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...](https://math.stackexchange.com/questions/734107/a-3-manifold-with-fundamental-group-isomorphic-to-a-surface-group) ), 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]



Noncompact three-manifold with fundamental group isomorphic to a surface group
I'll restate the question for the convenience of the reader.$\newcommand{\RR}{\mathbb{R}}$



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> 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 $F \times \RR$?
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The answer is "no". We can build a counterexample using the paper [Some examples of exotic non-compact three-manifolds](https://academic.oup.com/qjmath/article-abstract/40/4/481/1531319) 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.


