10
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ 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. $\endgroup$
    – Hunter Brooks
    Commented Aug 30, 2012 at 15:31
  • 5
    $\begingroup$ 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. $\endgroup$
    – John Klein
    Commented Aug 30, 2012 at 15:44
  • 4
    $\begingroup$ @John: Such a map would lift to the universal cover of $X$, so have degree divisible by 120. $\endgroup$
    – Oscar Randal-Williams
    Commented Aug 30, 2012 at 15:57
  • $\begingroup$ Yep. I was just posting it as an answer, but you beat me to the punch. $\endgroup$
    – John Klein
    Commented Aug 30, 2012 at 16:00
  • $\begingroup$ Merci beaucoup. $\endgroup$
    – Peter Muller
    Commented Aug 30, 2012 at 21:51

1 Answer 1

Reset to default
13
$\begingroup$

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.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .