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Question: Let $X$ and $Y$ be path-connected spaces that admit a contractible universal cover, with $\pi_1(X) \cong \pi_1(Y)$. Is $X$ homotopy equivalent to $Y$?

Comments: $X$ and $Y$ are both $K(\pi_1(X),1)$s. In particular, this implies that every homomorphism $\varphi: \pi_1(X) \rightarrow \pi_1(Y)$ (e.g., the isomorphism) is induced by a map $f: X \rightarrow Y$. If $X$ and $Y$ are both CW-complexes, Whitehead's theorem says that $f$ is a homotopy equivalence. (In general, by definition, $f$ is a weak homotopy equivalence.) So a counterexample requires that at least one of $X$ and $Y$ fails to have homotopy type a CW-complex.

If one removes the requirement that $X$ and $Y$ have contractible universal cover, in particular if one even relaxes it to $X$ and $Y$ have weakly contractible universal cover, the double comb space is a simply-connected counterexample, as it is not contractible. (A proof that it is not contractible can be found here.)

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    huh I have a friend called Mike Miller. – oxeimon May 28 '14 at 19:37
  • @oxeimon I'm an undergraduate, so probably not me. I was at Penn State in the fall, though, curiously enough. –  May 28 '14 at 19:45
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    Do you only consider spaces, which admit a universal covering? – archipelago May 28 '14 at 22:39
  • @archipelago That was my intent, thanks. I'll edit that in. (I've also removed the word 'aspherical' everywhere but the title because my usage appears to be nonstandard.) –  May 28 '14 at 22:43
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    Did you try $X=S^1$ and $Y$ the pseudo-circle http://en.wikipedia.org/wiki/Pseudocircle ? – Moishe Kohan May 28 '14 at 23:13
  • That might work by establishing the Khalimsky line (or Digital line) as its universal cover, since this line is contractible. – archipelago May 29 '14 at 00:01
  • @archipelago It's easy to see that it's a covering space of the pseudocircle; I've been trying to demonstrate a contraction, though haven't gotten one yet. –  May 29 '14 at 00:04
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    From http://arxiv.org/pdf/0901.2621.pdf follows the contractibility of the Khalimsky line. – archipelago May 29 '14 at 00:09
  • But maybe that's too much overload for you. At least it answers you question. – archipelago May 29 '14 at 00:10
  • @archipelago Yes, it's a great counterexample. (I would have liked an explicit deformation retract, but alas.) Feel free to add that as an answer. –  May 29 '14 at 00:20

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As proposed by studiosus in the comments, the standard unit circle and the pseudocircle (http://en.wikipedia.org/wiki/Pseudocircle) serve as a counterexample, since their universal covering spaces are the real line and the Khalimsky line, both of them contractible. (The contractibility of the latter follows from http://arxiv.org/pdf/0901.2621.pdf .)

archipelago
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    Good to know it works: I was trying to prove contractibility of the universal cover by hand and got stuck. – Moishe Kohan May 29 '14 at 16:13
  • And this leaves open the question: What happens if we assume in addition that $X, Y$ are, say, metrizable? My guess is that the answer is still negative, but one would play with something like Borsuk's examples of compact locally contractible metrizable spaces not h.e. to CW complexes. See http://mathoverflow.net/questions/167954/is-every-paracompact-hausdorff-locally-contractible-space-homotopy-equivalent. – Moishe Kohan May 29 '14 at 16:45