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Dear community,

assume that $f\colon X\to Y$ is a homology equivalence, i.e. induces isomorphisms on all integer homology groups. Is there a map $g\colon Y\to X$ which is also a homology equivalence?

Actually, I am only interested in the case if $f\colon X\to X^+$ is obtained by Quillen´s plus construction.

Thanks for any comments in advance.

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In extreme generality no - take the circle and map it onto the non-Hausdorff space "$[-1, 1]$ with two-origins" (by projecting vertically downward, mapping the north and south poles to the separate origins). This gives an isomorphism on all homology groups, but any map the other way is homotopic to a constant since the origins collapse. – Hunter Brooks Aug 30 '12 at 15:31
My guess is that the $X =$ the Poincare homology sphere should provide a counterxample. If there were a map $S^3 \to X$ which induced a homology isomorphism, then this map is necessarily of degree one and we see that the fundamental class of $X$ is spherical. I don't see it at the moment, but this is very unlikely to be true. – John Klein Aug 30 '12 at 15:44
@John: Such a map would lift to the universal cover of $X$, so have degree divisible by 120. – Oscar Randal-Williams Aug 30 '12 at 15:57
Yep. I was just posting it as an answer, but you beat me to the punch. – John Klein Aug 30 '12 at 16:00
Merci beaucoup. – Peter Muller Aug 30 '12 at 21:51
up vote 13 down vote accepted

As I was writing this answer, Oscar beat me to the punch. I will keep it posted anyway.

Let $X^3$ be the Poincare homology sphere. Let $\tilde X \to X$ be the universal cover (note: $\tilde X$ is $S^3$). As in my comment, if there were a map $S^3 \to X$ inducing a homology equivalence, then that map must necessarily factor through $\tilde X$ (by covering space theory). But this is impossible since $\tilde X \to X$ has degree 120, whereas the composite $S^3 \to \tilde X \to X$ is supposed to have degree one.

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