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The open ball is a manifold, and the closed ball is a compact manifold-with-boundary which extends the open ball, in which the open ball is dense, in which all the new points are boundary points. Is it the only one? can every manifold be compactified with a boundary?

What if we allow other spaces, like a closed polyhedron? There is a morphism (of locally ringed spaces) from a closed convex polyhedron to a closed ball, which is a homeomorphism on the topological spaces. What other non-smooth compactifications-by-boundary can we expect to smooth-out like this?

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If a manifold can be compactified with a boundary, then the original manifold and its compactification are homotopy equivalent, and the fundamental group is finitely generated. So the infinite ladder surface, with nonfinitely generated fundamental group, is a counterexample. –  Lee Mosher Jun 19 '12 at 21:35
You might want to peruse mathoverflow.net/questions/22441/… for some other answers. –  Lee Mosher Jun 19 '12 at 21:38
I think you will need to make your question more precise, or risk closure. For instance, you seem to be suggesting that every closed polyhedron is homeomorphic to a closed ball, which is patently false. –  Mark Grant Jun 19 '12 at 21:39
mark: oops. convex polyhedron. –  Larry D'Anna Jun 19 '12 at 22:25
lee: thanks! It looks like the other question is exactly the discussion I was looking for. –  Larry D'Anna Jun 19 '12 at 23:21
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Turning Lee's comment into an answer:

In the first paragraph you are asking if all manifolds are "tame" (http://en.wikipedia.org/wiki/Tame_manifold). The answer is "no" and there are even simply connected examples, due to Whitehead. The second half of your question is very different, and seems to be asking about smoothing corners.

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