I've read in a paper that if $M$ is a prime, reducible $3$-manifold, then $\pi_{1}(M) \cong \mathbb{Z}$. Can anyone explain why this is true?
Thanks in advance.
I've read in a paper that if $M$ is a prime, reducible $3$-manifold, then $\pi_{1}(M) \cong \mathbb{Z}$. Can anyone explain why this is true?
Thanks in advance.
I'm probably assuming $M$ to be orientable.
Reducible means that there is an essential sphere $S \subseteq M$. Two cases happen : either that sphere disconnects $M$, and $M$ is decomposable as a nontrivial connected sum (so it is not prime) or $M$ is homeomorphic to $S^2 \times S^1$.
So, $S^2 \times S^1$ is the only prime reducible $3$-manifold.