# Are homeomorphic differentiable manifolds actually diffeomorphic?

Let $M$ and $N$ be two n-dimensional smooth manifolds.Suppose their underlying topological spaces are homeomorphic through $f$. Does $f$ automatically become a diffeomorphism with respect to the given smooth structures? If not, can I adjust any of the smooth structures to make $f$ a diffeomorphism? What if I restrict the manifolds to be embedded manifolds in Euclidean space $\mathbb{E}^n$ with endowed topology and smooth structure?

• You might find this interesting: en.wikipedia.org/wiki/… – Brian Fitzpatrick Apr 21 '14 at 4:47
• In high dimensions, adjustment may not be possible. A summary of what is currently known can be found in this answer for a related question on MO. – achille hui Apr 21 '14 at 7:31
• The answer is no. Take exotic 7-spheres. – annie marie cœur May 22 at 15:50

Let $M=N=\mathbb R$, endowed with its usual structure of a smooth manifold, and let$f:M\to N$ be the map such that $f(x)=x^3$. This is a homeomorphism but not a diffeo!

Now, if $f:M\to N$ is a homeomorphism of smooth manifolds, you can always «adjust» the smooth structures so that $f$ becomes a diffeo: indeed, you should be able to prove the following:

if $M$ is a smooth manifold, $N$ a topological space and $f:M\to N$ a homeomorphism, then there is a structure of smooth manifold on $N$ such that $f$ becomes a diffeomorphism.

• In your second and third paragraphs, are you still assuming that $M = N = \mathbb{R}$ or are you now allowing them to be arbitrary smooth manifolds, as in the OP? – tparker Aug 21 '18 at 16:21

In higher dimensions there exist exotic spheres which are smooth manifolds that are homeomorphic but not diffeomorphic to the $n$-sphere.

If by "adjust any of the smooth structure" means throw away the smooth structure on one of the manifolds and install a new unrelated one, then obviously you can use $f$ to transfer the smooth structure of one manifold to another¸ and $f$ then trivially becomes diffeomorphism. But arguably the manifold whose structure you "adjusted" wont't be the same smooth manifold anymore.
If I'm understanding things correctly, the exotic spheres can even be embedded smoothly in $\mathbb R^m$ for sufficiently large $m$.
• Yes, you are understanding things correctly: Whitney proved that any connected (Hausdorff and second countable, of course) differential manifold of dimension $n$ can be embedded smoothly into $\mathbb R^{2n}$ . – Georges Elencwajg Apr 21 '14 at 7:52