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.)

Solution 1:

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 .)