TITLE: What closed 3-manifolds have fundamental group $\Bbb Z$?
QUESTION [24 upvotes]: For certain small groups, it is easy (and desirable) to classify closed (and orientable if necessary) 3-manifolds with that group as their fundamental group. (Essentially due to Waldhausen is that for "large" 3-manifold groups, indecomposable under free product, the 3-manifold is determined up to homeomorphism by its fundamental group, though this was only proved in full generality after geometrization. Large means, precisely, infinite and not $\Bbb Z$).
For finite groups, as a corollary of elliptization, $\Bbb Z/p$ only appears as the fundamental group of lens spaces $L(p,q)$, and one can classify the other finite fundamental groups (before elliptization this was harder and one could only classify up to connected sum with a homotopy sphere.) as having either at most two manifolds per group (and precisely what they are).
$\Bbb Z$ is the simplest infinite group I know. What's the classification of closed 3-manifolds with fundamental group $\Bbb Z$?

REPLY [2 votes]: In fact something significantly stronger is true. If $Y$ is closed orientable and $\Bbb Z$ injects into  the abelianization $\pi_1(Y)$, either $Y$ has an $S^1 \times S^2$ factor or $Y$ has a $\pi_1$-injective surface of genus at least one (An injection $\pi_1(\Sigma_g) \to \pi_1(Y^3)$ for $\Sigma_g$ a genus $g$ surface with $g\geq 1$ induced from an embedding $\Sigma_g \to Y$). 
Consider a prime factor of $Y$, $Y'$ with $\Bbb Z$ injecting into the abelianization of $\pi_1(Y')$. By Poincare duality $H_2(Y')\ne 0$, and by a very famous result of Gabai there exists a taut foliation on $Y'$ with a closed leaf $\Sigma$. By a theorem of Novikov, the induced map $\pi_1(\Sigma) \to \pi_1(Y')$ is injective and by another theorem of Novikov if $\Sigma$ is $S^2$ if and only if $Y'$ is $S^1\times S^2$.