# The Fundamental Group of the Circle, Part 5

Today we come to part 5 in the outline of the proof that $\pi_1(S^1)\cong \mathbb{Z}$.

Part 1: Set-up/observations
Part 2: Show $\Phi$ is well defined
Part 3: Show $\Phi$ is a group homomorphism
Part 4: Show $\Phi$ is surjective
Part 5: Show $\Phi$ is injective  (Note: parts 4 and 5 require two lemmas whose proofs we will defer until part 5)
Part 6: Prove the two lemmas used in parts 4 and 5

Last week we showed that the homomorphism $\Phi:\mathbb{Z}\to\pi_1(S^1)$ by $n\mapsto[\omega_n]$ where $\omega_n:[0,1]\to S^1$ is the loop given by $s\mapsto (\cos{2\pi n s}, \sin{2\pi ns})$ is surjective. Today we will show that it is injective.

## Claim: Φ is Injective

As in our proof that $\Phi$ is surjective, we begin with a lemma whose proof we defer until next time.

#### Lemma 2: For each homotopy $f_t:I\to S^1$ of paths starting at $f_t(0)=x_0$ and for each $\tilde{x}_0\in p^{-1}(x_0)$, there is a unique lift $\tilde{f}_t:I\to\mathbb{R}$ such that $\tilde{f}_t(0)=\tilde{x}_0$.

A few observations:

• To be careful, we should emphasize that $f_t$ isn't merely a homotopy, it is a path homotopy, i.e. it is a homotopy between two paths $f$ and $g$ that share the same initial point, $x_0$ (this is what we mean by $f_t(0)=x_0$) and ending point.
• We are fixing $t$ here (just momentarily) and are imagining that $f_t:I\to S^1$ is just a path, much like the path $f$ in Lemma 1. But in the back of our mind, we know that the statement of Lemma 2 holds for all $t\in[0,1]$ and hence for the family of maps - i.e. the homotopy - $f_t$.
• It's easy to see that a homotopy in the helix ($\mathbb{R}$) will project down to a homotopy in $S^1$. This lemma tells us that the converse is true too! That is, given a homotopy in $S^1$ of paths that start at $x_0$, you can find a lifted homotopy of paths in $\mathbb{R}$ that start at the lift of $x_0$.

We will assume Lemma 2 for now and will proceed to show $\Phi$ is one to one.

Suppose $\Phi(m)=\Phi(n)$ so that $\omega_m\simeq \omega_n$, for some $m,n\in\mathbb{Z}$. We wish to show $m=n$. Let $f_t:I\to S^1$ be the homotopy from $\omega_m$ to $\omega_n$ with $f_0=\omega_m$ and $f_1=\omega_n$ and notice that $f_t(0)=(1,0)\in S^1$. This follows since $(1,0)$ is the common starting point of $\omega_m$ and $\omega_n$ (i.e. $\omega_m(0)=\omega_n(0)=(1,0)$). Also note that $0\in p^{-1}(1,0)=\mathbb{Z}$.

By Lemma 2, there exists a unique lift $\tilde{f}_t:I\to\mathbb{R}$ such that $\tilde{f}_t(0)=0\in \mathbb{R}$. Moreover, since the lift $\tilde{f}_t$ is unique, we must have    $$\tilde{f}_0=\widetilde{\omega}_m \quad \text{and} \quad \tilde{f}_1=\widetilde{\omega}_n.$$ Indeed, we know that $\omega_m=f_0$ has a lift, namely $\widetilde{\omega}_m$ as defined in Part 1. But Lemma 2 tells us that the given lift $\tilde{f_t}$ of $f_t$ is unique. So if $\tilde{f}_0$ is any other lift of $f_0=\omega_m$, it must be that $\tilde{f}_0=\widetilde{\omega}_m$. A similar statement holds for the two lifts $\tilde{f}_1$ and $\widetilde{\omega}_n$ of $\omega_n=f_1$.

Since $\tilde{f}_t$ is a path homotopy, $\tilde{f}_t(1)$ (the shared endpoint of the loops that $\tilde{f}_t$ is "homotopy-ing") must be constant for all $t$. In particular, we must have    $$\tilde{f}_t(1)=\widetilde{\omega}_m(1)=\widetilde{\omega}_n(1)=\text{the common end point of \widetilde{\omega}_m and \widetilde{\omega}_n.}$$    But this common end point must be an integer, and it must be both $m$ and $n$! Hence $m=n$.

Just to recap: the key  was to show that $\widetilde{\omega}_m$ and $\widetilde{\omega}_n$ are path homotopic. This allows us to draw this picture:

(of course, the $\widetilde{\omega}$'s are really straight lines in $\mathbb{R}$, but you get the idea) and conclude that the two paths must end at the same point. (That's what a path homotopy requires!) But we know that $\widetilde{\omega}_m$ ends at $m$ and $\widetilde{\omega}_n$ ends at $n$. So it must be that $m=n$. Voila!

This essentially ends our proof that $\pi_1(S^1)\cong \mathbb{Z}$, but it still remains to prove Lemmas 1 and 2. We will do so next time.

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