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Prove that every local homeomorphism between compact, connected, topological spaces is a covering map of some finite degree.

If the spaces were Hausdorff, the proof is easy, since then the singleton is closed, hence so is its preimage, and as a closed subset of a compact subset it is compact, etc.

However, without the Hausdorff condition, I have not found a way to proceed.

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    $\begingroup$ I think the claim is false without the Hausdorff condition. $\endgroup$ – Zhen Lin Aug 13 '14 at 8:23
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Let $X$ be the interval $[-1,1]$ with the origin doubled, that is, $X$ is the non-Hausdorff space that you get by glueing two copies of $[-1,1]$ along $[-1,0) \cup (0, 1]$. Then $X$ is compact and connected and the obvious map $X \to [-1,1]$ is a local homeomorphism which is not a covering projection.

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