Show that the map $p : S^1 → S^1$ given by $p(z) = z^2$ is a covering map. Generalize to the map $p(z) = z^n$.
I know that there are already several posts offering an answer to this question as here and here, but my question is so that someone can explain an answer to the problem I found out there, the solution is the following:
There are several things that I do not understand, I know that in order to prove that a function is a covering map, we have to take a point in $S^1$ and show that there is a neighborhood of this point that when taking the inverse image can be written as a union of open sets of $S^1$ and disjoint I think that in the problem take the $z$ and the neighborhood to prove that there is $U$, now I guess the problem is to show that $p^{-1}(U)$ can be expressed as a union of disjoint openings of $S^1$, but I do not understand why $U$ is an open semicircle centered on $z$, could someone explain to me why? And what follows of the test I do not understand either. Thank you very much.
You can choose a different $U$ if you like, the point is that we need to only exhibit a neighborhood $U$ of each point $x \in S^1$ so that $p^{-1}(U)$ is two disjoint sets $V_1 \coprod V_2$ so that $V_i \cong U$ is an isomorphism.
So, take the interval $I_x=(x-\pi/2,x+\pi/2)$ under the map $\exp:x \mapsto e^{(ix)}$. Then explicitly:
$\exp(I_x)$ cconsists of points on the circle with angle $(x-\pi/2,x+\pi/2)$. This is a semicircle with "Center" at $x$. One way to see this is that $I_0 \mapsto \exp(-\pi/2,\pi/2)$, which I hope makes obvious that it is a semicircle.
Then, using some other $x$ is the same as first rotating the image by an angle of $x$ for all $x \in [0,2 \pi)$.
Looking at the preimage of $U_x$ associated to some $x$ is just considering the collection of points so that when you double their angle you end up in $U$. For $U_0$, those are all the points in between $(-\pi/4,\pi/4)$ but also those guys antipodal to each of those angles. In particular $(2 \cdot \pi)$ for example is $2 \pi$, which also lands you at zero. So we have two neighborhoods we can identify on the circle that are diametrically opposed, and are both quarter circles, which are open intervals, which are homeomorphic to $U_x$.