# The fundamental Group – Part 2

In the last post, we’ve discussed about the fundamental group, defined properly the functor between the category of pairs $(X,a)$ 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 $(X,\tau)$ – where $\tau$ is the topolgy on $X$.

I am interested in defining a new topological space $(Y,\sigma)$ which is what we get from $X$ 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 $f:X\to Y$.

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

In fact, $f$ induces an equivalence realtion $\sim_f$ on the space $X$, Where:

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

Therefore, the space $Y$ is actually the space $X/\sim_f$. Thus, we call $Y$ a quotient space.

However, we still don’t know what’s the topology on $Y$. First off all – the minimal requirement from $\sigma$ – the topology on $Y$ is to be one that makes $f$ 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 $X$ as much as we can. Therefore, we would want that if a pre-image of a subset$O\subseteq Y$ is open, then $O$ itself would be open as well. Formally:

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

Therefore, we will define $\sigma$ as:

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

i.e. A set $O$ is open in $Y$ if and only if $f^{-1}(O)$ is open in $X$.

The map $f$ is called a quotient map.

##### Important remark

if $q:X\to Y$ is a qutient map, it does not mean that if $O\in\tau$ then $q(O)\in\sigma$.

Recall that the map $q$ doesn’t have to be one-to-one, therefore – if $f(O)\in Y$, in most cases: $f^{-1}(f(O))\neq O$. Some elements outside of $O$ may be sent to $f(O)$ as well, therefore, the pre-image may be some (not open) set that contains $O$.

Make sure that you understand it!

However, if $O\in\tau \Rightarrow q(O)\in \tau$ (such a map is called an open map), and $q$ is onto then $f$ 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 $q:X\to Y$ is a qoutient map and $f:Y\to Z$. Then $f$ is continuous if and only if $f\circ q:X\to Z$ is continuous.

If we know that $f\circ q$ is continuous, then so as $f$ and vice versa.

First notice that if $f$ is continuous, then so as $f\circ q$ as a compostion of continuous maps. Conversely, suppose that $f\circ q$ is continuous. We need to show that $f$ is continuous as well. So, let $O\subseteq Z$ be an open set. We want to prove that $f^{-1}(O)$ is open as well. Notice that $(f\circ q)^{-1}(O)$ is open since $f\circ q$ is continuous, moreover:

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

(Convince yourself that the equation in the above is true). Now, $q$ is a quotient map, therefore $f^{-1}(O)$ 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 $q:X\to Y$ is continuous, onto and open / closed then $q$ is a quotient map
• If $f:X\to Y$ where $X$ is compact and $Y$ is hausdorff ($T_2$) then $f$ is a closed map.
• An onto function $f:X\to Y$ where $X$ is compact and $Y$ 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 $f:X\to Y$ and $\sim$ is an equivalence relation on $X$. We say that $f$ respects the relation if:

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

Now, for the last property I want to mention:

$f:X\to Y$ respects the relation $\sim$ obtained by the (quotient) map $q$ ($a\sim b \iff q(a)=q(b)$) 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 $f$ respects the relation and $[x]=[y]$, then $x\sim y\Rightarrow f(x)=f(y)$, thus: $\overline f([x])=f(x)=f(y)=\overline f([y])$. On the other hand, suppose that $\overline f$ is well defined. Therefore if $x\sim y$ then $[x]=[y]$ thus $\overline f([x])= \overline f([y])$, then $f(x)=f(y)$.

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 $f:S^1\to X$ is a continuous map. Then the following are equivalent:

1. $f$ is null-homotopic (homotopic to a constant function)
2. $f$ is null-homotopic with respect to some point in $S_1$.
3. If we will think of $S^1$ as a subset of $D^2$ (in fact, $S^1=\partial D_2$), then we can extend $f$ to a continuous map $F:D^2\to X$.

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 $S^1$ and $D^2$ – but for every other spaces that are homeomorphic to them. For example, I will use this theorem on the unit square $I\times I$ and it’s boundary.

Ok, let’s start proving:

#### Proof

$(2\Rightarrow 1)$ 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.

$(3\Rightarrow 2)$: We need to find a homotopy with respect to a point $s\in S^1$:

H:S^1\times I\to X

With $H(x,0)=f(x),H(x,1)=K_p$ for some $p\in X$.

Let’s pick some arbitrary point $m\in S^1$ 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 $K$ is a homotopy from $i:S^1\to D^2$ to $K_m:S^1\to D_2$ with respect to $m\in S^1$. Thus, $i\sim_m K_m$

By our assumption, there exists $F:D^2\to X$ that extends $f$. Using the statement from the previous post, we can conclude that:

F\circ i\sim_mF\circ K_m

However, $F\circ K_m$ is a constant map, and $F\circ i = f$ by definition. Therefore:

f\sim_mK_p

As we wanted.

We only need to prove that $(1\Rightarrow 3)$ in order to finish the proof. This is where our introduction is playing it’s part.

##### The interesting part of the proof

By given, $f$ is homotopic to a point. Therefore, there is a homotopy $H:S^1\times I\to X$ such that $H(x,0) = f(x), H(x,1) = K_p$ for some $p\in X$. Our goal is to define a map $F:D^2\to X$ that extends $f$. 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 $\rho$ is a homotopy from the inculsion map $i:S^1\to D^2$ to the constant map $K_0$.

However, I want to look on this map from a different aspect: What is the space $S^1\times I$? It is a cylinder. $\rho$ is mapping the cylinder into the disc $D^2$. How? The lower base of the cylinder $\rho(s,0)$ is staying at it’s place. As $t$ (which is the height) is increasing, the circle $\rho(s,t)$ is mapped to a smaller circle inside $D^2$, where the higher base – $\rho(s,1)$ 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, $\rho$ induces an equivalence relation, where two points are equivalent if and only if they are the same or they are both of the form $(s,1)$.

(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:

• $(s_1,t_1)=(s_2,t_2)\Rightarrow H(s_1,t_1)=H(s_2,t_2)$
• $H(s_1,1)=H(s_2,1)=p$

Therefore, $H$ respects the equivalence relation induced by $\rho$. Moreover:

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

Thus, a map $h:D^2\to X$ where $H=h\circ \rho$ is well defined and extends $f$ (since $h([(s,0)])=H(s,0)=f(s)$). We now have the diagram:

The only thing left to prove now it that $h$ is continuous. To prove that, I an going to use the facts that $S^1\times I$ is compact, $D^2$ is hausdorff and $\rho$ is contiuous and onto. Therefore, $\rho$ is a quotient map! Thus, we can use the ‘strong topology property’ and the fact that $H=h\circ \rho$ is continuous to conclude that $h$ 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 $H:I \times I\to X$ is a homotpy from $\varphi$ to $\psi$ where both of them are loops with base points $a, b$ respectively. such that $H(0,t)=H(1,t)$ for every $t\in I$. Let’s draw a square of this situation:

Recall that for every $t\in I$, $H(0,t)=H(1,t)$. Therefore, we can think of both of the vertical edges of the square, as the same path $\gamma$ 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 $H$). Using the theorem, we conclude that this path is null homotopic, thus it is homotopic to the point $\gamma(1)=b$ with respect to $\partial I$ (here we used the part $(3\Rightarrow 2)$ 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 $f,g:X\to Y$ and $f\sim g$, then $f_{*}, g_{*}$ sent the fundamental group at $a$ to different groups:

We can now solve this problem, it turns out that there exists some $\gamma$ with $\gamma(0)=f(a),\gamma(1)=g(a)$ such that: $g_{*}= F_\gamma\circ f_{*}$ (Recall that $F_\gamma([\psi])=[\overline\gamma][\psi][\gamma]$.

Let’s prove it:

Let $H:X\times I \to Y$ be a homotopy from $f$ to $g$. Define $\gamma(t):=H(a,t)$. Indeed, $\gamma(0)=H(a,0)=f(a)$ and $\gamma(1)=H(a,1)=g(a)$.

Now, let $[\varphi]\in\pi_1(X,a)$ 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 $K(s,t):=H(\varphi(s),t)$. The square in this case is:

Since $K(0,t)=H(\varphi(0),t)=H(a,t)=\gamma(t)$, and:$K(1,t)=H(\varphi(1),t)=H(a,t)=\gamma(t)$.

And we can now use the lemma to get that $[g\circ\varphi]=[\overline{\gamma}][f\circ\varphi][\gamma]$.

## 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 $f:X\to Y$ is an homotopy equivalence. Then, for every $a\in X$:

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)!