# Show that every covering map is a local homeomorphism.

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!

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