# The Fundamental Group

In the last post we’ve finally defined the fundamental group of a topological space $X$ at a point $a\in X$. It’s elements are equivalence classes of loops (that starts and ends at $a$). The equivalence realtion is homotopy with respect to $\partial I$ (which is homotopy with respect to the only endpoint).

The action in this group was:

[\gamma][\delta]=[\gamma*\delta]

Where:

\gamma*\delta=\begin{cases}
\gamma(t) & 0\leq t\leq\frac{1}{2}\\
\delta(t) & \frac{1}{2}\leq t\leq1
\end{cases}

We’ve seen that this action is well defined (and not only in a specific group – it is always well defined when $\gamma$ ends at the same point that $\delta$ starts in).

The inverse element of $[\gamma]$ was $[\overline{\gamma}]$ – Where $\overline{\gamma}=\gamma(1-t)$ (you can think on $\overline{\gamma}$ as ‘going backwards’ on $\gamma$).

Lastly, the identity element was $[K_a]$ where $K_a$ is the constant loop on $a$.

We’ve denote the fundamental group as $\pi_1(X,a)$.

In this post, we will start seeing some properties of the group, and find connections between group. First things first though, I still need to define formally how to ‘extract’ the group from a topological space with a point.

## Back to categories

Since we are interested not only in topological spaces – but also a point in them, I am going to focus on the category of pairs $(X,a)$ where $X$ is a topological space and $a\in X$. The morphisms in this category are going to be continuous maps $f:(X,a)\to (Y,b)$ where $f(a)=b$.

The requirement $f(a)=b$ actually makes sense – the goal of the morphisms in general is to find connections between two objects in a category. Therefore, this requierment helps us to find some connection between those two pairs.

On the other hand, the second category we are dealing with is the category of groups, and obviously, the morphisms in this category are going to be group homomorphisms.

Our goal is to map pairs into groups, and continuous maps to group homomorphisms. For that, we need a functor. So let’s define now.

Let’s denote this functor by $F$. The most natural thing to do is to map a pair $(X,a)$ to it’s fundamental group $\pi_1(X,a)$:

F((X,a))=\pi_1(X,a)

Great, what about morphisms now? We need to map a continuous function:

f:(X,a)\to(Y,b),\ \ f(a)=b

Into a group homomrphism:

f^\prime:\pi_1(X,a)\to \pi_2(X,b)

What I am going to define $F(f)=f_{*}$ where $f_{*}([\gamma])=[f\circ \gamma]$. It seems like the most natural way to define $f_{*}$, and this is the first thing that came to my head when I asked to think of a way to define $f_{*}$. However, there is a risk that this function won’t even be well defined! if $[\gamma]=[\delta]$ why would $[f\circ \gamma]$ wil be the same as $[\delta\circ \gamma]$. This question is equivalent to the statement: If $\gamma \sim_{\partial I} \delta$ and $f:(X,a)\to (Y,b)$ with $f(a)=b$ then $f\circ\gamma \sim_{\partial I}f\circ\delta$

##### A stronger statemenet

I shall prove now a stronger statement:

Suppose that $A\subseteq X, B\subseteq Y$, $h,h^\prime: X\to Y$ satisfy $h\sim_A h^\prime$. And, $k,k^\prime:Y\to Z$ satisfy $k\sim_B k^\prime$.

X\overset{h,h^\prime}{\longrightarrow} Y\overset{k,k^\prime}{\longrightarrow} Z

Moreover, $h(A)\subseteq B$. Then:

k\circ h\sim_Ak^\prime\circ h^\prime

In our case, $k=k^\prime= f$, $h=\gamma,h^\prime = \delta$ and $A=\{0,1\}, B=\{a\}$. We know that $f$ is homotopic to itself with respect to any set. In addition, $\gamma({\partial I})=\{a\}$, Then we can conclude what we wanted.

##### The proof

Let’s see what we’ve got.

• An homomtopy $H:X\times I\to Y$ from $h$ to $h^\prime$ with respect to $A$.
• An homomtopy $K:Y\times I\to Z$ from $k$ to $k^\prime$ with respect to $B$.

We are looking for a homotopy $J:X\times I\to Z$.

I am just going to define $J(x,t)=K(H(x,t),t))$.

J:X\times I\overset{(H,\rho_2)}{\longrightarrow} Y\times I\overset{K}{\longrightarrow} Z

It is indeed continuous. Let’s verify that it is indeed a homotpy with respect to $A$. For every $a\in A$:

J(a,t)=K(H(a,t),t)=K(h(a),t)

Since $h(a)\in B$ by our assumption, and $K$ is an homotpy with respect to $B$, we get:

J(a,t)=K(h(a),t)=h(a)

Indeed, this map is constant on $A$. Moreover:

J(x,0)=K(H(x,0),0)=K(h(x),0)=k(h(x))=k\circ h(x)
J(x,1)=K(H(x,1),1)=K(h^\prime(x),1)=k^\prime(h^\prime(x))=k^\prime\circ h^\prime(x)

To summarize, $J$ is indeed the homotopy we watned, and the proof is done.

#### Back to $f_{*}$$f_{*}$

Great, now we know that $f_{*}$ is well defined. We still need to show that $f_{*}$ is a group homorphism, i.e:

f_{*}([\gamma][\delta])=f_{*}([\gamma])f_{*}([\delta])

Which is equivalent to show:

[f\circ(\gamma*\delta)]=[(f\circ\gamma)*(f\circ\delta)]

However, notice that:

f\circ(\gamma*\delta)(t)=\begin{cases}
f\circ\gamma(t) & 0\leq t\leq\frac{1}{2}\\
f\circ\delta(t) & \frac{1}{2}\leq t\leq1
\end{cases}=(f\circ\gamma)*(f\circ\delta)(t)

Not only those function are homotopic, they are actually the same.

The only thing left to check is that the map $F(f)=f_{*}$ is indeed a functor.

##### Proving it is a functor

To prove that a map is a functor between categories $\mathcal{C},\mathcal{D}$, we need to show two things:

• $F(Id_A)=Id_{F(A)}$ where: $A\in \mathcal{C}$
• $F(\gamma\circ \psi)=F(\gamma)\circ F(\psi)$ where: $\psi\in\text{Mor}(A,B),\gamma\in\text{Mor}(B,C)$

Ok, In our case we have:

F(Id_{(X,a)})={Id_{(X,a)}}_*

Notice that:

{Id_{(X,a)}}_*([\gamma])=[Id_X\circ\gamma]=[\gamma]=Id_{\pi_1(X,a)}

Therefore, $F(Id_{(X,a)}=Id_{F(X,a)}$ as we wanted. In addition:

F(f\circ g)={(f\circ g)}_*\ \ , \ F(f)\circ F(g)=f_*\circ g_*

However:

(f\circ g)_*([\gamma])=[(f\circ g)\circ \gamma]=[f\circ(g\circ \gamma)]=f_*([g\circ\gamma])=f_*(g_*([\gamma]))=f_*\circ g_*([\gamma])

Thus, $F(f\circ g)=F(f)\circ F(g)$ and we’ve completed the proof!

## Some connections between groups

After we’ve constructed a functor between the category of topological spaces (with a point) to the category of groups, we have a real connection between the categrories. We can translate topological problems into group problems. However, there is still something that might be bothering you – we are limited to a specific point!

To overcome this problem we need to understand the fundamental groups of a given space a little better.

#### Path-connected spaces

Suppose that $X$ is not a path-connected space and $a\in X$ is some point in the space.

The elements of the group $\pi_1(X,a)$ are equivalence classes of loops that starts and ends at $a$.

Now, if we denote by $A$ the path-connected component of $a$ and focus only at it – the fundamental group is still the same, it is ‘not aware’ of the other components. We also have no way of finding a connection between the fundamental group at $a$ to another fundamental group at a point whose not in the same component of $a$.

Therefore, at least for now, we think of our topological spaces as path connected. By that, we can treat a non-path-connected space as many path-connected spaces and apply our results on path-connected spaces there.

#### Two homotopic maps

we’ve seen a way to translate a contiuous map into a group homomorphisms using our functor $F(f)=f_{*}$. Suppose that there are two maps $f,g:(X,a)\to(Y,b)$ which are homotopic with respect to $\{a\}$.

f\sim_{\{a\}}g

This means we can continuously change $f$ to $g$ without changing the output of $a$ (In that case, the output of $a$ is $b$).

What can we say about $f_{*},g_{*}$?

Let’s pick some class $[\gamma]\in \Pi_1(X,a)$. Of course, $\gamma\sim_{\partial I}\gamma$. Moreover, by out assumption, $f\sim_{\{a\}}g$. We can now applay the statement from before to conclude that:

f\circ\gamma\sim_{\partial I}g\circ\gamma

Therefore, $[f\circ\gamma] = [g\circ\gamma]$. However, $[f\circ\gamma]$ is exactly $f_{*}([\gamma])$ and $[g\circ\gamma]$ is exactly $g_{*}([\gamma])$. Thus, we get that $f_{*}=g_{*}$! That’s great – if we have two homotopic maps, they induce the same homomorphism between $\pi_1(X,a)$ and $\pi_1(Y,a)$.

What about $f,g$ which are just homtopic? Well… then we know that $f(a)$ may not be the same as $f(b)$. Thus, even though they both induce a homorphism from the group $\pi_1(X,a)$, the homomorphisms may not even be into the same group!

So that’s kind of a problem… We need to find a way to connect these groups somehow.

#### Isomorphisms between groups

Even though the maps send the group $\pi_1(X,a)$ to differenet group, yet, we can find some hope! If we will be able to find a connection between two fundamental groups $\pi_1(X,a),\pi_1(X,b)$, then, maybe the fact that $f,g$ range’s from before are different groups is not so bad after all…

Recall that we are dealing with path-connected spaces! So let $X$ be some path-connected space and $a,b\in X$. Then there is a path $\gamma:I\to X$ where $\gamma(0)=a,\gamma(1)=b$. Morover, pick some arbitrary loop $\delta$ with $a$ as it’s base-point.

From those path, we can create a loop with $b$ as it’s endpoint:

We begin by ‘walking backwards’ over $\gamma$, then ‘walking’ over $\delta$, and finally, ‘walk’ on $\gamma$ to return to $b$.

Formally, this is the path:

\overline{\gamma}*\delta*\gamma

This is acutally useful – with the help of $\gamma$, we can create a loop with base-point $b$ from a loop $\delta$ with base-point $a$.

Therefore, we can define a map:

F_\gamma:\pi_1(X,a)\to \pi_1(X,b)

Such that:

F_\gamma([\delta])=[\overline{\gamma}][\delta][\gamma]\ \ (=[\overline{\gamma}*\delta*\gamma])

This is a map between groups – we would be happy if this map was a homomorphism, let’s check if it is: To do so, we need to show that:

F_\gamma([\varphi])F_\gamma([\psi])=F_\gamma([\varphi][\psi])

Well:

F_\gamma([\varphi])F_\gamma([\psi])=([\overline{\gamma}][\varphi][{\gamma}])([\overline{\gamma}][\psi][{\gamma}])

Recall that the action between equivalence classes is associative, therefore:

([\overline{\gamma}][\varphi][{\gamma}])([\overline{\gamma}][\psi][\gamma])=[\overline\gamma][\varphi]([\gamma][\overline{\gamma}])[\psi][\gamma]=[\overline{\gamma}][\varphi][K_a][\psi][\gamma]=[\overline{\gamma}][\varphi][\psi][{\gamma}]
=[\overline{\gamma}]([\varphi][\psi])[{\gamma}]=F_\gamma([\varphi][\psi])

Great, now we know that $F_\gamma$ is a group homomorphism. However, maybe it is even more than that?

With similar process, we can also define the map:

F_{\overline{\gamma}}:\pi_1(X,b)\to\pi_1(X,a)

Right? $\overline{\gamma}$ is indeed a path that starts at $b$ and ends at $a$, then we can repeat (almost) the exact same process as before to get the map $F_{\overline{\gamma}}$.

This map is actually looks like a candidate of an inverse function to $F_\gamma$. Let’s verify it:

F_{\overline{\gamma}}\circ F_\gamma([\varphi])=F_{\overline{\gamma}}(F_\gamma([\varphi]))=F_{\overline{\gamma}}([\overline{\gamma}][\varphi][\gamma])=[\gamma]([\overline\gamma][\varphi][\gamma])[\overline\gamma]=([\gamma][\overline\gamma])[\varphi]([\gamma][\overline\gamma])
=[K_a][\varphi][K_a]=[\varphi]=Id_{\pi_1(X,a)}([\varphi])

With similar process, we get that $F_{{\gamma}}\circ F_{\overline\gamma}([\varphi])=Id_{\pi_1(X,b)}$.

Wow! So not only $F_\gamma$ is a group homomorphisms, it is actually a group isomorphism!

#### A fabulous corollary

This map proves us that in a path-connected space $X$ all the fundamental groups $\pi_1(X,a)$ are isomorphic to each other. This fact allows us to keep studying the space in a single point without worrying about the other point – since computing the fundamental group of the other point will yield an isomorphic group to the one we are dealing with.

We’ve also seen an explicit isomorphism between the group, however – it is definitely not unique! every path $\gamma$ from $a$ to $b$ yields a different isomorphism between the groups.

## Summary

In this post, we’ve met a useful functor that allows us to translate pairs of a topoligcal space and a point in it to a group!

Moreover, we have fonud out that the fundamental groups off all the points in a path-connected components are isomorphic to each other!

In the next post, I am going to keep exploring the fundemental group and it’s properties to find even stronger things that we can learn from the group.