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Let $M$ and $N$ be two connected, smooth (possibly non-compact), aspherical manifolds (that is, $\pi_k(-)=1$ for $k\geq 2$) of dimensions $n$ and $n-r$ respectively. Let $f:M\to N$ be a smooth map inducing a surjective homomorphism $f_*:\pi_1(M)\to \pi_1(N)$. What can we say about the kernel of $f_*$? Is the cohomological dimension of the kernel $r$?

Edit: Thanks to user19232801 for answering my question. Now, I would like to correct/update it to what exactly I wanted. Assume that the cohomological dimensions of $\pi_1(M)$ and $\pi_1(N)$ are $n$ and $n-r$, respectively, $0<r<n$.

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  • $\begingroup$ 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. $\endgroup$
    – RKS
    Commented Oct 18, 2023 at 9:33
  • $\begingroup$ But $f|_{M-f^{-1}(0)}:M-f^{-1}(0)\to N-\{0\}$ is a fibration. $\endgroup$
    – RKS
    Commented Oct 18, 2023 at 10:02

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Take $n = 2$,$r = 0$ and $M = \mathbb R^2 - \{0,1\}$ and $N = \mathbb R^2 - \{0\}$ and $M \to N$ to be the natural inclusion.

Then on fundamental groups we have the surjection $F_2 \to \mathbb Z$ taking one generator to $1$ and the other to $0$. The kernel is a nontrivial subgroup of $F_2$, hence free and of cohomlogical dimension $1 \neq 0$.

(Note that if $M,N$ were compact and $f$ is a submersion, then by Ehresmann the fiber of $M \to N$ would be a classifying space for the kernel-- hence the kernel would indeed have the desired cohomological dimension.)

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  • $\begingroup$ 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 $r > 0$. $\endgroup$
    – RKS
    Commented Oct 17, 2023 at 18:24
  • $\begingroup$ Also, in the compact case, does surjectivity of $f_*$ imply submersion of $f$? $\endgroup$
    – RKS
    Commented Oct 17, 2023 at 18:25
  • $\begingroup$ @RKS Sorry, I'm too used to AG where smooth means submersion. (I've edited for correctness). Do you want an example with $r> 0$? $\endgroup$
    – user19232801
    Commented Oct 17, 2023 at 19:00
  • $\begingroup$ Yes, if you kindly give some idea. I made an edited hypothesis. $\endgroup$
    – RKS
    Commented Oct 18, 2023 at 5:17
  • $\begingroup$ @RKS You could take $(\mathbb R^2 - \{0,1\}) \times K \to \mathbb R^2 - 0$ where $K$ is any aspherical manifold of cohomlogical dimension $k \geq 1$ and the map is projection onto the first factor. Then the cohomological dimension of the source is $k+1$, the cohomological dimension of the target is $1$, and the cohomological dimension of the kernel is again $k + 1$. $\endgroup$
    – user19232801
    Commented Oct 18, 2023 at 14:06

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