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A fibration of a manifold $\phi: M \to S^1$ gives rise to a short exact sequence $$ 1 \to \pi_1(N) \to \pi_1(M) = \mathbb{Z} \overset{f_\ast}{\to} 1 $$ where $N$ is the fiber.

I've heard that, if $M$ is a hyperbolic 3-manifold, $\pi_1(N)$ is finitely generated or even finitely presented. Is that true and if so, why?


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  • $\begingroup$ You do not need hyperbolicity but you should assume compactness. The question itself is more appropriate for Math Stack Exchange. $\endgroup$ Commented Jul 12, 2023 at 18:29
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    $\begingroup$ Originally posted 1 day ago on math.stackexchange math.stackexchange.com/questions/4734180/… $\endgroup$
    – Lee Mosher
    Commented Jul 12, 2023 at 19:23

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This isn't true. Here's one easy construction. For some $g \geq 2$, let $\Sigma_g$ be a closed oriented genus $g$ surface and let $f\colon \Sigma_g \rightarrow \Sigma_g$ be a homeomorphism representing a pseudo-Anosov mapping class that acts trivially on $H_1(\Sigma_g)$. There are many constructions of such pseudo-Anosov mapping classes (for example, this paper shows that suitable "random" mapping classes acting trivially on homology are pseudo-Anosov). For simplicity, isotope $f$ such that it fixes a basepoint.

By Thurston's hyperbolization theorem, the mapping torus $M_f$ is hyperbolic. Now let $S$ be the universal abelian cover of $\Sigma_g$. Using the fixed basepoint of $f$, you can lift $f$ to a homeomorphism $\tilde{f}\colon S \rightarrow S$. The mapping torus $M_{\tilde{f}}$ is then a cover of $M_f$, so it is also hyperbolic. The associated short exact sequence is $$1 \longrightarrow \pi_1(S) \longrightarrow \pi_1(M_{\tilde{f}}) \longrightarrow \mathbb{Z} \longrightarrow 1,$$ and $\pi_1(S)$ is an infinite rank free group.

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You are perhaps thinking of Stallings fibration theorem, contained in the paper On fibering certain 3-manifolds. You can find a short discussion of the paper on his Wikipedia page.

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