Let $q: S^1 \rightarrow S^1$ be given by $q(z)=z^2$. Identity $\mathbb{R}^2$ with $\mathbb{C}$ in the usual way and embed $S^1\subset \mathbb{R}^2$ as the unit circle. I want to show that $q$ is a covering map.
ATTEMPT:
Let $z \in S^1$. Then $z=e^{ix}$ and $q(z)=e^{2ix}$, where $x=\arg(z)\in [0, 2\pi)$. Now, define the map $f: [0, 2\pi)\rightarrow S^1$ by $f(y)=e^{iy}$. Let $U:= (x-\frac{\pi}{2}, x+\frac{\pi}{2})$, which is a neighborhood of $z$ in $S^1$.
I'm a bit stuck at this point. My goal is to find a partition of $q^{-1}(U)$ into slices, but I can't determine what it should be.
I'd appreciate some help that does not rely on differential methods.
Best Answer
You are almost done. I take a simpler $U$, namely $U = S^1 \backslash \{1\}$. The preimage is $q^{-1}(U) = U_1 \sqcup U_2$ where $U_1 = \{e^{i\theta} \mid \theta \in (0, \pi)\}$ and $U_2 = \{e^{i\theta} \mid \theta \in (\pi, 2\pi)\}$.
Similarly, taking $V = S^1 \backslash \{-1\}$ gives the desired trivialisation of the covering. As written in comments, any map between Riemann surface is a covering map assuming that the derivative does not vanish.