But $f|_{M-f^{-1}(0)}:M-f^{-1}(0)\to N-\{0\}$ is a fibration.

I have a concrete example. Let $N={\Bbb C}-\{\pm 1\}$, and ${\Bbb Z}_2$ is acting on $N$ in the obvious way (rotation). Consider $M=\{(a_1,a_2)\in N^2\ |\ {\Bbb Z}_2a_1\neq {\Bbb Z}_2a_2\}$ and let $f:M\to N$ be the first projection. I know $f_*$ is surjective. Is the kernel of $f_*$ free? Note that $f$ is not a fibration.

Yes, if you kindly give some idea. I made an edited hypothesis.

Also, in the compact case, does surjectivity of $f_*$ imply submersion of $f$?

Thank you! I was about to edit my question. I made a mistake. I meant cohomological dimension of $\pi_1(M)$ and $\pi_1(N)$ are $n$ and $n-r$, respectively with $r > 0$.

Yes, I see it now.

Oh! It is that simple. I was thinking in the real plane sense and thought that the line $xy=1$ will delete some part of $f^{-1}(V)$, away from $(0,0)$. Thank you!

Thanks a lot. Kindly share a reference for the cohomology non-vanishing result. Also I am curious to know if $0$ is the only troubling point. I mean if we did not add the point $(0,0)$, is the restriction a fibre bundle?

Ryan, if someone can edit his/her wrong answer then a question also can be edited for the discussion to continue in the desired direction.