The Fundamental Group of the Circle, Part 4

‍

Welcome back to our series on the fundamental group of the circle where we're following the outline below to prove 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 6

‍

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

Claim: Φ is Surjective

To prove the claim, we will need the following lemma whose proof we defer until Part 6:    Lemma 1 (The Path Lifting Property): For each path $f:I\to S^1$, with $f(0)=x_0$ and for each $\tilde{x_0}\in p^{-1}(x_0)$, there is a unique lift $\tilde{f}:I\to\mathbb{R}$ such that $\tilde{f}(0)=\tilde{x_0}$.  A couple of observations:

• It's easy to see that a path in the helix ($\mathbb{R}$) will project down to a path in $S^1$ (we've already used this idea in Part 3). Lemma 1 tells us that the converse is true too! That is, given a path $f$ 'downstairs' in $S^1$ that starts at $x_0$, and given a lift $\tilde{x_0}$ of $x_0$, there is a path $\tilde{f}$ 'upstairs' in $\mathbb{R}$ starting at $\tilde{x_0}$ that projects down to $f$.    
• Notice the set $p^{-1}(x_0)$ is infinite! So the lemma says that if you fix an $\tilde{x_0}\in p^{-1}(x_0)$ (we call this a point in the "fiber" above $x_0$), then there is a map $\tilde{f}:I\to\mathbb{R}$ which takes $0$ to that $\tilde{x_0}$.

We will assume Lemma 1 for now and will proceed to show $\Phi$ is onto.

Let $[f]\in\pi_1(S^1)$ be the path-homotopy class* of some loop in $S^1$ and pick a representative $f:I\to S^1$ with base point $(1,0)=f(0)=f(1)$. We wish to show that there is an $n\in\mathbb{Z}$ such that $\Phi(n)=[f]$, i.e. such that $f\simeq \omega_n$ for some $n$.  

To this end, begin by observing that $0\in p^{-1}(1,0)=\mathbb{Z}$. Hence by Lemma 1, there is a unique lift $\tilde{f}:I\to\mathbb{R}$ such that $\tilde{f}(0)=0$. Now since $$(p\circ \tilde{f})(1)=f(1)=(1,0)\in S^1,$$ we see that $\tilde{f}(1)$ must be an integer. In other words, $\tilde{f}$ is a path from $0$ to $n$ for some integer $n$. (In the picture below, $\tilde{f}$ is the red path and $n=3$.)         

Thus $\tilde{f}\underset{F}{\simeq} \widetilde{\omega}_n$ for some homotopy $F$** and so $$f= p\circ\tilde{f}\;\underset{p\circ F}{\simeq}\; p\circ\widetilde{\omega}_n=\omega_n.  $$ With this we conclude $[f]=[\omega_n]=\Phi(n)$ and so $\Phi$ is surjective.

Next time, we will prove $\Phi$ is one to one.

‍

Footnotes:

*Just as a reminder, two paths $f$ and $g$ are path homotopic if they are homotopic and if their starting/ending points are the same.

**They are homotopic and not equal because $\tilde{f}$ isn't necessarily the straight line $\widetilde{\omega}_n(s)=ns$ for $s\in[0,1]$.