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 be two topological spaces. A map is called a **covering map** if:

- is
**onto**. - For every , there exists an open neighborhood such that is a
**disjoint**union of open sets . Moreover, for every , is a**homeomorphism**.

Perhaps a sketch will make things clearer:

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

Of course, we call a **cover space** that covers .

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** – .

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 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 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 be a cover map. Suppose that is a path in such that . Then, for every such that there **exists** a **unique** path such that:

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

We are given a path , and the desired lift is . What makes unique is the pick of the starting point – If we would have picked a point , 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** is a number such that every set with diamenter less than is contained in one of the elements of the cover.

**Lebesgue’s number lemma** tells us that if 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** – .

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 is a *“good set”* if it satisfies the second property in the defintion of a cover map. That is, is a dijoint union of open sets in . Such that is an homeomorphism.

Now, since is a cover map, every point has a neighborhood which is a *“good set”*.

Therefore, the collection is an open cover of . Therefore, the collection is an open cover of the unit interval .

Now, is a metric compact space, thus the open cover has a lebesgue number .

Now I am going to pick large enough such that , and split the unit interval to 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 . So for every interval there exists some 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 this process is going to **end**.

##### Lifting the path

We begin with . Note that and thus:

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

Therefore:

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

However, is a *“good set”*, so it’s a union of disjoint open sets such that the restrection of to each one of them is a homeomorphism. Suppose that is an open set such that . Since 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 is indeed a lift of 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:

starts where ends, therefore, we know the new ‘initial point’. Formally, suppose that ends in . Moreover, denote . Note that .

Again, , thus:

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

is a *“good set”*, then,similar to before, there exists an open set such that . 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’ and is a lift of .

We can continue this process and create path , define:

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

And indeed, is a lift of .

#### Uniqueness

Great, after proved existence of the ‘lift’, we now need to prove uniqueness. Suppose that is a differnet ‘lift’ of such that .

We can look at those paths at . 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 is only depends on and and since , and paths are **conncected**, then the whole image of under is in .

However, 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 separatly and conclude that .

## 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 – the lifted homotopy will induce a homotopy of two path with respect to ! (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 is a cover map, and , then:

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

Where is ‘lifted’ to and starts at .

Note that since is a **loop**! However, it may not be – we’ll see it in a moment…

Since we can lift homotopies, this map is indeed well defined: If we can lift the homotopy with respect to the endpoints between them, and conclude that their ‘lifted’ paths have the same endpoint – .

Let’s ask some questions:

#### What if is **path connected**?

Then for every , there is a path from to . Now define . This is a path in that starts in .

However, is in fact a lift of to ! Therefore:

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

We just found out that for every , there exists a path such that . In other words, if is **path-connected**, then it is **onto**!

#### What if is **simply connected**?

So 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 . That is

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

Moreover, . Now, since is simply connected, we know that every two paths with the same endpoints are homotopic with respect to . 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 is simply connected, then 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 – .

However, 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 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 in has to be (up to homotopy) of the form:

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

Where . 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 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!