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?
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?
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 $S^1$.