if $p:\widetilde{X}\rightarrow X$ is a covering space and $\widetilde{X}$ is path connected ,show that $p^{-1}(A)$ is path connected.

Question

if $p:\widetilde{X}\rightarrow X$ is a covering space and $\widetilde{X}$ is path connected ,also $A\subset X$ is a path connected subset,show that $p^{-1}(A)$ is path connected.

I suppose that $p^{-1}(A)$ isn't path connected and I think it will come something contradiction to the covering space but I don't know how should I continue.please help me with your knowledge,thank you very much.

Answer 1

This is not true. Take any connected covering of degree $d>1$ and $A$ a point. $p^{-1}(A)$ has $d$ connected components, while $A$ is connected.

Answer 2

It is true in the trivial case of $d=1$, $p:X\rightarrow X$. We know that if $p:\tilde{X}\rightarrow X$ is a covering and $X$ is locally path-connected then so is $X$.

Answer 3

Massey's book on "Algebraic Topology" gave some information on the connectivity of a pullback of a covering map; a more complete account is given in this paper "Groupoids and the Mayer-Vietoris sequence" JPAA 30 (1983) 109-129. It uses the notion of covering morphism of groupoids, which encapsulates the notion of covering map of spaces. So from a pullback of a covering morphism you get a covering morphism, and by choosing base points you get a diagram involving various vertex groups and sets of components the exactness properties of which enables you to read of facts about connectivity etc. This result is also in Topology and Groupoids, Section 10.7, but is a bit too long to give here.