I have heard that an open, orientable 3-manifold $X$ (non-compact, without-boundary) that is homotopy equivalent to an orientable surface $S\_g$ must itself already be homemorphic to $S\_g \times \mathbb R$. There seems to be a very deep theorem behind it, however, I couldn't find any reference which would bring this up. Does anybody know a good reference for this ?


*Edit*: The original question has been answered well. I wonder in how far the assumptions have to be strengthened in order for this to be true. The *Tameness Theorem* shows that $X = S\_g \times \mathbb R$ whenever $X$ additionally admits a complete hyperbolic metric. However, for my purposes, it would be nice to have purely topological assumptions. I have the following suggestions:


$X$ is as above, but additionally we require $X$ to have two ends, such that for each end, there exists a sequence of end-neighborhoods $U\_1 \supset U\_2 \supset U\_3 \supset ....$ with $\bigcap cl(U\_i) = \emptyset$ and the following properties:


* $U\_1$ is homotopy equivalent to $S\_g$
* the natural map $i\_\*: \pi\_1(U\_{n+1}) \to \pi\_1(U\_n)$ is an isomorphism for all $n$.


Intuitively, this will prevent $X$ from "going crazy" at one end. Is this enough now ?



3-manifolds homotopy equivalent to a surface
The answer (by studiosus) to [this question on MSE](https://math.stackexchange.com/questions/734107/a-3-manifold-with-fundamental-group-isomorphic-to-a-surface-group) gives references AND a counterexample!


