# Submersion vs fiber bundle

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.

• 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. Feb 22 at 16:05