1
$\begingroup$

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.

  • 1
    http://en.wikipedia.org/wiki/Prime_manifold#Prime_manifolds_and_irreducible_manifolds2012-07-10

1 Answers 1

2

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.

  • 2
    The same technique of proof works in the non-orientable case. Either way, take a loop intersecting this non-separating sphere once transversally, then an open neighborhood is either a (punctured) $S^2\times S^1$ or its nonorientable cousin.2012-07-10