1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $p(x)=b$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.