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.



Reverse map of a homology equivalence.
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.


