I have been learning about covering maps, and I am having trouble proving something which my intuition is telling me should be true.

For this question, a covering map is a continuous, surjective map $q : E \to X$ where $E$ is locally path connected and connected, and every point in $X$ has an evenly covered open neighbourhood.

Assume that $A \subseteq X$ is simply connected. Then $q^{-1}(A)$ is a disjoint union of sets (just take connected components in the subspace topology). Let $W$ be one of these connected components.

Question: Does $q$ restrict from a homeomorphism from $W$ onto $A$?


Since $q$ is a covering, it follows that the restriction $q^{-1}(A) \to A$ is a covering map. But $A$ is simply connected, so this covering is trivial, which implies an affirmative answer to your question.

  • $\begingroup$ i dont think this works. $q^{-1}(A)$ is in general going to be a disconnected space. EDIT. maybe it does, depedning on what you mean by trivial $\endgroup$ – DBr Jun 1 '12 at 3:40
  • $\begingroup$ also, I am about to start graduate school at UM :) $\endgroup$ – DBr Jun 1 '12 at 3:41
  • $\begingroup$ I think the statement "$q^{-1}(A) \to A$ is a covering map" is only true when $A$ is locally path connected. I cant think of a counter example but I cant prove it unless $A$ is locally path connected $\endgroup$ – DBr Jun 1 '12 at 5:30
  • $\begingroup$ @DBr: By "trivial" I mean "isomorphic to the projection $A \times I \to A$, where $I$ is a discrete set." And you may be right that we must assume $A$ is locally path connected, but this is a totally harmless hypothesis when talking about covering spaces. $\endgroup$ – Justin Campbell Jun 1 '12 at 15:54

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