If one starts with a fiber bundle $f: X \to Y$ so that fibers having trivial integral homology by using spectral sequence one can get the induced map $f_*: H_*(X;\mathbb{Z}) \to H_*(Y;\mathbb{Z})$ is an isomorphism.

My question is as follows:

Given a surjective submersion (I am not considering it to be proper) $f:X \to Y$ (connected smooth manifolds) such that all the fibers have trivial integral homology does this mean the induced map $f_*: H_*(X;\mathbb{Z}) \to H_*(Y;\mathbb{Z})$ is isomorphism?

I will be grateful to see some counterexamples or proof.

Thanks in advance!!

  • 1
    $\begingroup$ No, that is not true. This has come up in MO before. I will include a link in the next comment. $\endgroup$ Feb 21 at 23:57
  • $\begingroup$ @JasonStarr that will be really helpful.. $\endgroup$
    – piper1967
    Feb 22 at 0:06
  • $\begingroup$ Actually I was thinking of another example where all fibers are diffeomorphic, but the morphism is not a fiber bundle. However, in the example that I know, the fibers have nontrivial integral cohomology (but the Kuenneth theorem does not hold, not even locally on the base). $\endgroup$ Feb 22 at 1:27
  • $\begingroup$ It's true under the assumption that each fibre has trivial homotopy, in which case $f$ is a fibration and a homotopy equivalence (this is due to G. Meigniez). This condition is sufficient but not necessary for $f$ to be a homology equivalence. $\endgroup$
    – Tyrone
    Feb 22 at 16:05


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