TITLE: When does homeomorphism imply diffeomorphism?
QUESTION [8 upvotes]: In $R^n$, suppose $U$ and $V$ and two homeomorphic open sets. Then, is $U$ diffeomorphic to $V$? 
If not, can we impose stronger conditions such that this true?

REPLY [6 votes]: Such examples exist already in dimension 4: Mike Freedman was the first to construct small exotic $R^4$, that is an open subset of the standard 4d space which is homeomorphic but not diffeomorphic to $R^4$. 
On the positive side, suppose that you have a manifold $M$ homotopy-equivalent to a finite CW-complex and $dim(M)\ne 4$, so that $H_3(M, Z_2)=0$. Then $M$ admits a unique PL (piecewise-linear) structure. A proof of this can be found in 
"The Hauptvermutung Book" or this Rudyak's paper. It is also in Kirby and Siebemnann's book, but it is virtually unreadable. Since in dimensions $\le 6$ the categories PL and DIFF are equivalent, it follows that for domains in $R^5$ and $R^6$, if 3rd homology vanishes then homeomorphism implies diffeomorphism. A similar statement holds in other dimensions, when you want to show that homeomorphic manifolds are diffeomorphic. However, the obstructions are more complicated, the sufficient condition is that if you have an $n$-manifold $M$ ($n\ne 4$) which is homotopy-equivalent to finite CW complex and $H_k(M)=0$ ($k\ge 3$) then $M$ admits a unique smooth structure. For this, the only reference I know is Kirby and Siebemnann's book, but, as I said, it is essentially unreadable (the result is not even stated there in this form, you have to slog your way through Essay IV to make this conclusion).