this map $p(z)=z^n$ is a covering map? For any positive integer $n$ the map $p:S^1\to S^1$, defined by $p(z)=z^n$ is a covering map.
I think it's easy, but as I'm a really beginner I'm struggling to prove it. By the way, what kind of operation is that $z^n$?
Thanks
 A: I'm in a rush, but here's a hint:
EDIT: Now that I have more time, let me see if I can explain this better.
Let's begin by thinking of the circle as a paramaterized curve. In particular, let's consider parameterizing $S^1$ as $\gamma(t)=e^{2\pi it}$ for $t=[0,1]$. The curve $\gamma$ can be thought of as traversing the unit circle once, counterclockwise. While this parameterization will not really make any of our work any easier, it will at least make it clear that the statement is true.
Ok, so note then that if $f:z\mapsto z^n$ is the $n^{\text{th}}$ power map, then we can get a handle of what $f$ does to $S^1$ by seeing how the map $f\circ \gamma$ works--for ease of notation let's call $f\circ\gamma=h$. Now, just unraveling this, we see that, by definition, $h$ is a parameterization of $S^1$ given by $h:[0,1]\to S^1:t\mapsto e^{2\pi i n t}$. Now, the important thing to notice about $h$, is that while $\gamma$ traversed $S^1$ once, we've (in essence) "sped up" so that in our allotted time of $[0,1]$ we actually round the circle $n$ times. In particular, we see that we round the circle once in the interval $[0,\frac{1}{n}]$, again in the interval $[\frac{1}{n},\frac{2}{n}]$, and in general, once in the interval $[\frac{m}{n},\frac{m+1}{n}]$ with $m\in\{0,n-1\}$. Moreover, if we restrict to open subsets of each of these intervals, our map is a homeomorphism onto its image.  Thus, we see that, at least intuitively, $h$ does cover $S^1$ "$n$-times".
Now, let's see what happens if we pick a point $p_0\in S^1$ and start looking at neighborhoods around it, and in particular, how about we consider $1\in S^1$. Now, while there are tons of neighborhoods, it makes sense to, if we are going to prove that $f$ is a covering map, restrict our attention to neighborhoods of $p_0=e^{\frac{\pi i}{3}}$ that "come" from $f$, or in our case $h$. In particular, we note that $p_0=h\left(\frac{1}{12n}\right)$. So, a good way to pick a neighborhood of $p_0$ would be to look at the image under $f$ of a neighborhood of $\frac{1}{12n}$--let's say we look at $\left(\frac{1}{13n},\frac{1}{11n}\right)$. Now, this is all valid since we know that on open subsets of each of our subintervals $[\frac{m}{n},\frac{m+1}{n}]$ the map $h$ is a homeomorphism so that $U=h\left(\frac{1}{13n},\frac{1}{11n}\right)$ really is a neighborhood of $p_0$
Now, that we have a very nice neighborhood of $p_0$, let's try to verify the conditions on $f$ that make it a covering map. Let's look at $f^{-1}(U)$. Well, let's actually being by looking at $h^{-1}(U)$, because this is easy to get a handle on. Namely, we know that $h$ just wraps around $S^1$ $n$-times and it will covering $U$ precisely during the time intervals in each interval $[\frac{m}{n},\frac{m+1}{n}]$ that "act like" $\left(\frac{1}{13n},\frac{1}{11n}\right)$. It is not hard to see that each of these subintervals are just going to be of the form $\left(\frac{1}{13n}+\frac{m}{n},\frac{1}{11n}+\frac{m}{n}\right)$ for $m\in\{0,\cdots,n-1\}$. So, now that we have a handle on what $h^{-1}(U)$ looks like, we can easily deal with $f^{-1}(U)$ by just "pushing forward" by $\gamma$. Namely, $f^{-1}(U)$ will just be the union of the sets $\gamma\left(\frac{1}{13n}+\frac{m}{n},\frac{1}{11n}+\frac{m}{n}\right)$ for each $m\in\{0,\cdots,n-1\}$ (where on the circle are these regions?). Moreover, it's clear that each of these intervals are disjoint, because $\gamma$ acts (except for the endpoints, which we aren't worrying about) injectively. Thus, we see that $f^{-1}(U)$ is a disjoint union of $n$ open subsets $\gamma\left(\frac{1}{13n}+\frac{m}{n},\frac{1}{11n}+\frac{m}{n}\right)$ of $S^1$. Moreover, since the action of $f$ on each of these subsets is just $h$ on the intervals $\left(\frac{1}{13n}+\frac{m}{n},\frac{1}{11n}+\frac{m}{n}\right)$ we know that $f$ restricted to each of these maps is a homeomorphism. Thus, we see that the covering map condition is at least at the point $p_0$.
Now, I went into way too much detail, and I'm sure you got the point half-way through, but I hope that was able to help, and I hope you can generalize this to prove that $f$ is a covering map! 
A: My answer is nothing more than an elaboration, based on the idea of the proof by Alex Youcis, of the one sketched by John M. Lee, Introduction to Topological Manifolds, 2.ed, p.279.
Let $p_n:S^1 \to S^1, z \mapsto z^n, \, n \in \mathbb N, \, n \geq 2 $, and for $ x \in S^1 $ chose the open neighborhood $ U= S^1 \backslash \{-x\} $ with the point $-x$ being the one opposite to $x$ on the unit circle. The preimage of $U$ is the set $ \{ z \in S^1: z^n \neq -x \} $ which is the complement of the set of solutions of $ z^n=-x $, the n roots of unitiy based at the point $-x$ on the unit circle. Thus, $ p_n^{-1}(U) $ is the disjoint union of n open, connected and equally long arcs of the circle. In order to show that everyone of them is homeomorphic to U, for $ k=1,\dots,n$ we look at the map $p_n \circ \epsilon: (\frac{k-1}{n}, \frac{k}{n}) \to U $ with the map $ \epsilon:\mathbb R \to S^1, t \mapsto e^{2 \pi \, i \, t} $ which for example in the intervall [0,1] travels once around the circle, counterclockwise, starting at 1. The map
$ p_n \circ \epsilon(t) = \left( e^{2 \pi \, i \, t} \right)^n = e^{2 \pi \, i \, n t} $
runs n times faster around the circle than $ \epsilon $. So no matter which value k has, for any of the open arc segments corresponding to the open inveralls $ (\frac{k-1}{n}, \frac{k}{n}) \subseteq \mathbb R $ it runs once around $ S^1 \backslash \{ 1 \} $. The only thing that remains to be done is to let it start at $-x$ instead of 1. So we give it an offset of $ \frac{arg (-x)}{2 \pi}$ with $ 0 \leq arg(-x) < 2 \pi $ and call it
$ \varphi: \, (\frac{k-1}{n} + \frac{\arg (-x)}{2 \pi}, \frac{k}{n} + \frac{\arg (-x)}{2 \pi}) \to U , \, t \mapsto e^{2 \pi \, i \, n \, t}, \quad k=1,\dots,n $.
This function is continuous everywhere, since we do not complete a full circle, but omit one point. It is bijective with the inverse function
$ \varphi^{-1}: \, U \to (\frac{k-1}{n} + \frac{\arg (-x)}{2 \pi}, \frac{k}{n} + \frac{\arg (-x)}{2 \pi}), \quad u \mapsto \frac{\ln u}{2 \pi \, i \, n} + \frac{(k-1) 2 \pi}{n} + \frac{\arg (-x)}{2 \pi}, \quad k=1,\dots,n. $
Again this is continuous everywhere, since we do not describe a full circle. Since $ \epsilon $ is itself a homeomorphism as long as we do not complete a full circle, we have proved that $ \varphi \circ \epsilon^{-1} = p_n \circ \epsilon \circ \epsilon^{-1} = p_n $ is a homeomorphism for each one of the n disjoint arc segments of the preimage of U.
If anyone sees a mistake in my proof, please let me know.
