In the last post, we’ve discussed about the fundamental group, defined properly the functor between the category of pairs to the category of groups.

In this post I want to keep talking about the fundamental group and show the first connection between a topological property to the group structure. However, I am going to use some stuff from topology first, so let’s do it:

## Quotient maps

What is a qutient map? If you have studied topology before, then I am pretty sure that you are already familiar with this term. If not, I’ll quickly go over it and present here the things that are necessary to us.

Suppose that we have a topological space – where is the topolgy on .

I am interested in defining a new topological space which is what we get from when we are ‘gluing’ points toghether.

How is this thing is done formally? Well, with a map of course. It is obvious that this map needs be **onto** / **surjective**. Let’s denote the map as .

Two point of are ‘glued’ together if they are sent to the same point. i.e. (If the function is **injective** / **one-to-one** then no points are glued together).

In fact, induces an **equivalence realtion** on the space , Where:

x\sim_fy\iff f(x)=f(y)

Therefore, the space is actually the space . Thus, we call a **quotient space**.

However, we still don’t know what’s the topology on . First off all – the minimal requirement from – the topology on is to be one that makes a continuous function. Therefore, we shall demand:

O\in\sigma \Rightarrow f^{-1}(O)\in\tau

On the other hand, we still want to keep the properties of as much as we can. Therefore, we would want that if a pre-image of a subset is open, then itself would be open as well. Formally:

f^{-1}(O)\in\tau\Rightarrow O\in\sigma

Therefore, we will define as:

\sigma:=\{O\subseteq Y:f^{-1}(O)\in\tau\}

i.e. A set is open in if and only if is open in .

The map is called a **quotient map**.

**Important remark**

if is a qutient map, it **does not** mean that if then .

Recall that the map doesn’t have to be one-to-one, therefore – if , in most cases: . Some elements outside of may be sent to as well, therefore, the pre-image may be some (not open) set that **contains **.

Make sure that you understand it!

However, if (such a map is called an **open map**), and is onto then is indeed a qutient space, you can try and prove it yourself

**note**: I am not going to go over some proofs here since quotient spaces will ‘get’ their own post when I’ll write about topology.

#### Strong Topology property

I am now going to prove the main property that I will mostly use for the rest of the post.

Suppose that is a qoutient map and . Then is continuous **if and only if** is continuous.

If we know that is continuous, then so as and vice versa.

First notice that if is continuous, then so as as a compostion of continuous maps. Conversely, suppose that is continuous. We need to show that is continuous as well. So, let be an open set. We want to prove that is open as well. Notice that is open since is continuous, moreover:

(f\circ q)^{-1}(O)=q^{-1}(f^{-1}(O))

(Convince yourself that the equation in the above is true). Now, is a quotient map, therefore is open as we wanted.

#### Some more properties

I am going to list here some more properties without a proof that I am going to use here. Again, I **will** prove them, however – not in this post.

- If is continuous, onto and open / closed then is a quotient map
- If where is compact and is hausdorff () then is a
**closed**map. - An onto function where is compact and is hausdorff is a qutient map

The third one follows immediately from the first two – and this is the one I will use.

#### Respecting the relation

Suppose that and is an equivalence relation on . We say that **respects** the relation if:

a\sim b\Rightarrow f(a)=f(b)

Now, for the last property I want to mention:

respects the relation obtained by the (quotient) map () **if and only if** the function:

\overline{f}:X/_\sim \to Y, \overline f([x])=f(x)

is well defined.

The proof is pretty straightforward: Suppose that respects the relation and , then , thus: . On the other hand, suppose that is well defined. Therefore if then thus , then .

## Back to our business

That was kind of a long intro, however, I am going to use everything from there – so it wasn’t for nothing.

I would like to prove a major theorem, that will help us a lot in this post, and also in the future.

Suppose that is a continuous map. Then the following are equivalent:

- is
**null-homotopic**(homotopic to a constant function) - is null-homotopic
**with respect**to some point in . - If we will think of as a subset of (in fact, ), then we can extend to a continuous map .

We will see right after the proof how we can use the theorem to conclude a non-trivial reslut. However, I want to note one thing: The theorem holds not just for and – but for every other spaces that are homeomorphic to them. For example, I will use this theorem on the unit square and it’s boundary.

Ok, let’s start proving:

#### Proof

is trivial, since 2 is **stornger** than 1, it requires not only a homotopy – but a homotopy with respect to a point on the circle.

: We need to find a homotopy with respect to a point :

H:S^1\times I\to X

With for some .

Let’s pick some arbitrary point and define a homomtpy:

K:S^1\times I\to D^2,K(s,t)=(1-t)\cdot i(s) + t\cdot K_m=(1-t)x+t\cdot m

Notice that is a homotopy from to with respect to . Thus,

By our assumption, there exists that extends . Using the statement from the previous post, we can conclude that:

F\circ i\sim_mF\circ K_m

However, is a constant map, and by definition. Therefore:

f\sim_mK_p

As we wanted.

We only need to prove that in order to finish the proof. This is where our introduction is playing it’s part.

##### The interesting part of the proof

By given, is homotopic to a point. Therefore, there is a homotopy such that for some . Our goal is to define a map that extends . To do so, I am going to define a map:

\rho:S^1\times I\to D^2,\rho(s,t)=(1-t)\cdot s

The first thing you might have noticed is that is a homotopy from the inculsion map to the constant map .

However, I want to look on this map from a different aspect: What is the space ? It is a **cylinder**. is mapping the cylinder into the disc . How? The lower base of the cylinder is staying at it’s place. As (which is the height) is increasing, the circle is mapped to a smaller circle inside , where the higher base – is mapped to a single point – 0.

Clearly this function is **onto**, and it is one-to-one everywhere **exept** the higher base – which is sent to zero. (Recall that we are only considering the boundary of the cylender – the shape we are dealing with is like a toilet paper cylinder).

Therefore, induces an equivalence relation, where two points are equivalent if and only if they are the same or they are both of the form .

(s_1,t_1)\sim(s_2,t_2)\iff(s_1,t_1)=(s_2,t_2)\lor t_1=t_2=1

Notice that:

Therefore, **respects** the equivalence relation induced by . Moreover:

S^1/_{\sim_\rho}=D^2

Thus, a map where is well defined and extends (since ). We now have the diagram:

The only thing left to prove now it that is continuous. To prove that, I an going to use the facts that is **compact**, is **hausdorff **and is **contiuous **and **onto**. Therefore, is a **quotient map**! Thus, we can use the ‘strong topology property’ and the fact that is continuous to conclude that is continuous.

That’s it!

#### Proof review

I don’t know how about you, but I find this proof **extraordinary**. By changing our point of view and realizing what **shapes** and the **spaces** we are dealing with – we were able to prove this theorem in such a simple way using abstract arguments that fits this case in such a pefrect way!

## What’s next?

Using this theorem allows us to prove a really cool lemma:

Suppose that is a homotpy from to where both of them are loops with base points respectively. such that for every . Let’s draw a square of this situation:

Recall that for every , . Therefore, we can think of both of the **vertical** edges of the square, as the **same path** such that:

\gamma(0)=\varphi(0)=\varphi(1)=a,\ \ \gamma(1)=\psi(0)=\psi(1)=b

Notice the red arrows I’ve added to the image. I am now going to define a new path – a one where I am ‘walking’ on the boundary of the rectangle, The path is gonna be:

\overline\psi*\overline\gamma*\varphi*\gamma

We know that this path has a continuous extension to the whole square (the extension is just ). Using the theorem, we conclude that this path is null homotopic, thus it is homotopic to the point with respect to (here we used the part picking a point from the circle is the same thing as picking the endpoints of the unit inerval [why?]):

\overline\psi*\overline\gamma*\varphi*\gamma\sim_{\partial I}K_b

Thus:

[\overline\psi][\overline\gamma][\varphi][\gamma]=[K_b]\Rightarrow[\overline\gamma][\varphi][\gamma]=[\psi]

Remember this fact, we are going to use it right away.

## Solving the problem from the previous post

If you remember, we’ve seen that if and , then sent the fundamental group at to different groups:

We can now solve this problem, it turns out that there exists some with such that: (Recall that .

Let’s prove it:

Let be a homotopy from to . Define . Indeed, and .

Now, let be some element. We need to show that:

g_{*}([\varphi])=F_\gamma\circ f_{*}([\varphi])=F_\gamma(f_{*}([\varphi]))

Which is exactly the same as proving:

[g\circ\varphi]=[\overline{\gamma}][f\circ\varphi][\gamma]

We can now define a homotopy . The square in this case is:

Since , and:.

And we can now use the lemma to get that .

## A wonderful result

I am now going to present the theorem I was building up to. I am not going to prove it in this post since it’s already kind of long and the proof is not that short. Moreover, the theorem is really importnat and deserves a separate post. Ok, l shall present it now:

Suppose that is an **homotopy equivalence**. Then, for every :

f_{*}:\pi_1(X,a)\to\pi_1(Y,f(a))

Is a **group isomorphism**!

This theorem allows us to what we wanted from the first place: Studying a space through a group.

We now know that two homotopic equivalent spaces have the same (more precisely, isomorphic) fundamental groups. In other words, if two group doesn’t have isomorphic fundemental groups – then they are not homotopic equivalent (thus, not homeomorphic)!

## One thought on “The fundamental Group – Part 2”