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Sam Nead
Sam Nead
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Your question ("is $M$ homeomorphic to $T_f$?") is answered in the affirmative by Theorem 2 of Stallings' paper *On fibering certain 3-manifolds". You will also need his Theorem 1. Here are the statements (slightly simplified).

Theorem 1: Suppose that $M$ is a compact connected three-manifold. Suppose that $\Gamma$ is a finitely generated normal subgroup of $\pi_1(M)$ whose quotient group is $\mathbb{Z}$. Then there is a surface $F$ properly embedded in $M$ so that $\Gamma = \pi_1(F)$.

Theorem 2: With hypotheses as in Theorem 1. Suppose that $M$ is irreducible. Suppose that $\Gamma$ is not $\mathbb{Z}/2\mathbb{Z}$. Then $M$ is a surface bundle over the circle, with $F$ isotopic to a fibre.

AlsoThat is, your hypothesis on the short exact sequence (plus Theorem 1) gives the surface $F$. Your hypothesis that the manifold $M$ is hyperbolic then gives the additional hypotheses of Theorem 2.

Note that Stallings does not cite Waldhausen. Consulting my memory,I suppose that this is because oncehis situation is a very very simple case of a Haken hierarchy. Once you have $F$ in your hands (and all the group theory hypotheses), it is "easy" to show that $M - F$ is homeomorphic to a product.

Your question ("is $M$ homeomorphic to $T_f$?") is answered in the affirmative by Theorem 2 of Stallings' paper *On fibering certain 3-manifolds". You will also need his Theorem 1. Here are the statements (slightly simplified).

Theorem 1: Suppose that $M$ is a compact connected three-manifold. Suppose that $\Gamma$ is a finitely generated normal subgroup of $\pi_1(M)$ whose quotient group is $\mathbb{Z}$. Then there is a surface $F$ properly embedded in $M$ so that $\Gamma = \pi_1(F)$.

Theorem 2: With hypotheses as in Theorem 1. Suppose that $M$ is irreducible. Suppose that $\Gamma$ is not $\mathbb{Z}/2\mathbb{Z}$. Then $M$ is a surface bundle over the circle, with $F$ isotopic to a fibre.

Also, Stallings does not cite Waldhausen. Consulting my memory, this is because once you have $F$ in your hands (and all the group theory hypotheses), it is "easy" to show that $M - F$ is homeomorphic to a product.

Your question ("is $M$ homeomorphic to $T_f$?") is answered in the affirmative by Theorem 2 of Stallings' paper *On fibering certain 3-manifolds". You will also need his Theorem 1. Here are the statements (slightly simplified).

Theorem 1: Suppose that $M$ is a compact connected three-manifold. Suppose that $\Gamma$ is a finitely generated normal subgroup of $\pi_1(M)$ whose quotient group is $\mathbb{Z}$. Then there is a surface $F$ properly embedded in $M$ so that $\Gamma = \pi_1(F)$.

Theorem 2: With hypotheses as in Theorem 1. Suppose that $M$ is irreducible. Suppose that $\Gamma$ is not $\mathbb{Z}/2\mathbb{Z}$. Then $M$ is a surface bundle over the circle, with $F$ isotopic to a fibre.

That is, your hypothesis on the short exact sequence (plus Theorem 1) gives the surface $F$. Your hypothesis that the manifold $M$ is hyperbolic then gives the additional hypotheses of Theorem 2.

Note that Stallings does not cite Waldhausen. I suppose that this is because his situation is a very very simple case of a Haken hierarchy. Once you have $F$ in your hands (and all the group theory hypotheses), it is "easy" to show that $M - F$ is homeomorphic to a product.

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Sam Nead
Sam Nead
  • 28.1k
  • 5
  • 72
  • 131

Your question ("is $M$ homeomorphic to $T_f$?") is answered in the affirmative by Theorem 2 of Stallings' paper *On fibering certain 3-manifolds". You will also need his Theorem 1. Here are the statements (slightly simplified).

Theorem 1: Suppose that $M$ is a compact connected three-manifold. Suppose that $\Gamma$ is a finitely generated normal subgroup of $\pi_1(M)$ whose quotient group is $\mathbb{Z}$. Then there is a surface $F$ properly embedded in $M$ so that $\Gamma = \pi_1(F)$.

Theorem 2: With hypotheses as in Theorem 1. Suppose that $M$ is irreducible. Suppose that $\Gamma$ is not $\mathbb{Z}/2\mathbb{Z}$. Then $M$ is a surface bundle over the circle, with $F$ isotopic to a fibre.

Also, Stallings does not cite Waldhausen. Consulting my memory, this is because once you have $F$ in your hands (and all the group theory hypotheses), it is "easy" to show that $M - F$ is homeomorphic to a product.