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I am reading Munkres' Topology, and on pg 337, he gives the proof of the theorem that $p:\mathbb{R}\to S^1$ given by $p(x) = (\cos 2\pi x, \sin 2\pi x)$ is a covering map. Part of the proof says:

Consider the subset $U$ of $S^1$ consisting of those points having positive first coordinate. The set $p^{-1}(U)$ consists of those points $x$ for which $\cos 2\pi x$ is positive; that is, it is the union of the intervals $$ V_n = (n-\tfrac{1}{4}, n+\tfrac{1}{4}) $$ for all $n\in\mathbb{Z}$. Now, restricted to any closed interval $\overline{V}_n$, the map $p$ is injective... Since $\overline{V}_n$ is compact, $p|\overline{V}_n$ is a homeomorphism of $\overline{V}_n$ with $\overline{U}$.

I don't understand the last part. What does the compactness of $\overline{V}_n$ have to do with the fact that $p|\overline{V}_n$ is a homeomorphism?

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1 Answer 1

up vote 7 down vote accepted

There is a theorem in point-set topology that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

The proof is quite lovely (one of my favorites in point-set topology!): let $f$ be such a map. We wish to show that $f$ is an open map. This means, we want to show that $f(O)$ is open for any open set $O$. Well, if $O$ is open, $O^C$ is closed, and any closed subset of a compact set is compact, so $O^C$ is compact, and hence $f(O^C)$ is compact. But in a Hausdorff space, compact implies closed, so $f(O^c)$ is closed. Then $f(O^c)^c$ is open. By bijectivity, we have

$$f(O) = f(O^c)^c$$

so we are done.

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I wish Munkres had put that lemma in his book, it's extremely useful. –  Justin Young Jan 6 '13 at 6:19

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