3
$\begingroup$

Let $A\subset X$ and $p: E \to X$ be a covering space. Assume that $A,X,E$ are all locally path connected, path connected, Hausdorff. Suppose that $p^{-1}(A)$ is path connected.

I want to show that $p^{-1}(A)$ is locally path connected. It is a preimage of a locally path connected space, also a subspace of a locally path connected, but I think these do not help. To check directly if $x\in U \subset p^{-1}(A)$, then $p(x) \in p(U) \subset A$ but $p(U)$ need not be an open set.

How can I prove it?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ what is your definition of covering space? $\endgroup$ – Camilo Arosemena-Serrato Dec 13 '12 at 21:04
  • $\begingroup$ @CamiloArosemena surjective, continuous map, every point has a neighborhood whose preimage consists of disjoint union of open sets homeomorphic to the neighborhood. $\endgroup$ – Gobi Dec 14 '12 at 4:01
  • $\begingroup$ I think every subspace of a locally (path) connected space is locally (path) connected. $\endgroup$ – Arun Kumar Dec 5 '14 at 4:56
2
$\begingroup$

Every point $x \in p^{-1}(A)$ has an open neighborhood $U$ which maps homeomorphically onto an open neighborhood $V$ of $f(x) \in A$. As an open subset of either $A$ or $X$, this set $V$ is locally path-connected, and that property is invariant under homeomorphisms.

| cite | improve this answer | |
$\endgroup$

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.