# Is the homotopy type of an aspherical space determined by its fundamental group?

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.)

• 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. – user98602 May 28 '14 at 19:45
• Do you only consider spaces, which admit a universal covering? – archipelago May 28 '14 at 22:39
• Did you try $X=S^1$ and $Y$ the pseudo-circle en.wikipedia.org/wiki/Pseudocircle ? – Moishe Kohan May 28 '14 at 23:13
• From arxiv.org/pdf/0901.2621.pdf follows the contractibility of the Khalimsky line. – archipelago May 29 '14 at 0:09

• 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 mathoverflow.net/questions/167954/…. – Moishe Kohan May 29 '14 at 16:45