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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 $z^n$?


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Consider $S^1$ as $\{z\in \Bbb C:|z|=1\}$ –  Dominik Dec 10 '12 at 20:50
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up vote 4 down vote accepted

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!

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Thank you for your answer! –  user42912 Dec 17 '12 at 5:45
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