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Let $Y^3$ be a closed 3-manifold with $\pi_1(Y)=\mathbb{Z}$.

Is it true that $Y$ is homeomorphic to $S^1\times S^2$ from the prime decomposition of 3-manifold?

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It is possible to classify all compact 3-manifolds with free fundamental group, and in this case, if $Y^3$ is closed, it is an $S^2$ bundle over $S^1$, so either $S^2\times S^1$ or the non-orientable version. If it has boundary, It can also be the solid torus or solid Klein bottle (and any punctured version of the above four). – user641 Jun 25 '11 at 16:55
up vote 2 down vote accepted

I am pretty sure the answer is yes if it is orientable, as $S^1 \times S^2$ the only orientable prime 3-manifold with $\pi_1(M)=\mathbb{Z}$. If you are uncertain to it's orientability things get more complicated, as it could be equivalent to a non-orientable fiber of $S^2$ over an $S^1$.

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True in the "boundary-less" case. – user641 Jun 25 '11 at 16:56

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