Stallings' fibration theorem states that if we have a compact irreducible $3$-manifold $M^3,$ with $G\rightarrow \pi_1(M^3) \rightarrow \mathbb{Z},$ and $G$ is finitely generated and is not of order $2,$ then $M^3$ fibers over a circle. The question is whether the last condition (that $G\neq \mathbb{Z}/2\mathbb{Z}$) is actually necessary (the answer is probably in Stallings' original paper, but I can't find it online).
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asked Nov 15, 2011 at 18:01
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This should follow from geometrization. The fundamental group of the manifold is $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}$, and geometrization should tell you that the manifold is then $\mathrm{RP}^2 \times S^1$.
I looked in Stallings's paper, and he says that it is a hard open problem, so this might be the only way.
answered Nov 15, 2011 at 18:28
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$\begingroup$ That's what I was afraid of :( $\endgroup$ Commented Nov 15, 2011 at 20:04
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$\begingroup$ Without geometrization, you can show it is homotopy equivalent to $\mathrm{RP}^2 \times S^1$, but I agree that you probably need the full force of geometrization to get diffeomorphism. $\endgroup$– Steve DCommented Nov 16, 2011 at 0:49