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I came across the following interesting problem while studying covering spaces for the first time, and was trying to prove it, but was not really able to get a proof off of the ground (i.e., beyond recalling the hypotheses of the problem, I was not really able to gain momentum in my written attempts after an hour and a half):

A convex body is a compact, convex subset of $\mathbb{R^n}$ that contains an interior point. Suppose $p: Y \rightarrow X$ is a covering, $Z$ is a convex body, and $f: Z \rightarrow X$ a continuous mapping. Show that for any $z \in Z$, and any $y \in Y$ such that $p(y) = f(z)$, there is a unique continuous mapping $\tilde{f}: Z \rightarrow Y$ such that $p \circ \tilde{f} = f$ and $\tilde{f}(z) = y$.

I am confident the remark about $Z$ having an interior point really matters here (perhaps suggesting that $Z$ is homeomorphic to another [perhaps nice!] topological subspace of Euclidean space), but I am not sure how. I expect that map lifting results (seen in connection with covering spaces) will factor in. I would like to know if anyone visiting would be up for walking me through a proof of this neat looking exercise.

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Note that $Z$ is simply connected (it is actually homeomorphic to the closed unit ball in $\mathbb R^n$). Now try to use the map lifting property of coverings. –  Alexander Thumm Jan 31 '12 at 6:21
    
Since Z is simply connected, it is both connected and path-connected. But could anyone explain why having an interior point would imply local path-connectedness (assuming this is legit)? If this is true then I am aware of a map lifting theorem that will do the job, but I need local path-connectedness. –  Vulcan Feb 1 '12 at 0:28
    
$Z$ is locally path-connected. This follows from the fact, that $\mathbb R^n$ is locally convex. –  Alexander Thumm Feb 1 '12 at 6:15
    
Thank you so much. Now I should be able to use the map lifting theorem to prove this result! –  Vulcan Feb 1 '12 at 6:40

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