# Homotopy Equivalence To Isomorphism

In the last post I have presented a great theorem that allows us to translate homotopy equivalence to isomorphism. Let’s give a quick reminder:

If $f:X\to Y$ is a homotopy equivalence then for every $a\in X$:

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

is a group isomorphism.

In this post, I am going to prove this theorem, and present some conclusions. Ok, so I shall begin now.

## Proving The Theorem

Before I’ll start proving this theorem I want to give two quick reminders from elementary set theory.

#### Quick preparations

• If $f:A\to B, g:B\to C$ such that $g\circ f: A\to C$ is one to one, then $f$ is one-to-one.
• The proof is pretty straightforward: Suppose that $f(a_1)=f(a_2)$ where $a_1,a_2\in A$. Then $g(f(a_1))=g(f(a_2))$. However, $g\circ f$ is one-to-one, therefore $a_1 = a_2$. Thus, $f$ is one to one as well.
• With the same functions $f,g$, we now assume that $g\circ f$ is onto, then $g$ is onto as well.
• Again, this is also pretty easy. Let $c\in C$ be some element, since $g\circ f$ is onto, there is some $a\in A$ such that $g\circ f(a)=g(f(a))=c$. Thus, $f(a)\in B$ is an element such that $g(f(a))=c$. Thus $g$ is onto.

The last thing I want to remind is a theorem from the previous post:

If $f,g:X\to Y$ such that $f\sim g$ and $a\in X$, Then there is a path $\gamma$ from $f(a)$ to $g(a)$ such that $g_{*}=F_\gamma\circ f_{*}$. We also proved that $F_\gamma$ is an isomorphism (here).

Ok, we are now ready to the proof

## The proof

We know that $f:X\to Y$ is an homotopic equivalence (The definition is here). By definition, there is some $g:Y\to X$ such that:

• $g\circ f\sim Id_X$.
• $f\circ g \sim Id_Y$.

We are going to use both of this facts. Let’s pick some $a\in X$ and start with the first:

#### Using Fact 1

Since $g\circ f\sim Id_X$ then $(g\circ f)_{*}=F_\gamma\circ (Id_X)_{*}$ for some $\gamma$. Notice that:

• $(g\circ f)_{*} = g_{*}\circ f_{*}$ (A proof can be found here)
• $F_\gamma\circ (Id_X)_{*} = F_\gamma$.

Moreover:

• $f_{*}: \pi_1(X,a)\to \pi_1(Y,f(a))$
• $g_{*}: \pi_1(Y,f(a))\to \pi_1(X,g(f(a)))$
• $F_\gamma: \pi_1(X,a)\to \pi_1(X,g(f(a)))$

It’s easier to look on a diagram of the situation:

That’s a bit better right? notice that I’ve added to $f_{*}$ a little $a$ to it, and a little $f(a)$ to $g_{*}$. Those are just symbols that will make things clearer.

Recall that $F_\gamma$ is an isomorphism thus it is onto. Therefore:

F_\gamma=g_{*}^{f(a)}\circ f_{*}^a\text{ is onto}\Rightarrow g_{*}^{f(a)}\text{ is onto}

Great, that’s enough for part 1.

#### Using Fact 2

I am going to extend the diagram using the second fact:

f\circ g \sim Id_Y

Therefore, there exists some $\delta$ such that:

f_*\circ g_*=(f\circ g)_*=F_\delta\circ Id_Y = F_\delta

However, we are going to use the same $g$ we used in the last part, i.e. $g_{*}^{f(a)}:\pi_1(Y,f(a))\to \pi_1(X,g(f(a)))$.

Therefore, $f_{*}$ now maps the group $\pi_1(Y,g(f(a)))$ to the group (get ready for the confusing base-point) $\pi_1(X,f(g(f(a))))$.

The diagram is now looks like:

Pause and ponder on the diagram for a few seconds. Explain yourself what is going on here, even out loud if necessary. Once you understand what’s going on here, you will realize that everything here makes perfect sense!

Ok, So the second part of the diagram yields:

f_*^{g(f(a))}\circ g_*^{f(a)}=F_\delta

Recall that $F_\delta$ is an isomorphism and in particular, is one-to-one. Therefore:

F_\delta=f_*^{g(f(a))}\circ g_*^{f(a)}\text{ is one-to-one }\Rightarrow g_*^{f(a)}\text{ is one-to-one}


#### Putting the parts together

From both parts we conclude that:

• $g_*^{f(a)}$ is onto.
• $g_*^{f(a)}$ is one-to-one.

In other words, $g_*^{f(a)}$ is an isomorphism! In addition, we also know that:

g_{*}^{f(a)}\circ f_{*}^{a}=F_\gamma

Since $g_*^{f(a)}$ has an inverse map (which is also an isomorphism) we get:

 f_{*}^{a}=( g_{*}^{f(a)})^{-1}\circ F_\gamma

Therefore, $f_{*}^a$ is a composition of isomorphisms, thus it is an isomorphism itself, and that completes the proof!

## Conclustions

Yes! we’ve finally found some real connection between groups and spaces that actually teaches us about the space itself!

You want to prove that two spaces are not homotopic (and in particular – homeomorphic)? then you can calculate the fundamental groups of them, check whether they are isomorphic or not – if not, then that’s it!

Ok, let’s try to find some concrete exapmles:

What is the fundemental group of a single point space $\{p\}$? Since there is only one point in the space, there is only one loop in it – the constant loop $K_p$. Therefore, it’s fundemental group is just the set $\{[K_p]\}$. i.e. it is the trivial group.

Now, recall that a Contractible space $X$ is a space that is homotopic to a single point space. Therefore, the fundamental group of every contractible space is trivial.

What contractible spaces have we met so far? $\mathbb{R}^n$ for every $n\in \mathbb{N}$. Therefore, thier fundamental group is trivial.

We’ve also seen that $S^{n-1}$ is a deformation retract of the space $\mathbb{R}^n-\{0\}$. Therefore, those spaces are homotopic equivalent to each other, thus, they have the same fundamental group!

But what is this group exactly? is trivial? isn’t it? we don’t know yet (we will though). We still don’t even know if those space are contractible or not? If we will manage to prove that their fundamental group is not trivial, then we automatically prove that there exist non-contractible spaces!

To end this post, I just want to tell you the plan for the future. My main goal now is to calculate the fundamental group of a circle $S^1$. We will have to work hard in order to do that, so prepare yourself!