# The Fundamental Group Of a Circle

After discussing about simply connected spaces, the next definition that is going to be useful for us is the one of covering spcae. Intuitivley speaking, just by the name we can kind of understand what a covering space is – it is a space that covers another space.

Though the intuition is simple, the definition is a little complicated. Let’s see what is it exaclty:

## Covering spaces & maps

Let $E,B$ be two topological spaces. A map $\rho:E\to B$ is called a covering map if:

1. $\rho$ is onto.
2. For every $x\in B$, there exists an open neighborhood $U\subseteq B$ such that $\rho^{-1}(U)$ is a disjoint union of open sets $\{V_\alpha\}_{\alpha\in J}$. Moreover, for every $\alpha\in J$, $\rho|_{V_\alpha}:V_{\alpha}\to U$ is a homeomorphism.

Perhaps a sketch will make things clearer:

In simpler words, we can find a neighborhood for every point in $U$, such that it’s pre-image is made up of many disjoint ‘copies’ of it.

Of course, we call $E$ a cover space that covers $B$.

Now the question is, how is this useful for us? what spaces are we going to use? short answer: we will cover the circle.

With what space are we going to cover it? with the real line$\mathbb{R}$.

It’s really easy to imagine such a covering. We can think of the real line as a long wire that we can wrap around the circle many times.

There is a really simple map that does exactly that:

\rho:\mathbb{R}\to S^1 \\ \rho(t)=e^{2\pi i t}

Recall that $e^{2\pi i n}= 1$ for every integer. And I’ll give you a little spoiler and note I am going to focus on:

\rho^{-1}(1)=\mathbb{Z}

I am not going to prove that $\rho$ is indeed a cover map, but that’s not hard at all, it can easilly be proven that any open neighborhood of a point on the circle will satisfy the second property…

Great, now that we understand what covering maps are, it’s time to prove the major theorem that will be exactly what we need!

## Lifting paths

Since covering spaces are… covering a space. It would be nice to have a way to ‘lift’ a path from the covered space into the covering space. As it turns out, it is indeed possible. And not only it’s possible, we can lift a path in a unique way (under a specific condition)!

Let’s present this proposition formally:

Let $\rho:E\to B$ be a cover map. Suppose that $\gamma:[0,1]\to B$ is a path in $B$ such that $\gamma(0)=b$. Then, for every $e\in E$ such that $\rho(e)=b$ there exists a unique path $\delta:[0,1]\to E$ such that:

• $\delta(0)=e$
• $\rho\circ\delta = \gamma$

Let’s look at a sketch of the statement to make thing a little clearer:

We are given a path $\gamma$, and the desired lift is $\delta$. What makes $\delta$ unique is the pick of the starting point – If we would have picked a point $e^\prime$, we would have get a different ‘lift’.

In order to prove this theorem I am going to use a lemma from topology, that I will discuss on in the future (when I’ll write about topology).

### Lebesgue’s number lemma

Recall that a Lebesgue number of an open cover of a metric space $X$ is a number $\delta>0$ such that every set with diamenter less than $\delta$ is contained in one of the elements of the cover.

Lebesgue’s number lemma tells us that if $X$ is a metric and compact space, then every open cover has a Lebesgue number.

How it that going to be useful? with what metric comact space are we dealing with?

The space is going to be the unit interval$I$.

Before I’ll start proving the theorem I just want to note one thing: I think that the proof is beautiful! Using lebesgue’s number lemma is a brilliant idea, that will make the proof almost trivial!

Ok, let’s start proving:

#### Proof of the lifting theorem

We’ll say that an open set $U\subseteq B$ is a “good set” if it satisfies the second property in the defintion of a cover map. That is, $\rho^{-1}(U)$ is a dijoint union of open sets ${V_\alpha}_{\alpha\in J}$ in $E$. Such that $\rho^{}|_{V_\alpha}:V\alpha\to U$ is an homeomorphism.

Now, since $\rho$ is a cover map, every point $x\in B$ has a neighborhood $U_x$ which is a “good set”.

Therefore, the collection $\{U_x\}_{x\in B}$ is an open cover of $B$. Therefore, the collection $\{\gamma^{-1}(U_x)\}_{x\in B}$ is an open cover of the unit interval $I$.

Now, $I$ is a metric compact space, thus the open cover $\{\gamma^{-1}(U_x)\}_{x\in B}$ has a lebesgue number $\delta >0$.

Now I am going to pick $n\in\mathbb{N}$ large enough such that $\frac{1}{n}<\delta$, and split the unit interval to$n$ parts:

I_1=[0,\frac{1}{n}],I_2=[\frac{1}{n},\frac{2}{n}],I_3=[\frac{2}{n},\frac{3}{n}],\dots,I_n=[\frac{n-1}{n},1]

Each one of those intervals has a diameter $\frac{1}{n}<\delta$. So for every interval $I_i$ there exists some $x_i\in B$ such that:

I_i\subseteq \gamma^{-1}(U_{x_i})

Great, we are now ready to start lifting! we are going to ‘lift’ the path part-by-part. One interval at a time. Since there is only a finite number of intervals $I_i$ this process is going to end.

##### Lifting the path

We begin with $I_1$. Note that $\gamma(0)=b$ and $I_1\sub \gamma^{-1}(U_{x_1})$ thus:

b\in \gamma(I_1)\sub U_{x_1}

Therefore:

e\in\rho^{-1}(b)\sub\rho^{-1}(U_{x_1})

However, $U_{x_1}$ is a “good set”, so it’s a union of disjoint open sets such that the restrection of $\rho$ to each one of them is a homeomorphism. Suppose that $V_1\sub\rho^{-1}(U_{x_1})$ is an open set such that $e\in V_1$. Since $\rho|_{V_1}$ is a homeomorphism – it has an inverse map, then we can define:

\delta_1:I_1\to V_1\sub E \\
\delta_1=(\rho|_{V_1})^{-1}\circ\gamma|_{I_1}

And $\delta_1$ is indeed a lift of $\gamma|_{I_1}$ where:

\delta_1(0)=(\rho|_{V_1})^{-1}\circ\gamma|_{I_1}(0)=(\rho|_{V_1})^{-1}(\gamma|_{I_1}(0))=(\rho|_{V_1})^{-1}(b)=e

As we wanted! However, we still need to lift the rest of the path, but that’s not a problem at all:

$I_2$ starts where $I_1$ ends, therefore, we know the new ‘initial point’. Formally, suppose that $\delta_1$ ends in $e_2$. Moreover, denote $b_2=\rho|_{V_1}(e_2)$. Note that $\gamma(\frac{2}{n})=b_2$.

Again, $I_2\sub \gamma^{-1}(U_{x_2})$, thus:

b_2\in \gamma(I_2)\sub U_{x_2}

$U_{x_2}$ is a “good set”, then,similar to before, there exists an open set $V_2\sub \rho^{-1}(U_{x_2})$ such that $e_2\in V_2$. We can now define:

\delta_2:I_2\to V_2\sub E \\
\delta_2=(\rho|_{V_2})^{-1}\circ\gamma|_{I_2}

which is a path that ‘continues’ $\delta_1$ and is a lift of $\gamma|_{I_2}$.

We can continue this process and create path $\delta_1,\delta_2,\dots,\delta_n$, define:

\delta:=\delta_1*\delta_2*\cdots*\delta_n

And indeed, $\delta$ is a lift of $\gamma$.

#### Uniqueness

Great, after proved existence of the ‘lift’, we now need to prove uniqueness. Suppose that $\delta^\prime$ is a differnet ‘lift’ of $\gamma$ such that $\delta\prime(0)=e$.

We can look at those paths at $I_1$. Both of them satisfy:

\rho|_{V_1}\circ\delta|_{I_1}=\gamma_1=\rho|_{V_1}\circ\delta^{\prime}|_{I_1}

Why? Recall that the set $V_1$ is only depends on $I_1$ and $\gamma$ and since $\delta^{\prime}(0)=e\in V_1$, and paths are conncected, then the whole image of $I_1$ under $\delta$ is in $V_1$.

However, $\rho|_{V_1}$ is a homeomorphism, then:

\rho|_{V_1}\circ\delta|_{I_1}=\rho|_{V_1}\circ\delta^{\prime}|_{I_1}\ \ /\circ(\rho|_{V_1})^{-1} \\\Downarrow \\ \delta|_{I_1}=\delta^\prime|_{I_1}

We can prove it for each $I_i$ separatly and conclude that $\delta=\delta^\prime$.

## Lifting homotopies

Notice that the exact same proof works when we are trying to ‘lift’ a map from the unit square. It is also a metric and compact space, thus it has a lebesgue number, and we can split the square into small squares, and define the ‘lift’ for each little square at a time:

Moreover, if we decide to lift a homotopy with respect to $\partial I$ – the lifted homotopy will induce a homotopy of two path with respect to $\partial I$! (convince yourself why it’s true!)

#### Now what?

So we now know how to lift homotopies and paths. Those facts allows me to define a function. If $\rho:E\to B$ is a cover map, $b\in B$ and $e\in\rho^{-1}(b)$, then:

F:\pi_1(B,b)\to\rho^{-1}(b) \\ F([\varphi])=\hat{\varphi}^{e}(1)

Where $\hat{\varphi}^{e}$ is $\varphi$ ‘lifted’ to $E$ and starts at $e$.

Note that $\hat{\varphi}^{e}(1)\in \rho^{-1}(b)$ since $\varphi$ is a loop! However, it may not be $e$– we’ll see it in a moment…

Since we can lift homotopies, this map is indeed well defined: If $[\varphi]=[\psi]$ we can lift the homotopy with respect to the endpoints between them, and conclude that their ‘lifted’ paths have the same endpoint – $\hat{\varphi}^{e}(1)$.

#### What if $E$$E$ is path connected?

Then for every $x\in\rho^{-1}(b)$, there is a path $\gamma$ from $e$ to $x$. Now define $\varphi=\rho\circ\gamma$. This is a path in $B$ that starts in $b$.

However, $\gamma$ is in fact a lift of $\varphi$ to $E$! Therefore:

F([\varphi])=\hat{\varphi}^e(1)=\gamma(1)=x

We just found out that for every $x\in\rho^{-1}(b)$, there exists a path $\varphi$ such that $F([\varphi])=x$. In other words, if $F$ is path-connected, then it is onto!

#### What if $E$$E$ is simply connected?

So $E$ is in particular path connected, thus it is onto. However, maybe it’s even more than that? Maybe it is also one-to-one? Let’s try to prove it:

Suppose that $F([\psi])=F([\gamma])$. That is

\hat{\psi}^e(1)=\hat{\varphi}^e(1)

Moreover, $\hat{\psi}^e(0)=\hat{\varphi}^e(0)=e$. Now, since $E$ is simply connected, we know that every two paths with the same endpoints are homotopic with respect to $\partial I$. We can now conclude:

\hat{\psi}^e\sim_{\partial I}\hat{\varphi}^e \\ \Downarrow \\ \psi=\rho\circ\hat{\psi}^e\sim_{\partial I}\rho\circ\hat{\varphi}^e =\varphi \\ \Downarrow \\
[\psi]=[\varphi]

As we wanted.

To summarize, we know that when $E$ is simply connected, then $F$ is a bijection (one-to-one & onto). Can you see why it’s useful?

## Finally – calculating the fundamental group of a circle

I mentioned in the beginning that we are going to ‘cover’ the circle with the real line – $\mathbb{R}$.

However, $\mathbb{R}$ is contractible, thus it has a trivial fundamental group, which implies that he’s simply connected!

Moreover, we’ve seen that the covering map was:

\rho:\mathbb{R}\to S^1 \\ \rho(t)=e^{2\pi i t}

And:

\rho^{-1}(1)=\mathbb{Z}

So we can conclude that:

F:\pi_1(S^1,1)\to\rho^{-1}(1)=\mathbb{Z}
\\
F([\varphi])=\hat{\varphi}^0(1)

Is a bijection!

The only thing left to prove now is that $F$ is not only a bijection, but it is also a group homomorphism. That is:

F([\varphi][\psi])=F([\varphi])\overbrace{+}^{\text{The action in }\mathbb{Z}}F([\psi])

Every loop that starts at $1$ in $S^1$ has to be (up to homotopy) of the form:

\gamma_n(t)=e^{2\pi in\cdot t}

Where $n\in\mathbb{Z}$. it follows from the fact that:

\gamma_n(1)=e^{2\pi in}=1\iff n\in\mathbb{Z}

It’s not hard to verify from the definition that:

\gamma_n*\gamma_m(t)=\begin{cases}
e^{2\pi int} & 0\leq t\leq0.5\\
e^{2\pi imt} & 0.5\leq t\leq1
\end{cases}\sim_{\partial I} e^{2\pi i (n+m)t}=\gamma_{n+m}(t)

Thus:

[\gamma_n][\gamma_m]=[\gamma_{n+m}]

The only thing left to mention is that the lift of $\gamma_n$ that starts in 0 is:

\hat{\gamma}_n^0(t)=nt

Since:

\rho\circ\hat{\gamma}_n^0(t)=\rho(\hat{\gamma}_n^0(t))=\rho(nt)=e^{2\pi i n t}=\gamma_n(t)

Therefore:

F([\gamma_n])=\hat{\gamma}_n^0(1)=n

Finally:

F([\gamma_n][\gamma_m])=F([\gamma_{n+m}])=\overbrace{n}^{F([\gamma_n])}+\overbrace{m}^{F([\gamma_m])}={F([\gamma_n])}+{F([\gamma_m])}

So it’s indeed a bijection and homomorphism. Thus, it is an isomorpihsm.

## The result we’ve been waiting for

\pi_1(S^1,a)=\mathbb{Z}

Yes! Finally not a trivial fundamental group! From this fact we can conclude a lot!

First, we now know that the circle is not contractible! Moreover it is not simply connected! And there are a lot of stuff that we can learn from this result.

In fact, there are so many, that they desereve a separate post – and even more than that. One result is so great that it deserves it’s own separate post! This result is the fundametal theorem of algebra.

This theorem states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

At first sight, it seems like this theorem is not even related to what we’ve done so far! We were dealing with path, and not with polynomials! That’s what makes the proof so unique and unexpected!

Ok, so that’s going to be it for now… So prepare yourself for some unexpected results!