## Smooth bijection between non-diffeomorphic smooth manifolds?

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?)

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I guess the word "manifold" forbids me from just rolling the half-open interval around the circle, and perhaps also from sticking continuum-many discrete points on the line.... – Theo Johnson-Freyd Oct 13 2010 at 8:00
Alternately, maybe you don't want to rest that much on the word "manifold", and instead mean to ask for smooth homeomorphisms that are not diffeomorphisms? – Theo Johnson-Freyd Oct 13 2010 at 8:01
Hi Theo, yes, "manifold" with (what I think of as) the default meaning -- second countable, without boundary, etc.... – D. Savitt Oct 13 2010 at 8:07
(and of course a smooth homeomorphism would be even better, but I'd be happy with just a smooth bijection) – D. Savitt Oct 13 2010 at 8:09
In the compact case, every continuous bijection is a homeomorphism. – Greg Kuperberg Oct 13 2010 at 8:36

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 PoincarĂ© 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.

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Perhaps a side issue, but I don't understand what you mean by pseudofunctor here. To me, a pseudofunctor is a type of map between 2-category or bicategory structures, but I am not aware of any interesting 2-category structure on the category of PL manifolds. – Todd Trimble Oct 13 2010 at 13:43
It looks like I used the wrong word; I just didn't check the definition. What I meant is a "mock functor", a partial functor defined for objects and isomorphisms that has no reasonable behavior for general morphisms. And even compositions of isomorphisms are somewhat problematic. – Greg Kuperberg Oct 13 2010 at 14:07
Nice. Thanks, Greg! – D. Savitt Oct 13 2010 at 14:26
Thanks for that response, Greg. – Todd Trimble Oct 13 2010 at 15:52