TITLE: Smooth bijection between non-diffeomorphic smooth manifolds?
QUESTION [29 upvotes]: The "textbook" example of a smooth bijection between smooth manifolds that is not a diffeomorphism is the map $\mathbb{R} \rightarrow \mathbb{R}$ sending $x \mapsto x^3$.  However, in this example, the source and target manifolds are diffeomorphic -- just not by the given map.  Is there an example of a smooth bijection $X \rightarrow Y$ of smooth manifolds where $X,Y$ are not diffeomorphic at all?  (and if so, what?)
(For instance, is it possible to arrange a smooth bijection from a sphere to an exotic sphere, failing to be a diffeomorphism because of the existence of critical points? or do homeomorphisms between different smooth structures on spheres fail to be everywhere smooth in some catastrophic way?)

REPLY [22 votes]: Every smooth manifold has a smooth triangulation, which yields a pseudofunctor from the category of smooth manifolds to the category of PL manifolds.  (There is no actual functor; that would be crazy.)  If two smooth manifolds are PL isomorphic, then the answer is yes.  You can start with the PL isomorphism, and then build a homeomorphism that follows it and that has the property that all derivatives vanish in all directions perpendicular to every simplex.  You can build the homeomorphism by induction from the $k$-skeleton to the $(k+1)$-skeleton using bump functions.
The PL Poincaré conjecture is true in dimensions other than 4, so all exotic spheres in the same dimension $n \ge 5$ are PL homeomorphic.  (High-dimensional examples of exotic spheres start in dimension 7, it was calculated.)  In dimension 4, by contrast, every PL manifold has a unique smooth structure, and it is not known whether there are any exotic spheres.
On the other hand, if the manifolds are homeomorphic but not even PL homeomorphic, then I don't know.  It is known that every manifold of dimension $n \ge 5$ has a unique Lipschitz structure, but I do not know a Lipschitz version of the above argument.  On the positive side, passing from smooth to Lipschitz is an actual functor, so the answer to a modified question, is there a Lipschitz-smooth homeomorphism, is yes, and you can even make it bi-Lipschitz.