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