K).

Lemma. Let

K → L

be a simplicial map. Then

induces a chain map

: C

(K) → C

(L). Hence it also induces f

∗

: H

(K) → H

(L).

Proof.

This is fairly obvious, except that simplicial maps are allowed to “squash”

simplices, so

might send an

-simplex to an (

n −

1)-simplex, which is not in

(L). We solve this problem by just killing these troublesome simplices.

Let

be an oriented

-simplex in

, corresponding to a basis element of

(K). Then we define

(σ) =

(

f(σ) f(σ) is an n-simplex

0 f(σ) is a k-simplex for k < n

More precisely, if σ = (a

, · · · , a

), then

(σ) =

(

(f(a

), · · · , f (a

)) f(a

), · · · , f (a

) spans an n-simplex

0 otherwise

We then extend f

linearly to obtain f

: C

(K) → C

(L).

It is immediate from this that this satisfies the chain map condition, i.e.

commutes with the boundary operators.

Definition (Cone). A simplicial complex is a cone if, for some v

∈ V

|K| = St

) ∪ Lk

We see that a cone ought to be contractible — we can just squash it to the point

. This is what the next lemma tells us.

Lemma. If

is a cone with cone point

, then inclusion

} → |K|

induces

a chain homotopy equivalence i

: C

({v

}) → C

(K). Therefore

(K) =

(

Z n = 0

0 n > 0

The homotopy inverse to

is given by

, where

k 7→ v

is the only map.

It is clear that

◦ i

, and we need to show that

◦ r

' id

. This chain

homotopy can be defined by

(

)

→ C

n+1

(

), where

associates to

any simplex σ in Lk

) the simplex spanned by σ and v

. Details are left to

the reader.

Corollary. If ∆

is the standard

-simplex, and

consists of ∆

and all its

faces, then

(L) =

(

Z k = 0

0 k > 0

Proof. K is a cone (on any vertex).

What we would really like is an example of non-trivial homology groups.

Moreover, we want them in higher dimensions, and not just the examples we got

for fundamental groups. An obvious candidate is the standard n-sphere.

Corollary. Let

be the standard (

n −

1)-sphere (i.e. the proper faces of

from above). Then for n ≥ 2, we have

(K) =











Z k = 0

0 0 < k < n − 1

Z k = n − 1

Proof. We write down the chain groups for K and L.

0 C

(L) C

(L) · · · C

n−1

(L) C

(L)

0 C

(K) C

(K) · · · C

n−1

(K) C

(K) = 0

n−1

For

k < n −

1, we have

(

) =

(

) and

k+1

(

) =

k+1

(

). Also, the

boundary maps are equal. So

(K) = H

(L) = 0.

We now need to compute

n−1

(K) = ker d

n−1

= ker d

n−1

= im d

We get the last equality since

ker d

n−1

im d

= H

n−1

(L) = 0.

We also know that

(

) is generated by just one simplex (

, · · · , e

). So

(

)

∼

. Also

is injective since it does not kill the generator (

, · · · , e

n−1

(K)

∼

im d

∼

This is very exciting, because at last, we have a suggestion for a non-trivial

invariant of

n−1

for

n >

2. We say this is just a “suggestion”, since the simplicial

homology is defined for simplicial complexes, and we are not guaranteed that if

we put a different simplicial complex on

n−1

, we will get the same homology

groups. So the major technical obstacle we still need to overcome is to see that

are invariants of the polyhedron

|K|

, not just

, and similarly for maps.

But this will take some time.

We will quickly say something we’ve alluded to already:

Lemma (Interpretation of

(

)

∼

, where

is the number of path

components of K.

Proof.

Let

be our simplicial complex and

v, w ∈ V

. We note that by definition,

v, w

represent the same homology class in

(

) if and only if there is some

such that

w − v

. The requirement that

w − v

is equivalent to saying

is a path from

. So [

] = [

] if and only if

and

are in the same path

component of K.

Most of the time, we only care about path-connected spaces, so

(

)

∼

7.4 Mayer-Vietoris sequence

The Mayer-Vietoris theorem is exactly like the Seifert-van Kampen theorem

for fundamental groups, which tells us what happens when we glue two spaces

together.

Suppose we have a space K = M ∪ N.

M N

We will learn how to compute the homology of the union

M ∪ N

in terms of

those of M, N and M ∩ N.

Recall that to state the Seifert-van Kampen theorem, we needed to learn

some new group-theoretic notions, such as free products with amalgamation.

The situation is somewhat similar here. We will need to learn some algebra in

order to state the Mayer-Vietoris theorem. The objects we need are known as

exact sequences.

Definition (Exact sequence). A pair of homomorphisms of abelian groups

A B C

f g

is exact (at B) if

im f = ker g.

A collection of homomorphisms

· · · A

i+1

i+2

· · ·

i−1

i+1

i+2

is exact at A

ker f

= im f

i−1

We say it is exact if it is exact at every A

Recall that we have seen something similar before. When we defined the

chain complexes, we had

= 0, i.e.

im d ⊆ ker d

. Here we are requiring exact

equivalence, which is something even better.

Algebraically, we can think of an exact sequence as chain complexes with

trivial homology groups. Alternatively, we see the homology groups as measuring

the failure of a sequence to be exact.

There is a particular type of exact sequences that is important.

Definition (Short exact sequence). A short exact sequence is an exact sequence

of the form

0 A B C 0

f g

What does this mean?

–

The kernel of

is equal to the image of the zero map, i.e.

{

}

. So

injective.

–

The image of

is the kernel of the zero map, which is everything. So

surjective.

– im f = ker g.

Since we like chain complexes, we can produce short exact sequences of chain

complexes.

Definition (Short exact sequence of chain complexes). A short exact sequence

of chain complexes is a pair of chain maps i

and j

0 A

such that for each k,

0 A

is exact.

Note that by requiring the maps to be chain maps, we imply that

and

commute with the boundary maps of the chain complexes.

The reason why we care about short exact sequences of chain complexes

(despite them having such a long name) is the following result:

Theorem (Snake lemma). If we have a short exact sequence of complexes

0 A

then a miracle happens to their homology groups. In particular, there is a long

exact sequence (i.e. an exact sequence that is not short)

· · · H

(A) H

(B) H

(C)

n−1

(A) H

n−1

(B) H

n−1

∗

∂

∗

where

∗

and

∗

are induced by

and

, and

∂

∗

is a map we will define in the

proof.

Having an exact sequence is good, since if we know most of the terms in an

exact sequence, we can figure out the remaining ones. To do this, we also need

to understand the maps

∗

and

∂

∗

, but we don’t need to understand all, since

we can deduce some of them with exactness. Yet, we still need to know some of

them, and since they are defined in the proof, you need to remember the proof.

Note, however, if we replace

in the definition of chain groups by a field

(e.g.

), then all the groups become vector spaces. Then everything boils down

to the rank-nullity theorem. Of course, this does not get us the right answer in

exams, since we want to have homology groups over

, and not

, but this helps

us to understand the exact sequences somewhat. If, at any point, homology

groups confuse you, then you can try to work with homology groups over

and

get a feel for what homology groups are like, since this is easier.

We are first going to apply this result to obtain the Mayer-Vietoris theorem.

Theorem (Mayer-Vietoris theorem). Let

K, L, M, N

be simplicial complexes

with K = M ∪ N and L = M ∩ N. We have the following inclusion maps:

L M

N K.

Then there exists some natural homomorphism

∂

∗

(

)

→ H

n−1

(

) that

gives the following long exact sequence:

· · · H

(L) H

(M) ⊕ H

(N) H

(K)

n−1

(L) H

n−1

(M) ⊕ H

n−1

(N) H

n−1

(K) · · ·

· · · H

(M) ⊕ H

(N) H

(K) 0

∂

∗

−`

∗

∂

∗

−`

∗

−`

∗

Here

A ⊕ B

is the direct sum of the two (abelian) groups, which may also be

known as the Cartesian product.

Note that unlike the Seifert-van Kampen theorem, this does not require the

intersection

M ∩ N

to be (path) connected. This restriction was needed

for Seifert-van Kampen since the fundamental group is unable to see things

outside the path component of the basepoint, and hence it does not like non-path

connected spaces well. However, homology groups don’t have these problems.

Proof.

All we have to do is to produce a short exact sequence of complexes. We

have

0 C

(L) C

(M) ⊕ C

(N) C

(K) 0

−`

Here

(

)

→ C

(

)

⊕ C

(

) is the map

x 7→

(

x, x

), while

− 

(

)

⊕ C

(

)

→ C

(

) is the map (

a, b

)

7→ a − b

(after applying the

appropriate inclusion maps).

It is easy to see that this is a short exact sequence of chain complexes. The

image of

is the set of all elements of the form (

x, x

), and the kernel of

− 

is also these. It is also easy to see that

is injective and

− 

surjective.

At first sight, the Mayer-Vietoris theorem might look a bit scary to use, since

it involves all homology groups of all orders at once. However, this is actually

often a good thing, since we can often use this to deduce the higher homology

groups from the lower homology groups.

Yet, to properly apply the Mayer-Vietoris sequence, we need to understand

the map ∂

∗

. To do so, we need to prove the snake lemma.

Theorem (Snake lemma). If we have a short exact sequence of complexes

0 A

then there is a long exact sequence

· · · H

(A) H

(B) H

(C)

n−1

(A) H

n−1

(B) H

n−1

∗

∂

∗

where

∗

and

∗

are induced by

and

, and

∂

∗

is a map we will define in the

proof.

The method of proving this is sometimes known as “diagram chasing”, where

we just “chase” around commutative diagrams to find the elements we need.

The idea of the proof is as follows — in the short exact sequence, we can think

as a subgroup of

, and

as the quotient

B/A

, by the first isomorphism

theorem. So any element of

can be represented by an element of

. We

apply the boundary map to this representative, and then exactness shows that

this must come from some element of

. We then check carefully that this is

well-defined, i.e. does not depend on the representatives chosen.

Proof.

The proof of this is in general not hard. It just involves a lot of checking

of the details, such as making sure the homomorphisms are well-defined, are

actually homomorphisms, are exact at all the places etc. The only important

and non-trivial part is just the construction of the map ∂

∗

First we look at the following commutative diagram:

0 A

n−1

To construct

∂

∗

(

)

→ H

n−1

(

), let [

]

∈ H

(

) be a class represented

x ∈ Z

(

). We need to find a cycle

z ∈ A

n−1

. By exactness, we know the

map

→ C

is surjective. So there is a

y ∈ B

such that

(

) =

Since our target is

n−1

, we want to move down to the next level. So consider

(

)

∈ B

n−1

. We would be done if

(

) is in the image of

n−1

. By exactness,

this is equivalent to saying

(

) is in the kernel of

n−1

. Since the diagram is

commutative, we know

n−1

◦ d

(y) = d

◦ j

(y) = d

(x) = 0,

using the fact that

is a cycle. So

(

)

∈ ker j

n−1

im i

n−1

. Moreover, by

exactness again,

n−1

is injective. So there is a unique

z ∈ A

n−1

such that

n−1

(z) = d

(y). We have now produced our z.

We are not done. We have

∂

∗

[

] = [

] as our candidate definition, but we

need to check many things:

(i) We need to make sure ∂

∗

is indeed a homomorphism.

(ii) We need d

n−1

(z) = 0 so that [z] ∈ H

n−1

(A);

(iii)

We need to check [

] is well-defined, i.e. it does not depend on our choice

of y and x for the homology class [x].

(iv) We need to check the exactness of the resulting sequence.

We now check them one by one:

(i)

Since everything involved in defining

∂

∗

are homomorphisms, it follows

that ∂

∗

is also a homomorphism.

(ii) We check d

n−1

(z) = 0. To do so, we need to add an additional layer.

0 A

n−1

0 A

n−2

n−1

n−2

We want to check that

n−1

(

) = 0. We will use the commutativity of the

diagram. In particular, we know

n−2

◦ d

n−1

(z) = d

n−1

◦ i

n−1

(z) = d

n−1

◦ d

(y) = 0.

By exactness at

n−2

, we know

n−2

is injective. So we must have

n−1

(z) = 0.

(iii) (a)

First, in the proof, suppose we picked a different

such that

(

) =

(

) =

. Then

(

− y

) = 0. So

− y ∈ ker j

im i

. Let

a ∈ A

be such that i

(a) = y

− y. Then

) = d

− y) + d

(y)

= d

◦ i

(a) + d

(y)

= i

n−1

◦ d

(a) + d

(y).

Hence when we pull back

(

) and

(

) to

n−1

, the results differ

by the boundary

(

), and hence produce the same homology class.

(b)

Suppose [

] = [

]. We want to show that

∂

∗

[

] =

∂

∗

[

]. This time,

we add a layer above.

0 A

n+1

0 A

n−1

n+1

n−1

By definition, since [x

] = [x], there is some c ∈ C

n+1

such that

= x + d

n+1

(c).

By surjectivity of

n+1

, we can write

n+1

(

) for some

b ∈ B

n+1

By commutativity of the squares, we know

= x + j

◦ d

n+1

(b).

The next step of the proof is to find some

such that

(

) =

Then

(y + d

n+1

(b)) = x

So the corresponding

n+1

(

). So

(

) =

(

), and

hence ∂

∗

[x] = ∂

∗

(iv)

This is yet another standard diagram chasing argument. When reading

this, it is helpful to look at a diagram and see how the elements are chased

along. It is even more beneficial to attempt to prove this yourself.

(a) im i

∗

⊆ ker j

∗

: This follows from the assumption that i

◦ j

= 0.

(b) ker j

∗

⊆ im i

∗

: Let [

]

∈ H

(

). Suppose

∗

([

]) = 0. Then there is

some

c ∈ C

n+1

such that

(

) =

n+1

(

). By surjectivity of

n+1

there is some

∈ B

n+1

such that

n+1

(

) =

. By commutativity,

we know j

(b) = j

◦ d

n+1

), i.e.

(b − d

n+1

)) = 0.

By exactness of the sequence, we know there is some

a ∈ A

such

that

(a) = b − d

n+1

Moreover,

n−1

◦ d

(a) = d

◦ i

(a) = d

(b − d

n+1

)) = 0,

using the fact that

is a cycle. Since

n−1

is injective, it follows that

(a) = 0. So [a] ∈ H

(A). Then

∗

([a]) = [b] − [d

n+1

)] = [b].

So [b] ∈ im i

∗

⊆ ker ∂

∗

: Let [

]

∈ H

(

). To compute

∂

∗

(

∗

([

])), we first

pull back

(

) to

b ∈ B

. Then we compute

(

) and then pull it

back to

n+1

. However, we know

(

) = 0 since

is a cycle. So

∂

∗

([b])) = 0, i.e. ∂

∗

◦ j

∗

= 0.

(d) ker ∂

∗

⊆ im j

∗

: Let [

]

∈ H

(

) and suppose

∂

∗

([

]) = 0. Let

b ∈ B

be such that

(

) =

, and

a ∈ A

n−1

such that

n−1

(

) =

(

By assumption,

∂

∗

([

]) = [

] = 0. So we know

is a boundary,

say

(

) for some

∈ A

. Then by commutativity we know

(b) = d

◦ i

). In other words,

(b − i

)) = 0.

So [b − i

)] ∈ H

(B). Moreover,

∗

([b − i

)]) = [j

(b) − j

◦ i

)] = [c].

So [c] ∈ im j

∗

(e) im ∂

∗

⊆ ker i

∗

: Let [

]

∈ H

(

). Let

b ∈ B

be such that

(

) =

and a ∈ A

n−1

be such that i

(a) = d

(b). Then ∂

∗

([c]) = [a]. Then

∗

([a]) = [i

(a)] = [d

(b)] = 0.

So i

∗

◦ ∂

∗

= 0.

(f) ker i

∗

⊆ im ∂

∗

: Let [

]

∈ H

(

) and suppose

∗

([

]) = 0. So we can

find some

b ∈ B

n+1

such that

(

) =

n+1

(

). Let

n+1

(

). Then

n+1

◦ j

n+1

(b) = j

◦ d

n+1

(b) = j

◦ i

(a) = 0.

So [

]

∈ H

(

). Then [

] =

∂

∗

([

]) by definition of

∂

∗

. So [

]

∈

im ∂

∗

7.5 Continuous maps and homotopy invariance

This is the most technical part of the homology section of the course. The goal is

to see that the homology groups

∗

(

) depend only on the polyhedron, and not

the simplicial structure on it. Moreover, we will show that they are homotopy

invariants of the space, and that homotopic maps

f ' g

|K| → |L|

induce

equal maps

∗

(

)

→ H

∗

(

). Note that this is a lot to prove. At this point, we

don’t even know arbitrary continuous maps induce any map on the homology.

We only know simplicial maps do.

We start with a funny definition.

Definition (Contiguous maps). Simplicial maps

f, g

K → L

are contiguous if

for each

σ ∈ K

, the simplices

(

) and

(

) (i.e. the simplices spanned by the

image of the vertices of σ) are faces of some some simplex τ ∈ L.

The significance of this definition comes in two parts: simplicial approxima-

tions of the same map are contiguous, and contiguous maps induce the same

maps on homology.

Lemma. If

f, g

K → L

are simplicial approximations to the same map

then f and g are contiguous.

Proof.

Let

σ ∈ K

, and pick some

s ∈ ˚σ

. Then

(

)

∈ ˚τ

for some

τ ∈ L

Then the definition of simplicial approximation implies that for any simplicial

approximation f to F , f(σ) spans a face of τ.

Lemma. If f, g : K → L are continguous simplicial maps, then

∗

= g

∗

: H

(K) → H

(L)

for all n.

Geometrically, the homotopy is obvious. If

(

) and

(

) are both faces of

then we just pick the homotopy as

. The algebraic definition is less inspiring,

and it takes some staring to see it is indeed what we think it should be.

Proof. We will write down a chain homotopy between f and g:

((a

, · · · , a

)) =

i=0

(−1)

[f(a

), · · · , f (a

), g(a

), · · · , g(a

)],

where the square brackets means the corresponding oriented simplex if there are

no repeats, 0 otherwise.

We can now check by direct computation that this is indeed a chain homotopy.

Now we know that if

f, g

K → L

are both simplicial approximations to the

same

, then they induce equal maps on homology. However, we do not yet know

that continuous maps induce well-defined maps on homologies, since to produce

simplicial approximations, we needed to perform barycentric subdivision, and

we need to make sure this does not change the homology.

Given a barycentric subdivision, we need to choose a simplicial approximation

to the identity map

→ K

. It turns out this is easy to do, and we can do

it almost arbitrarily.

Lemma. Each vertex

ˆσ ∈ K

is a barycenter of some

σ ∈ K

. Then we choose

(

ˆσ

) to be an arbitrary vertex of

. This defines a function

→ V

This a is a simplicial approximation to the identity. Moreover, every simplicial

approximation to the identity is of this form.

Proof. Omitted.

Next, we need to show this gives an isomorphism on the homologies.

Proposition. Let

be the barycentric subdivision of

, and

→ K

a simplicial approximation to the identity map. Then the induced map

∗

) → H

(K) is an isomorphism for all n.

Proof.

We first deal with

being a simplex

and its faces. Now that

is just

a cone (with any vertex as the cone vertex), and

is also a cone (with the

barycenter as the cone vertex). Therefore

(K)

∼

)

∼

(

Z n = 0

0 n > 0

So only

= 0 is (a little) interesting, but it is easy to check that

∗

is an

isomorphism in this case, since it just maps a vertex to a vertex, and all vertices

in each simplex are in the same homology class.

To finish the proof, note that

is built up by gluing up simplices, and

is built by gluing up subdivided simplices. So to understand their homology

groups, we use the Mayer-Vietoris sequence.

Given a complicated simplicial complex

, let

be a maximal dimensional

simplex of

. We let

K \ {σ}

(note that

includes the boundary of

We let S = {σ and all its faces} ⊆ K and T = L ∩ S.

We can do similarly for

, and let

, S

, T

be the corresponding barycentric

subdivisions. We have

L ∪ S

and

∪ S

(and

∩ S

). By the

previous lemma, we see our construction of

gives

(

)

⊆ L

(

)

⊆ S

and

a(T

) ⊆ T . So these induce maps of the corresponding homology groups

) H

) ⊕ H

) H

n−1

) H

n−1

) ⊕ H

n−1

)

(T ) H

(S) ⊕ H

(L) H

(K) H

n−1

(T ) H

n−1

(S) ⊕ H

n−1

(L)

∗

⊕a

∗

⊕a

∗

By induction, we can assume all but the middle maps are isomorphisms. By

the five lemma, this implies the middle map is an isomorphism, where the five

lemma is as follows:

Lemma (Five lemma). Consider the following commutative diagram:

If the top and bottom rows are exact, and

a, b, d, e

are isomorphisms, then

also an isomorphism.

Proof. Exercise in example sheet.

We now have everything we need to move from simplical maps to continuous

maps. Putting everything we have proved so far together, we obtain the following

result:

Proposition. To each continuous map

|K| → |L|

, there is an associated

map f

∗

: H

(K) → H

(L) (for all n) given by

∗

= s

∗

◦ ν

−1

K,r

where

(r)

→ L

is a simplicial approximation to

, and

K,r

(

(r)

)

→

(

) is the isomorphism given by composing maps

(

(i)

)

→ H

(

(i−1)

)

induced by simplical approximations to the identity.

Furthermore:

(i) f

∗

does not depend on the choice of r or s.

(ii) If g : |M | → |K| is another continuous map, then

(f ◦ g)

∗

= f

∗

◦ g

∗

Proof. Omitted.

Corollary. If

|K| → |L|

is a homeomorphism, then

∗

(

)

→ H

(

) is

an isomorphism for all n.

Proof. Immediate from (ii) of previous proposition.

This is good. We know now that homology groups is a property of the space

itself, not simplicial complexes. For example, we have computed the homology

groups of a particular triangulation of the

-sphere, and we know it is true for

all triangulations of the n-sphere.

We’re not exactly there yet. We have just seen that homology groups are

invariant under homeomorphisms. What we would like to know is that they are

invariant under homotopy. In other words, we want to know homotopy equivalent

maps induce equal maps on the homology groups.

The strategy is:

(i) Show that “small” homotopies don’t change the maps on H

(ii) Note that all homotopies can be decomposed into “small” homotopies.

Lemma. Let

be a simplicial complex (with

|L| ⊆ R

). Then there is an

(

)

0 such that if

f, g

|K| → |L|

satisfy

(

)

− g

(

)

k < ε

, then

∗

= g

∗

: H

(K) → H

(L) for all n.

The idea of the proof is that if

(

)

− g

(

)

is small enough, we can

barycentrically subdivide

such that we get a simplicial approximation to both

f and g.

Proof.

By the Lebesgue number lemma, there is an

ε >

0 such that each ball of

radius 2ε in |L| lies in some star St

(w).

Now apply the Lebesgue number lemma again to

−1

(

))

}

y∈|L|

, an open

cover of |K|, and get δ > 0 such that for all x ∈ |K|, we have

f(B

(x)) ⊆ B

(y) ⊆ B

2ε

(y) ⊆ St

(w)

for some y ∈ |L| and St

(w). Now since g and f differ by at most ε, we know

g(B

(x)) ⊆ B

2ε

(y) ⊆ St

(w).

Now subdivide r times so that µ(K

(r)

) <

δ. So for all v ∈ V

(r)

, we know

(r)

(v) ⊆ B

(v).

This gets mapped by both

and

(

) for the same

w ∈ V

. We define

s : V

(r)

→ V

sending v 7→ w.

Theorem. Let f ' g : |K| → |L|. Then f

∗

= g

∗

Proof.

Let

|K| × I → |L|

. Since

|K| × I

is compact, we know

is uniformly

continuous. Pick

(

) as in the previous lemma. Then there is some

such

that |s − t| < δ implies |H(x, s) − H(x, t)| < ε for all x ∈ |K|.

Now choose 0 =

< t

< · · · < t

= 1 such that

− t

i−1

< δ

for all

Define

|K| → |L|

(

) =

(

x, t

). Then we know

− f

i−1

k < ε

for all

i. Hence (f

)

∗

= (f

i−1

)

∗

. Therefore (f

)

∗

= (f

)

∗

, i.e. f

∗

= g

∗

This is good, since we know we can not only deal with spaces that are

homeomorphic to complexes, but spaces that are homotopic to complexes. This

is important, since all complexes are compact, but we would like to talk about

non-compact spaces such as open balls as well.

Definition (

-triangulation and homology groups). An

-triangulation of a

space

is a simplicial complex

and a homotopy equivalence

|K| → X

We define H

(X) = H

(K) for all n.

Lemma. H

(X) is well-defined, i.e. it does not depend on the choice of K.

Proof. Clear from previous theorem.

We have spent a lot of time and effort developing all the technical results

and machinery of homology groups. We will now use them to do stuff.

Digression — historical motivation

At first, we motivated the study of algebraic topology by saying we wanted to

show that

and

are not homeomorphic. However, historically, this is not

why people studied algebraic topology, since there are other ways to prove this

is true.

Initially, people were interested in studying knots. These can be thought of

as embeddings

S → S

. We really should just think of

∪ {∞}

, where

the point at infinity is just there for convenience.

The most interesting knot we know is the unknot U :

A less interesting but still interesting knot is the trefoil T .

One question people at the time asked was whether the trefoil knot is just the

unknot in disguise. It obviously isn’t, but can we prove it? In general, given two

knots, is there any way we can distinguish if they are the same?

The idea is to study the fundamental groups of the knots. It would obviously

be silly to compute the fundamental groups of

and

themselves, since they

are both isomorphic to

and the groups are just

. Instead, we look at the

fundamental groups of the complements. It is not difficult to see that

\ U)

∼

and with some suitable machinery, we find

\ T )

∼

ha, b | a

−2

Staring at it hard enough, we can construct a surjection

(

\ T

)

→ S

. This

tells us

(

\ T

) is non-abelian, and is certainly not

. So we know

and

are genuinely different knots.

7.6 Homology of spheres and applications

Lemma. The sphere S

n−1

is triangulable, and

n−1

)

∼

(

Z k = 0, n − 1

0 otherwise

Proof.

We already did this computation for the standard (

n −

1)-sphere

∂

∆

where ∆

is the standard n-simplex.

We immediately have a few applications

Proposition. R

∼

for m 6= n.

Proof. See example sheet 4.

Theorem (Brouwer’s fixed point theorem (in all dimensions)). There is no

retraction

onto

∂D

∼

n−1

. So every continuous map

→ D

has a

fixed point.

Proof.

The proof is exactly the same as the two-dimensional case, with homology

groups instead of the fundamental group.

We first show the second part from the first. Suppose

→ D

has no

fixed point. Then the following g : D

→ ∂D

is a continuous retraction.

f(x)

g(x)

So we now show no such continuous retraction can exist. Suppose

→ ∂D

is a retraction, i.e. r ◦ i ' id : ∂D

→ ∂D

n−1

i r

We now take the (n − 1)th homology groups to obtain

n−1

) H

n−1

) H

n−1

∗

Since

r ◦ i

is homotopic to the identity, this composition must also be the identity,

but this is clearly nonsense, since

n−1

(

n−1

)

∼

while

n−1

(

)

∼

0. So

such a continuous retraction cannot exist.

Note it is important that we can work with continuous maps directly, and

not just their simplicial approximations. Otherwise, here we can only show

that every simplicial approximation of

has a fixed point, but this does not

automatically entail that f itself has a fixed point.

For the next application, recall from the first example sheet that if

is odd,

then the antipodal map

→ S

is homotopic to the identity. What if

even?

The idea is to consider the effect on the homology group:

∗

: H

) → H

By our previous calculation, we know

∗

is a map

∗

Z → Z

. If

is homotopic

to the identity, then

∗

should be homotopic to the identity map. We will now

compute a

∗

and show that it is multiplication by −1 when n is even.

To do this, we want to use a triangulation that is compatible with the

antipodal map. The standard triangulation clearly doesn’t work. Instead, we

use the following triangulation h : |K| → S

The vertices of K are given by

= {±e

, ±e

, · · · , ±e

This triangulation works nicely with the antipodal map, since this maps a vertex

to a vertex. To understand the homology group better, we need the following

lemma:

Lemma. In the triangulation of

given by vertices

{±

, ±

, · · · , ±

}

the element

x =

ε∈{±1}

n+1

· · · ε

(ε

, · · · , ε

)

is a cycle and generates H

Proof.

By direct computation, we see that

= 0. So

is a cycle. To show it

generates

(

), we note that everything in

(

)

∼

is a multiple of the

generator, and since

has coefficients

1, it cannot be a multiple of anything

else (apart from −x). So x is indeed a generator.

Now we can prove our original proposition.

Proposition. If n is even, then the antipodal map a 6' id.

Proof.

We can directly compute that

∗

= (

−

n+1

. If

is even, then

∗

−

but id

∗

= 1. So a 6' id.

7.7 Homology of surfaces

We want to study compact surfaces and their homology groups. To work with

the simplicial homology, we need to assume they are triangulable. We will not

prove this fact, and just assume it to be true (it really is).

Recall we have classified compact surfaces, and have found the following

orientable surfaces Σ

We also have non-compact versions of these, known as

, where we take the

above ones and cut out a hole:

We want to compute the homology groups of these Σ

. One way to do so is

to come up with specific triangulations of the space, and then compute the

homology groups directly. However, there is a better way to do it. Given a Σ

we can slice it apart along a circle and write it as

= F

g−1

∪ F

Then, we need to compute

∗

(

). Fortunately, this is not too hard, since it

turns out

is homotopic to a relatively simple space. Recall that we produced

by starting with a 4g-gon and gluing edges:

Now what is

is Σ

with a hole cut out of it. We’ll cut the hole out at

the center, and be left with

We can now expand the hole to the boundary, and get a deformation retraction

from F

to its boundary:

Gluing the edges together, we obtain the rose with 2

petals, which we shall call

Using the Mayer-Vietoris sequence (exercise), we see that

)

∼

)

∼











Z n = 0

n = 1

0 n > 1

We now compute the homology groups of Σ

. The Mayer-Vietoris sequence gives

the following

0 H

) H

g−1

) ⊕ H

) H

(Σ

)

) H

g−1

) ⊕ H

) H

(Σ

)

) H

g−1

) ⊕ H

) H

(Σ

) 0

We can put in the terms we already know, and get

0 H

(Σ

) Z Z

(Σ

) Z Z

Z 0

By exactness, we know

(Σ

) =

ker{Z → Z

}

. We now note that this map

is indeed the zero map, by direct computation — this map sends

to the

boundary of the hole of

g−1

and

. If we look at the picture, after the

deformation retraction, this loop passes through each one-cell twice, once with

each orientation. So these cancel, and give 0. So H

(Σ

) = Z.

To compute

(Σ

), for convenience, we break it up into a short exact

sequence, noting that the function Z → Z

is zero:

0 Z

(Σ

) ker(Z → Z

) 0.

We now claim that

Z → Z

is injective — this is obvious, since it sends 1

7→

1).

So the kernel is zero. So

(Σ

)

∼

This is a typical application of the Mayer-Vietoris sequence. We write down

the long exact sequence, and put in the terms we already know. This does not

immediately give us the answer we want — we will need to understand one or

two of the maps, and then we can figure out all the groups we want.

7.8 Rational homology, Euler and Lefschetz numbers

So far, we have been working with chains with integral coefficients. It turns out

we can use rational coefficients instead. In the past,

(

) was an abelian group,

or a

-module. If we use rational coefficients, since

is a field, this becomes a

vector space, and we can use a lot of nice theorems about vector spaces, such as

the rank-nullity theorem. Moreover, we can reasonably talk about things like

the dimensions of these homology groups.

Definition (Rational homology group). For a simplicial complex

, we can de-

fine the rational

-chain group

(

K, Q

) in the same way as

(

) =

(

K, Z

That is,

(

K, Q

) is the vector space over

with basis the

-simplices of

(with a choice of orientation).

We can define d

, Z

, B

as before, and the rational nth homology group is

(K; Q)

∼

(K; Q)

Now our homology group is a vector space, and it is much easier to work with.

However, the consequence is that we will lose some information. Fortunately,

the way in which we lose information is very well-understood and rather simple.

Lemma. If H

(K)

∼

⊕ F for F a finite group, then H

(K; Q)

∼

Proof. Exercise.

Hence, when passing to rational homology groups, we lose all information

about the torsion part. In some cases this information loss is not very significant,

but in certain cases, it can be. For example, in RP

, we have

(RP

)

∼











Z n = 0

Z/2 n = 1

0 n > 1

If we pass on to rational coefficients, then we have lost everything in

(

and RP

looks just like a point:

(RP

; Q)

∼

(∗; Q).

Example. We have

; Q)

∼

(

Q k = 0, n

0 otherwise

We also have

(Σ

, Q)

∼











Q k = 0, 2

k = 1

0 otherwise

In this case, we have not lost any information because there was no torsion part

of the homology groups.

However, for the non-orientable surfaces, one can show that

) =











Z k = 0

n−1

× Z/2 k = 1

0 otherwise

; Q) =











Q k = 0

n−1

k = 1

0 otherwise

This time, this is different from the integral coefficient case, where we have

an extra Z

term in H

The advantage of this is that the homology “groups” are in fact vector spaces,

and we can talk about things like dimensions. Also, maps on homology groups

are simply linear maps, i.e. matrices, and we can study their properties with

proper linear algebra.

Recall that the Euler characteristic of a polyhedron is defined as “faces

−

edges + vertices”. This works for two-dimensional surfaces, and we would like to

extend this to higher dimensions. There is an obvious way to do so, by counting

higher-dimensional surfaces and putting the right signs. However, if we define it

this way, it is not clear that this is a property of the space itself, and not just a

property of the triangulation. Hence, we define it as follows:

Definition (Euler characteristic). The Euler characteristic of a triangulable

space X is

χ(X) =

i≥0

(−1)

dim

(X; Q).

This clearly depends only on the homotopy type of

, and not the triangula-

tion. We will later show this is equivalent to what we used to have.

More generally, we can define the Lefschetz number.

Definition (Lefschetz number). Given any map

X → X

, we define the

Lefschetz number of f as

L(f) =

i≥0

(−1)

tr(f

∗

: H

(X; Q) → H

(X; Q)).

Why is this a generalization of the Euler characteristic? Just note that the

trace of the identity map is the number of dimensions. So we have

χ(X) = L(id).

Example. We have

χ(S

) =

(

2 n even

0 n odd

We also have

χ(Σ

) = 2 − 2g, χ(E

) = 2 − n.

Example. If

→ S

is the antipodal map, we saw that

∗

(

)

→

) is multiplication by (−1)

n+1

. So

L(α) = 1 + (−1)

(−1)

n+1

= 1 − 1 = 0.

We see that even though the antipodal map has different behaviour for

different dimensions, the Lefschetz number ends up being zero all the time. We

will soon see why this is the case.

Why do we want to care about the Lefschetz number? The important thing

is that we will use this to prove a really powerful generalization of Brouwer’s

fixed point theorem that allows us to talk about things that are not balls.

Before that, we want to understand the Lefschetz number first. To define the

Lefschetz number, we take the trace of

∗

, and this is a map of the homology

groups. However, we would like to understand the Lefschetz number in terms of

the chain groups, since these are easier to comprehend. Recall that the homology

groups are defined as quotients of the chain groups, so we would like to know

what happens to the trace when we take quotients.

Lemma. Let

be a finite-dimensional vector space and

W ≤ V

a subspace.

Let

V → V

be a linear map such that

(

)

⊆ W

. Let

W → W

and C : V/W → V/W the induced map on the quotient. Then

tr(A) = tr(B) + tr(C).

Proof. In the right basis,

A =



B A

0 C



What this allows us to do is to not look at the induced maps on homology,

but just the maps on chain complexes. This makes our life much easier when it

comes to computation.

Corollary. Let f

: C

(K; Q) → C

(K; Q) be a chain map. Then

i≥0

(−1)

tr(f

: C

(K) → C

(K)) =

i≥0

(−1)

tr(f

∗

: H

(K) → H

(K)),

with homology groups understood to be over Q.

This is a great corollary. The thing on the right is the conceptually right

thing to have — homology groups are nice and are properties of the space itself,

not the triangulation. However, to actually do computations, we want to work

with the chain groups and actually calculate with chain groups.

Proof. There is an exact sequence

0 B

(K; Q) Z

(K; Q) H

(K; Q) 0

This is since

(

K, Q

) is defined as the quotient of

over

. We also have

the exact sequence

0 Z

(K; Q) C

(K; Q) B

i−1

(K; Q) 0

This is true by definition of

i−1

and

. Let

, f

be the various

maps induced by f on the corresponding groups. Then we have

L(|f|) =

i≥0

(−1)

tr(f

)

i≥0

(−1)

(tr(f

) − tr(f

))

i≥0

(−1)

(tr(f

) − tr(f

i−1

) − tr(f

)).

Because of the alternating signs in dimension, each

appears twice in the sum

with opposite signs. So all f

cancel out, and we are left with

L(|f|) =

i≥0

(−1)

tr(f

Since

(

id |

(K;Q)

) =

dim

(

;

), which is just the number of

-simplices,

this tells us the Euler characteristic we just defined is the usual Euler character-

istic, i.e.

χ(X) =

i≥0

(−1)

number of i-simplices.

Finally, we get to the important theorem of the section.

Theorem (Lefschetz fixed point theorem). Let

X → X

be a continuous

map from a triangulable space to itself. If L(f) 6= 0, then f has a fixed point.

Proof.

We prove the contrapositive. Suppose

has no fixed point. We will show

that L(f ) = 0. Let

δ = inf{|x − f(x)| : x ∈ X}

thinking of

as a subset of

. We know this is non-zero, since

has no fixed

point, and X is compact (and hence the infimum point is achieved by some x).

Choose a triangulation L : |K| → X such that µ(K) <

. We now let

g : K

(r)

→ K

be a simplicial approximation to

. Since we picked our triangulation to be so

fine, for x ∈ σ ∈ K, we have

|f(x) − g(x)| <

since the mesh is already less than

. Also, we know

|f(x) − x| ≥ δ.

So we have

|g(x) − x| >

So we must have

(

)

6∈ σ

. The conclusion is that for any

σ ∈ K

, we must have

g(σ) ∩ σ = ∅.

Now we compute L(f) = L(|g|). The only complication here is that g is a map

from

(r)

, and the domains and codomains are different. So we need to

compose it with

(

;

)

→ C

(

(r)

;

) induced by inverses of simplicial

approximations to the identity map. Then we have

L(|g|) =

i≥0

(−1)

tr(g

∗

: H

(X; Q) → H

(X; Q))

i≥0

(−1)

tr(g

◦ s

: C

(K; Q) → C

(K; Q))

Now note that

takes simplices of

to sums of subsimplices of

. So

◦ s

takes every simplex off itself. So each diagonal terms of the matrix of

◦ s

0! Hence the trace is

L(|g|) = 0.

Example. If

is any contractible polyhedron (e.g. a disk), then

(

) = 1 for

any

X → X

, which is obvious once we contract the space to a point. So

has a fixed point.

Example. Suppose

is a path-connected topological group, i.e.

is a group

and a topological space, and inverse and multiplication are continuous maps.

If g 6= 1, then the map

: G → G

γ 7→ γg

has no fixed point. This implies

(

) = 0. However, since the space is path

connected,

' id

, where the homotopy is obtained by multiplying elements

along the path from e to r

. So

χ(G) = L(id

) = L(r

) = 0.

So if G 6= 1, then in fact χ(G) = 0.

This is quite fun. We have worked with many surfaces. Which can be

topological groups? The torus is, since it is just

× S

, and

is a topological

group. The Klein bottle? Maybe. However, other surfaces cannot since they

don’t have Euler characteristic 0.