# Simply connected spaces

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 $S^1$. 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 $\gamma$ is a (continuous) function:

\gamma:[0,1]\to X

( $X$ is a topological space) such that $\gamma(0)=\gamma(1)$.

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:

1. $f:S^1\to X$ is null-homotopic
2. $f:S^1\to X$ is null-homotopic with respect to any point on $S^1$.

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 $\mathbb{R}^2$. 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 $t\in I$. 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 $e^{2\pi i\cdot 0}=e^{2\pi i\cdot 1}=1$. Therefore, if $\gamma(0)\neq\gamma(1)$ then $\gamma^\prime$ is not even well-defined!

But we are dealing with loops! Therefore $\gamma(0)$ is indeed the same as $\gamma(1)$, so $\gamma^\prime$ 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 $\gamma$ 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 $\partial I=\{0,1\}$. Suppose that $H:I\times I\to X$ is a homotopy with respect to $\partial I=\{0,1\}$ from $\gamma_1$ to $\gamma_2$.

Now, consider the representations of $\gamma_1,\gamma_2$ as maps from the circle (I’ll denote them by $\gamma_1^\prime,\gamma_2^\prime$). 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 $\gamma_1^\prime$ to $\gamma_2^\prime$.

But there is one problem though – same as before, since $e^{2\pi i\cdot 0}=e^{2\pi i\cdot 1}$ 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 $\{0,1\}$. Therefore, this map is indeed well defined. In addition, this is an homotopy with respect to the point $e^{2\pi i\cdot 0}=e^{2\pi i\cdot 1}=1$ (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 $S^1$ or from $I$. Moreover, I can think of two homotopic paths with respect to $\{0,1\}$ to be equivalent to two homotopic paths (with $S^1$ 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 $X$ 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 $X$ is path-connected, then the following are equivalent:

1. $X$ is simply connected.
2. For every $a\in X$, $\pi_1(X,a)=\{1\}$.
3. For every two points $a,b\in X$ – every two paths from $a$ to $b$ are homotopic with respect to $\partial I=\{0,1\}$.
4. There are two points $a,b\in X$ such that every two paths from $a$ to $b$ are homotopic with respect to $\partial I=\{0,1\}$.

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 $a=b$, 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. A really nice illustration of this property that was taken from the wikipedia. It kind of gives us a spoiler though – we can see that the sphere is simply connected

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!

#### $1 \Rightarrow 2$ $1 \Rightarrow 2$

Suppose that $X$ is simply connected, thus, every map:

f:S^1\to X

Is null-homotopic.

Let $a\in X$ be an arbitrary point and $\gamma:S^1\to X$ a loop with $a$ as it’s base point.

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

Therefore, $\gamma$ is homotopic with respect to 1 to the constant function $K_{\gamma(1)}=K_{\gamma(e^{2\pi i 0})}=a=K_a$. Thus:

[\gamma]=[K_a]

So the only equivalence class in $\pi_1(X,a)$ is the trivial one. In other words, $\pi_1(X,a)=\{1\}$ as we wanted.

#### $2 \Rightarrow 3$ $2 \Rightarrow 3$

Suppose that $a,b\in X$, $\gamma_1,\gamma_2:[0,1]\to X$ are paths from $a$ to $b$.

Thus, the path $\gamma_1*\overline{\gamma_2}$ is a loop with base point $a$. Since we are assuming that $\pi_1(X,a)=\{1\}$ 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 $\partial{I}$ (though we worked really hard to prove properties of the action between classes). Therefore:

\gamma_1\sim_{\partial I}\gamma_2

As required.

#### $3 \Rightarrow 4$ $3 \Rightarrow 4$

This one is trivial…

#### $4 \Rightarrow 1$ $4 \Rightarrow 1$

Suppose that $f:S^1\to X$ is some map. We need to show it is null homotopic.

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

By the assumption, there are two point $a,b$ such that every two paths between them are homotopic with respect to $\partial I$.

Since $X$ is path-connected, there exists a path $\gamma_1$ from $a$ to $p$. Similarly, there exists a path $\gamma_2$ from $p$ to $b$. And finally, there exists a path $\gamma_3$ from $a$ to $b$.

Note that $\gamma_1*f*\gamma_2$ Is a path from $a$ to be $b$. Thus, it is homotopic to $\gamma_3$ with respect to $\partial I$. 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 $\gamma_1*f*\gamma_2*\overline{\gamma_3}$ is homotopic to the constant loop $K_a$. Let $H$ 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:

Simply connected spaces

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

And we proved that $f$ 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.