A proof of $\pi_1(S^1)=\mathbb{Z}$ without universal covers

Question

I'm reading "A Guide to the Classification Theorem for Compact Surfaces" by Jean Gallier adn at page 45 is this proposition:

I don't understand how $\beta_k$ and $\delta_k$ are defined. Can someone explain better or reformulate this part? Thanks!

Answer 1

Consider any closed path $\beta : [0,1] \to S^1$ based at $1$. Since $[0,1]$ is compact, $\beta$ is uniformly continuous. Hence there exists $\epsilon > 0$ such that $\lvert \beta(s) - \beta(t) \rvert < 2$ if $\lvert s - t \rvert < \epsilon$. Let $n > 1/\epsilon$. Then for each subinterval $I_k = [\frac{k-1}{n},\frac{k}{n}]$, $k=1,\dots,n$, it is impossible that both $-1,1 \in \beta(I_k)$ because $\lvert -1 - 1 \rvert = 2$. Now define $\beta_k(t) = \beta(\frac{k-1+t}{n})$. This is essentially the map $\beta \mid_{I_k}$.

Given two points $z_1,z_2 \in S^1$, let $q = z_2/z_1 \in S^1$ and write $q = e^{it_q}$ with $t_q \in [0,2\pi)$. There are two circular arcs $\delta^\pm : [0,1] \to S^1$ from $z_1$ to $z_2$: $$\delta^+(t) = z_1e^{it_qt}, \delta^-(t) = z_1e^{i(t_q - 2\pi)t} .$$ $\delta^+$ goes counterclockwise and $\delta^-$ clockwise.

If $\zeta \in S^1$ such that $z_1, z_2 \in S^1 \setminus \{ \zeta \}$, then exactly one of the paths $\delta^\pm$ does not pass $\zeta$.

If $\beta_k$ does not pass through $1$, let $\delta_k$ denote the unique circular arc from $\beta_k(0)$ to $\beta_k(1)$ not passing through $1$. The map $h : (0,2\pi) \to S^1 \setminus \{ 1\}, h(\tau) = e^{i\tau}$, is a homeomorphism. Hence the map $$H^k : [0,1] \times[0,1] \to S^1, H^k(t,s) = h((1-s)h^{-1}(\beta_k(t))+sh^{-1}(\delta_k(t))$$ is a homotopy from $\beta_k$ to $\delta_k$ which keeps the endpoints fixed.

The case that $\beta_k$ does not pass through $-1$ is treated similarly.

I think the rest of the proof should be clear, but I do not believe that the proof is elegant.