TITLE: Is every Compact $n$-Manifold a Compactification of $\mathbb{R}^n$?
QUESTION [30 upvotes]: I read the result that every compact $n$-manifold is
a compactification of $\mathbb{R}^n$.
Now, for surfaces, this seems clear: we take
an n-gon, whose interior (i.e., everything in
the n-gon except for the edges) is homeo. to $\mathbb{R}^n$,
and then we identify the edges to end up with
a surface that is closed and bounded.
We can do something similar with the $S^n$'s ; by
using a "1-gon" (an n-disk), and identifying
the boundary to a point. Or we can just use the
stereo projection to show that $S^n-\{{\rm pt}\}\sim\mathbb{R}^n$;
$S^n$ being compact (as the Alexandroff 1-pt. -compactification
of $\mathbb{R}^n$, i.e., usual open sets + complements of compact subsets of
$\mathbb{R}^n$). And then some messy work helps
us show that $\mathbb{R}^n$ is densely embedded in $S^n$. 
But I don't see how we can generalize this statement
for any compact n-manifold. Can someone suggest any
ideas?
Thanks in Advance.

REPLY [5 votes]: Here is the answer in the context of topological manifolds:

For smooth manifolds you already have completely satisfactory answers.
If a compact topological manifold $M$ has dimension $n\ne 4$ then it admits a handlebody decomposition (see the discussion and references here). Once you have a handlebody decomposition $M=H_1\cup_S H_2$, you use the fact that the complements to some $n-1$-dimensional disjoint disks $\Delta_{1,j}, \Delta_{2j}$ in the interiors of $H_1, H_2$ are homeomorphic to $R^n$. Now, find an open $n-1$-disk $D\subset S$ disjoint from the boundaries of the disks disks $\Delta_{1,j}, \Delta_{2j}$. Then the union
$$
int(H_1)\cup D \cup int(H_2)
$$
is homeomorphic to $R^n$. This union is clearly dense in $M$. 
There are 4-dimensional manifolds which do not admit a handle decomposition, but Frank Quinn in

Ends of Maps III: Dimensions 3 and 4, Journal of Differential Geometry vol. 17 (1982) 
proved that all noncompact 4-dimensional manifolds are smoothable. Hence, removing a point from such a manifold results in a smoothable manifold which 
admits an open and dense subset homeomorphic to $R^4$. 
The conclusion is then that every compact topological $n$-manifold is homeomorphic to a compactification of $R^n$.
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