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We know the fact that

If $p:E\longrightarrow B$ is a covering map, then $p$ is a local homeomorphism.

However, it is hard to find the proof of this statement. I only found one proof here: https://topospaces.subwiki.org/wiki/Covering_map_implies_local_homeomorphism, but it seems that it uses a really weird definition of covering map.

I tried to prove this statement but I got stuck.

Below is my attempt:

Let $e\in E$ and set $x=p(e)\in B$. Since $p$ is a covering map, we can choose a neighborhood $U$ of $x$ that is evenly covered by $p$.

Let $(V_{\alpha})$ be a partition of $p^{-1}(U)$ into slices, that is $p^{-1}(U)$ is a disjoint union of $V_{\alpha}$ and $p|_{V_{\alpha}}:V_{\alpha}\longrightarrow U$ is a homeomorphism onto $U$ for each $\alpha$.

Then, how could I argue that $e$ must be in one of $V_{\alpha}$?

If this was true, then it follow immediately since then the point $e$ has a neighborhood $V_{\alpha}$ that is mapped homomorphically by $p$ on to an open subset $U$ of $B$.

Thank you!

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1 Answer 1

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Since $U$ is a neighborhood of $x$, we have $p(e)=x\in U$, and thus $e\in p^{-1}(U) =\bigcup_\alpha V_\alpha$.

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