# Quick question about covering maps

Let $p:E\rightarrow B$ be a covering map and $b \in B$ so there exists a neighborhood $U$ of $b$ such that $$p^{-1}(U)=\bigcup V_\alpha \text{ (disjoint union)}$$ and each $p\restriction_{V_\alpha}:V_\alpha\rightarrow U$ is a homeomorphism. Does it follow that each $V_\alpha \cap p^{-1}(b)\neq\varnothing?$

This is a substep in a problem I am working through right now. We showed that if $V_\alpha \cap p^{-1}(b)\neq\varnothing$ then the intersection has only one element, which more or less led to $p^{-1}(b)$ having the discrete topology.

It follows directly from $p\restriction_{V_\alpha}:V_\alpha\rightarrow U$ being a homeomorphism. It is onto so for any $b\in U$ there must be an $x\in V_\alpha$ so that $p(x)=b$.