TITLE: the criteria for 3-dim manifolds diffeomorphic to $\mathbb{R}^3$
QUESTION [15 upvotes]: In Schoen and Yau's paper "Complete three-dimensional manifolds with positive Ricci curvature and scalar curvature", they mentioned “If $M^3$ is contractible, to prove it is diffeomorphic to $\mathbb{R}^3$, from a topological result by Stallings "Group Theory and Three DimensionalManifolds", it suffices to prove that $M^3$ is simply connected at infinity and irreducible.” 
I checked Stallings’ book, did not find the exact result they referred to. Can anyone tell me which theorem in Stallings’ book they referred to or any other place to find the detailed proof of the above statement? Thanks.

REPLY [10 votes]: Husch and Price is the right reference, but I thought I could give a quick sketch of a proof. 
Let $X^3$ 
be simply connected at infinity and contractible. For every compact
$C \subset X$, there exists a compact $D$ containing $C$ such
that $π_1(X-D)\to π_1(X-C)$ is trivial. Taking a regular
neighborhood, we may assume that $D$ is a compact
submanifold with boundary. Compress $∂D$ as much
as possible in the complement of $C$ to obtain a 
compact incompressible surface $S \subset X-C$. If $S$ is not a union of 
2-spheres, then each non-trivial component is
$π_1$-injective in $X-C$ by the loop theorem. 
But the fundamental group of this surface is in
the image of $π_1(∂D) \to π_1(X-C)$, which is
trivial, a contradiction. 
            Thus, $S$ must consist of a collection of
2-spheres. Since $X$ is contractible, each such
sphere bounds a contractible submanifold, 
which is a 3-ball by the Poincaré conjecture
(and I suppose van Kampen's theorem). 
Hence $C$ is contained in a 3-ball. Thus, $X$
is a nested union of 3-balls, and is therefore homeomorphic
to $R^3$.