TITLE: Noncompact three-manifold with fundamental group isomorphic to a surface group
QUESTION [9 upvotes]: Let $M$ be an open orientable three-manifold such that $\pi_1 (M)$ is isomorphic to the fundamental group of a closed orientable surface $S\ncong \mathbb{S}^2$. Furthermore, suppose that $\tilde{M} \cong \mathbb{R}^3$. Is it true that $M \cong S \times \mathbb{R}$?
Without making any assumptions about $\tilde{M}$, there are counterexamples to the question, (see for instance... ), but in those cases the universal cover is not very nice.
[As remarked by several commentators, you need $S$ to be orientable.  I added this hypothesis and fixed some of the notation. - Sam]

REPLY [8 votes]: I'll restate the question for the convenience of the reader.$\newcommand{\RR}{\mathbb{R}}$

Suppose that $M$ is a non-compact, connected, oriented three-manifold without boundary, with universal cover homeomorphic to $\RR^3$. Suppose that $F$ is a compact, connected, oriented surface with genus at least one.  Suppose that the fundamental groups of $M$ and $F$ are isomorphic. Is $M$ homeomorphic to $F \times \RR$?

The answer is "no". We can build a counterexample using the paper Some examples of exotic non-compact three-manifolds by Scott and Tucker.  Here I closely follow their wording and notation, starting at the bottom of page 488.
Consider their example manifold $M_5$.  This is a three-manifold with boundary a closed surface $F$.  The universal cover of $M_5$ is homeomorphic to $\RR^2 \times \RR_{\geq 0}$, and the inclusion of $F$ into $M_5$ is a homotopy equivalence.  However, $M_5$ is not "almost compact" (that is, it is not tame), and $M_5$ is not homeomorphic to $F \times \RR_{\geq 0}$.
Now double $M_5$ across its boundary to obtain $M = D(M_5)$.  This does not change the fundamental group.  Note that $M$ has two ends, separated by (the image of) $F$.  By Scott-Tucker, neither end is tame.  Thus $M$ is not homeomorphic to $F \times \RR$.  However, the universal cover of the double is (in this case) the double of the universal cover; thus it is $\RR^2 \times \RR$, as desired.