TITLE: 3-manifolds homotopy equivalent to a surface
QUESTION [6 upvotes]: 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 ?

REPLY [2 votes]: This paper gives some other very nice counter-examples. As I recall, for open manifolds you really need to think about proper homotopies, not just homotopies.
MR1033220 (91b:57021) Reviewed 
Scott, Peter(1-MI); Tucker, Thomas(1-COLG)
Some examples of exotic noncompact 3-manifolds. 
Quart. J. Math. Oxford Ser. (2) 40 (1989), no. 160, 481–499. 
57N10 (57M10)