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I have a very concrete question. Let $N={\Bbb C}-\{\pm 1\}$, and ${\Bbb Z}_2$ is acting on $N$ by rotation around $0$. Consider $M=\{(x,y)\in N^2\ |\ {\Bbb Z}_2x\neq {\Bbb Z}_2y\}$. Let $p:M\to N$ be the first projection. I know the following.

  1. $f$ is not a fibration. But if we also remove $0$ from $N$, then the first projection is a fibration with fiber homeomorphic to ${\Bbb C}$ minus four points.

  2. $f_*:\pi_1(M)\to \pi_1(N)$ is a surjection.

  3. $\pi_1(M)$ is torsion free. In fact, $M$ is aspherical.

Question: What is ker$f_*$? Is it free? If not, what is its' cohomological dimension?

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  • $\begingroup$ Have you looked at arxiv.org/abs/2111.06159? Theorem 1.3 seems to apply here. $\endgroup$
    – Mark Grant
    Commented Oct 19, 2023 at 8:28
  • $\begingroup$ Thanks Mark! The paper you referred is under revision. My question is the simplest possible case. So, I am hoping for a direct argument. $\endgroup$
    – RKS
    Commented Oct 20, 2023 at 3:36

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