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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?

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  • $\begingroup$ what is your definition of covering space? $\endgroup$
    – Camilo Arosemena-Serrato
    Dec 13, 2012 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, 2012 at 4:01
  • $\begingroup$ I think every subspace of a locally (path) connected space is locally (path) connected. $\endgroup$
    – Arun Kumar
    Dec 5, 2014 at 4:56

1 Answer 1

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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.

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