II Algebraic Topology - Homotopy and the fundamental group

2Homotopy and the fundamental group

II Algebraic Topology

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.