In the last post we’ve proved a great theorem: If two spaces are homotopy equivalent, then they have the same fundamental group (up to isomorphism). So far, all the fundamental groups we were able to calculate turned out to be trivial.

I’ve also said that my goal now is to calculate the fundamental group of the **circle** – . To do so, I want to present an important term that will help us calculate the group. However, this is a standalone term that has lots of other applications, so it’s something worth a standalone post.

Before I’ll present the defintion, I want to ponder about loops for a moment:

## The domain of loops

As you know, a loop is just a special kind of a path – it is a path that starts and ends at the same point. Formally, a loop is a (continuous) function:

\gamma:[0,1]\to X

( is a topological space) such that .

Intuitively, since 0 and 1 are being mapped to the same point – it feels like we can think of them as the **same** point, like we can actually bring them together:

This fact makes us wonder: “Can we describe loop as a function from the **circle** instead of the unit interval?”

The answer is **yes** and I’ll prove it, but, why do we even need it?

Recall that in this post, we’ve proved a really strong theorem. We’ve proved that 3 statement are equivalent. Two of them were:

- is null-homotopic
- is null-homotopic with respect to
**any**point on .

This theorem motivates us to prefer functions with the circle as their domain – The second statement gives us so much freedom! As we will see, I am going to use this fact a lot.

#### Describing loops as maps from the circle

A quick note first, when I’ll refrer the circle, I mean the circle on the **complex plane**:

S^1=\{z\in\mathbb{C}:|z|=1\}

If you want, you can think of the circle in . I prefer to think about the complex circle since There is a really nice way to represent it’s element. All of them are of the form:

e^{2\pi it}

Where . This way allows me to easily define a map:

\gamma^\prime:S^1\to X \\ \gamma^\prime(e^{2\pi it})=\gamma(t)

However, that’s not so simple – notice that . Therefore, if then is not even **well-defined**!

But we are dealing with **loops**! Therefore is indeed the same as , so is well-defined and we’ve found a way to describe a loop as a map from the circle.

The other direction is pretty similar. If I want to present a map from the circle as a map from the unit interval, I can easily define:

\gamma^\prime:I\to X \\\gamma^\prime(t)=\gamma(e^{2\pi it})

And this map has to be a loop since:

\gamma^\prime(0)=\gamma(e^{2\pi i\cdot0})=\gamma(e^{2\pi i\cdot1})=\gamma^\prime(1)

Great, now since that’s clear, let’s take it one step forward:

#### What about homotopy of loops?

Now that we have a way to represent loops. I want to apply this knowledge to homotopies. Suppose that:

\gamma_1,\gamma_2:[0,1]\to X

Are two loops that are homotopic with respect to . Suppose that is a homotopy with respect to from to .

Now, consider the representations of as maps from the circle (I’ll denote them by ). We can now define a new homotopy:

K:S^1\times I\to X \\ K(e^{2\pi is},t)=H(s,t)

It not hard to see that it a homotopy from to .

But there is one problem though – same as before, since we need to verify that:

K(e^{2\pi i\cdot 0},t)=K(e^{2\pi i\cdot 1},t)

Which is equivalent to check if:

H(0,t)=H(1,t)

However, we are dealing with **loops** that are homotopic with respect to . Therefore, this map is indeed well defined. In addition, this is an homotopy with respect to the point (notice how this fact fits prefectly to the theorem I mentioned earlier)!

Great, so from now on, I am going to think of loops as maps from or from . Moreover, I can think of two homotopic paths with respect to to be equivalent to two homotopic paths (with as their domain) with respect to 1.

You have no idea how useful those facts are going to be soon!

## What is a simply connected space?

After that long intro, it’s now finally time to define a really useful term:

A topological space is said to be **simply connected** if it is path-connected and every map:

f:S^1\to X

Is **null-homotopic**.

I think that you can already understand from the definition how the intro is going to be handy for us. However, the definition doesn’t reveal what simply connected spaces are.

Therefore, I am now going to present **3** more equivalent definitions.

Suppose that is path-connected, then the following are equivalent:

- is simply connected.
- For every , .
- For every two points – every two paths from to are homotopic with respect to .
- There are two points such that every two paths from to are homotopic with respect to .

Before I am going to prove why all of these are equivalent, let’s discuss about them a little bit.

The second satement gives us a full description of a simply connected space. It’s fundamental group is **trivial**. Therefore, all the spaces that we’ve calculated their fundamental groups are simply connected.

From the third statement we can conclude something pretty cool – If we pick , then the paths between them are **loops**. Moreover, the statement tells us that all of the loops are homotopic, therefore, any loop is homotpic to the **constant loop**, which is just the base-point!

This fact allows us to intuitivley describe simply connected spaces: Those are the spaces where we can “shrink” a loop to a single point.

Finally, the fourth statement shows us that it is enough to find only **2** points with this property in order to prove that all pairs of points fulfill it!

#### Proving the equivalence

So I am going to prove this theorem In the following order:

- 1 implies 2
- 2 implies 3
- 3 implies 4
- 4 implies 1

And that’s gonna be enough!

Suppose that is simply connected, thus, every map:

f:S^1\to X

Is null-homotopic.

Let be an arbitrary point and a loop with as it’s base point.

So by the assumption, we know that is null homotopic. Moreover, the theorem from the tells us that is also null-homotopic with respect to whatever point we want on the circle. I am going to pick the point **1**.

Therefore, is homotopic with respect to 1 to the constant function . Thus:

[\gamma]=[K_a]

So the only equivalence class in is the trivial one. In other words, as we wanted.

Suppose that , are paths from to .

Thus, the path is a **loop** with base point . Since we are assuming that we can conclude that:

[\gamma_1][\overline{\gamma_2}]=[\gamma_1*\overline{\gamma_2}]=[K_a]\ \ /\circ[\gamma_2] \\ \Downarrow \\ [\gamma_1]=[\gamma_1][\overline{\gamma_2}][\gamma_2]=[K_a][\gamma_2]=[\gamma_2]

And by definition, we know that the equivalence classes are with respect to (though we worked really hard to prove properties of the action between classes). Therefore:

\gamma_1\sim_{\partial I}\gamma_2

As required.

This one is trivial…

Suppose that is some map. We need to show it is null homotopic.

I am going to think of as a loop from the unit interval with a base point . This is going to make my life a lot easier…

By the assumption, there are two point such that every two paths between them are homotopic with respect to .

Since is path-connected, there exists a path from to . Similarly, there exists a path from to . And finally, there exists a path from to .

Note that Is a path from to be . Thus, it is homotopic to with respect to . That is:

[\gamma_1*f*\gamma_2]=[\gamma_3] \\ \Downarrow \\ [\gamma_1][f][\gamma_2]=[\gamma_3]\ \ /\circ[\overline{\gamma_3}] \\ \Downarrow \\ [\gamma_1][f][\gamma_2][\overline{\gamma_3}]=[K_a]

So we know that is homotopic to the constant loop . Let be the homotopy between them. We can now define a new homotopy:

K:I\times I\to X \\ K(s,t)=H(0.25+0.25\cdot s,t)

Notice that by definition:

\gamma_{1}*f*\gamma_{2}*\overline{\gamma_{3}}(s)=\begin{cases} \gamma_{1}(4s) & 0\leq s\leq0.25\\ f(4s-1) & 0.25\leq s\leq0.5\\ \gamma_{2}(4s-2) & 0.5\leq s\leq0.75\\ \overline{\gamma_{3}}(4s-3) & 0.75\leq s\leq1 \end{cases}

Hence:

K(s,0)=H(0.25+0.25s,0)=\gamma_{1}*f*\gamma_{2}*\overline{\gamma_{3}}(0.25+0.25s)

=f(4(0.25+0.25s)-1)=f(s)

And:

K(s,1)=H(0.25+0.25s,1)=K_a

And we proved that is null homotopic!

## Summary

So we’ve met the term of simply connected domain and proved equivalent definitions of it. We are now one step closer to calculate the fundamental group of a circle! However, the term of simply connected domain will show up many times in the future, so it’s important to be familiar with it regardless our goal.