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This is a basic issue of PL topology that I assume must be true, but I can't find a written reference: is a convex open PL subset of $\mathbb R^n$ PL homeomorphic to $\mathbb R^n$? I've scanned through Rourke-Sanderson, Hudson, Zeeman, and Stallings, but can't quite find it in there. I'd appreciate a quotable reference if possible.

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I do not know a reference, but it seems to be easy to prove. –  Anton Petrunin Aug 9 '12 at 23:42
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This can be proven by the "standard mistake" in PL topology (radial projection to the simplex). Standard mistake argument is, I think, in Rourke-Sanderson. Incidentally, you do not need the boundary to be PL since every convex solid is a limit of convex polytopes and so you can view your solid as a union of a convex polytope and infinitely many polyhedral annuli. –  Misha Aug 10 '12 at 6:51
    
@Misha: why is this called the standard mistake? –  Igor Rivin Aug 10 '12 at 15:29
    
@Igor: Because everybody makes it at first, I guess. It is also easily correctable (after you take a subdivision the projection induces an isomorphism of simplicial complexes). –  Misha Aug 24 '12 at 9:15
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