# What can we learn from the fundamental group of the circle?

In the last post we finally revealed the true identity of the fundamental group! It is $\mathbb{Z}$ – aka the group of integers / the infinite cyclic group.

This group is first non-trivial group that we’ve met so far. While The circle, is first non-contractible space we’ve met.

In this post we are going to see how many things we can learn just from the circle and it’s group, and trust me, we can learn a lot.

## The circle as a deformation retract

Recall that a deformation retract of a space $X$ is a subspace $A\subseteq X$ such that there exists a map $r:X\to A$ where:

• $r\circ i=Id_A$
• $i\circ r \sim Id_X$

($i:A\to X$ is the inclustion map, $i(a)=a$). If $i\circ r$ is not homotopic to $Id_X$, then the space is called a retract.

In particular a deformation retract is a subspace which is homotopic equivalent to the space. Since we know that homotopic equivalent spaces have the same fundamental group (I’ve proved it here), so it’s enough to know the fundamental group of the deformation retract in order to know the fundamental group of the space.

Recall that our example for a deformation retract (right here) was a circle, and the original space was $\mathbb{R}^n-\{0\}$.

Thus, we can conclude that:

\pi_1(\mathbb{R}^2-\{0\})=\pi_1(S^1)=\mathbb{Z}

(Sometimes I won’t mention the base point, it doesn’t matter what it is when the space is path-connected)

Great, but know even more than that – we know the elemnts of the group. They are (equivalence classes) loops around the circle. If we think about the loops as in $\mathbb{R}^2-\{0\}$, then the elements are loops that surround the point $0$. Those who aren’t will be homotopic to the constant loop – which represent the identity in the group.

This is actually pretty intuitive – Think about that if a path surround the origin, then we can’t “shrink” it to a point!

Are there other spaces that $S^1$ is a deformation retract of them? Well, of course, I’ll present the 2 famous examples:

### Cylinder

Cylinder is the space $S^1\times I$. I think it’s pretty clear why $S^1$ is a deformation retract of it (you can think of the cylinder as a streched circle). I won’t even bother defining a homotopy equivalence $r:S^1\to S^1\times I$. It’s a really simple exercise – We can construct a map to $S^1\times \{\frac{1}{2}\}$ for example, which is obviously (homeomoprihc to) a circle.

Nice, we now know that:

\pi_1(S^1\times I)=\mathbb{Z}

What else?

## Möbius strip

What is a Möbius strip? Formally it’s the quotient space of $I\times I$ obtained by the “gluing”

\forall t\in I:(0,t)\sim (1,1-t)

After folding the square and streching it a bit we get this strip:

It’s kind of a famous shape and it has some really special properties (such as: it only has one ‘side’). However, if you notice, in the middle of the Möbius strip, there is a circle hiding there!

Note that $I\times\{\frac{1}{2}\}$ is being mapped to itself and the endpoint are glued together since:

(0,\frac{1}{2})\sim(1,1-\frac{1}{2})=(1,\frac{1}{2})

And we know that an interval where it’s endpoints are glued together is a circle.

Let’s try to prove that this circle is a deformation retract.

First, let’s denote by $M$ the Möbius strip. Moreover, suppose that:

\rho:I\times I\to M

Is the quotient map from $I\times I$ to the Möbius strip. We’ll also denote:

S=\rho(I\times\{\frac{1}{2}\})\sube M

And we’ve already found out that $S\cong S^1$.

Notice that $I\times\{\frac{1}{2}\}\sub I\times I$ is a deformation retract. Indeed, we can define:

r:I\times I\to I\times \{\frac{1}{2}\} \\ r(s,t)=r(s,\frac{1}{2})

It is indeed continuous and $r\circ i = Id_{I\times\{\frac{1}{2}\}}$. Moreover, we can define an explicit homotopy:

H:(I\times I)\times I\to I\times I \\
H((a,b),t)=(a,(1-t)\cdot b+\frac{1}{2}\cdot t)

Note that:

H((a,b),0)=(a, b)=Id_{I\times I} \\
H((a,b),1)=(a,\frac{1}{2})=i\circ r

Thus, $i\circ r \sim Id_{I\times I}$, and $I\times\{\frac{1}{2}\}$ is a deformantion retract.

Moreover, for every $t\in I$:

H((0,b),t)=(0,(1-t)\cdot b+\frac{1}{2}\cdot t)\sim (1,1-[(1-t)\cdot b+\frac{1}{2}\cdot t])
=(1,1-(1-t)\cdot b-\frac{1}{2}\cdot t)=(1,(1-t)-(1-t)\cdot b+\frac{1}{2}\cdot t)=(1,(1-t)(1- b)+\frac{1}{2}\cdot t)
=H((1,1-b),t)

This fact shows us that $H$ “respects” the equivalence realtion. In other words, we can define a well-defined map:

\overline{H}:M\times I\to M \\
\overline{H}(\rho(a,b),t)=\rho(H((a,b),t))

It is well defined, since $\rho$ is onto thus every $m\in M$ has a pre-image. If $\rho(a_1,b_1)=\rho(a_2,b_2)$, since $H$ “respects” the equivalence realtion we get:

\overline{H}(\rho(a_1,b_1),t)=\rho(H((a_1,b_1),t))=\rho(H((a_2,b_2),t))=\overline{H}(\rho(a_2,b_2),t)

So it’s indeed well defined. In addition:

\overline{H}(\rho(a,b),0)=\rho(H((a,b),0))=\rho(a,b)\\ \Downarrow\\
\overline{H}(\rho(a,b),0)=Id_{M}

And:

\overline{H}(\rho(a,b),1)=\rho(H((a,b),1))=\rho(a,\frac{1}{2})\in S

So the map $\overline{r}: M\to S$ defined as:

\overline{r}(\rho(a,b))=\rho(a,\frac{1}{2})

Is an homotopic equivalence.

Great, so now we know that the Möbius strip is homotopic equivalent to a circle, therefore:

\pi_1(M)=\mathbb{Z}

## The circle as a retract

After all the ‘good news’ it’s time for a ‘bad’ one:

The boundary $\partial D^2$ is not a retract of $D^2$.

This means that we can’t continuously map the disc into it’s boundary in a continuous map where the boundary is fixed under the map. Let’s prove it:

Aiming for contradiction, suppose that there is a map:

r:D^2\to\partial D^2

Such that $r\circ i = Id_{\partial D^2}$. Let’s pick some arbitrary $a\in\partial D^2$. Recall that:

i_* :\pi_1(\partial D^2,a)\to \pi_1(D^2,a) \\
r_* :\pi_1(D^2,a)\to \pi_1(\partial D^2,a) 

(Iv’e defined those maps here, basically: $f_*([\gamma])=[f\circ\gamma]$).

Moreover, $r\circ i = Id_{\partial D^2}$. Thus:

(r\circ i)_* = {Id_{\partial D^2}}_* \\ \Downarrow \\
r_*\circ i_*=Id_{\pi_1(\partial D^2,a)}

Now:

Id_{\pi_1(\partial D^2,a)}:\pi_1(\partial D^2,a)\overset{i_*}{\to} \pi_1(D^2,a) \overset{r_*}{\to}\pi_1(\partial D^2,a)

However, $\pi_1(D^2,a)=\{1\}$ (it’s contractible), and $\pi_1(\partial D^2,a)=\mathbb{Z}$ thus:

Id_{\pi_1(\partial D^2,a)}:\mathbb{Z}\to\{1\}\to\mathbb{Z}

And that’s a contradicition! There is no way that $Id_{\pi_1(\partial D^2,a)}$ can be the identity, it has to be the trivial map!

## Brouwer fixed-point theorem

Suppose that $f:D^2\to D^2$, then there is some point $x\in D^2$ such that $f(x)=x$ ($x$ is called a fixed-point)

This is actually a pretty famous and important theorem. It can be proved for $f:D^n\toD^n$, however, not with our current tools…

My plan for this proof is to assume that there isn’t a fixed point, and by I’ll show that the boundary is a retract of the disk, which is a contradiction to the previous theorem.

Ok, let’s do it: Aiming for contradiction, suppose that for every $x\in D^2$:

f(x)\neq x

I am now going to define a function $r:D^2\to\partial D^2$ as follows:

• Since $f(x)\neq x$, those point form a line. Thus, we can define a ray that starts at $f(x)$ and goes through $x$.
• This ray intersects the disk’s boundary.
• We define $r(x)$ to be the intersection point (But not $f(x)$ if it turns out to be an intersection point as well).

It is a continuous function, you’re gonna have to trust me on that one since $r(x)$ is an intersection of a line and circle, so the explicit formula will be made of elementary functions (you can calculate it yourself, but trust, it’s not so pleasant…)

Moreover, points on the boundary are mapped to themselves:

Suppose that $x\in \partial D^2$, so the ray coming out of $f(x)$ towards $x$ intersects the boundary exactly at $x$.

Well, $r$ satisfies the properties that allow us to conclude that $\partial D^2$ is a retract of $D^2$ and that’s a contradiction to the previous theorem!

## Summary

So we’ve seen some pretty great results here, however, There is still one more theorem that I would like to prove – The fundamental theorem of algebra. But it deserves a separate post.