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I'm starting studying differential topology and I somehow arrived to the following (which I feel is wrong).

If $M,N$ are $C^r$ manifolds and $f:M\to N$ is a $C^r$ diffeomorphism then $f$ is actually a $C^\infty$ diffeomorphism (*) ($r>0$)

I couldn't find an error in my proof but the statement looks wrong because this would imply that a differentiable function $f:V\to W$ between open subsets of $\mathbb{R}$ with continuous derivative actually has all derivatives and this doesn't look right.

So the question is:

Is (*) right or is there a counterexample?

The definition of $C^\infty$ we are using is: A $C^\infty$ map is a function $f:M\to N$ such that for each $p\in M$ there is an atlas $(U,x)$ of $M$, an atlas $(V,y)$ of $N$ such that $p\in U$, $f(p)\in V$ and $y\circ f\circ x^{-1}$ is a $C^\infty$ map (between open subsets of some $\mathbb{R}^n$). A $C^\infty$ diffeomorphism is a bijective function $f:M\to N$ such that both $f$ and $f^{-1}$ are $C^\infty$ maps.

Zero
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    I wonder if this is related to the fact that $C^r$ atlases can always be enlarged to a $C^{\infty}$ atlas. – Justin Benfield Aug 07 '16 at 23:14
  • ^ I got that from here, after bullet pointed list: https://en.wikipedia.org/wiki/Differentiable_manifold – Justin Benfield Aug 07 '16 at 23:19
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    You probably need to specify exactly what you mean by a $C^\infty$ diffeomorphism between $C^r$ manifolds - how do you define $k$-times differentiable for $k>r$ in a way that's not chart-dependent? Regardless, I expect $f : \Bbb R \to \Bbb R : x \mapsto |x|x$ is a counterexample. – Anthony Carapetis Aug 07 '16 at 23:23
  • How about, say, a homeomorphism between two non-diffeomorphic exotic spheres? – Neal Aug 07 '16 at 23:25
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    @JustinBenfield Restricted to a $C^\infty$ atlas, not enlarged. –  Aug 07 '16 at 23:46
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    @AnthonyCarapetis I just realized that when $k>r$, this definition of $C^\infty$ diffeomorphism becomes chart-dependent and somehow isn't equivalent to the usual notion of $C^\infty$ differentiabilty between open subsets of $\mathbb{R}^n$. You could put this into an answer together with your example :) – Zero Aug 07 '16 at 23:46
  • @Neal I assume here what's meant is $r>0$. It is true that every $C^r$ manifold, $r>0$, has a unique compatible $C^\infty$ structure up to diffeomorphism (which of course fails for $r=0$, as you point out). But it looks like the question has now been resolved. –  Aug 07 '16 at 23:48
  • @MikeMiller, it was resolved in a strange way :-| – Mariano Suárez-Álvarez Aug 08 '16 at 08:21

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If your manifolds are only endowed with $C^r$ atlases, it doesn't make sense to talk about derivatives of higher order than $r$ - since you can't differentiate the transition maps enough, different charts will not necessarily agree on these derivatives.

If you require $M,N$ to be $C^\infty$ so that the question makes more sense, the answer is in the negative. A simple counterexample is the $C^1$ diffeomorphism $f:\mathbb R \to \mathbb R$ given by $f(x) = x(|x|+1)$, which has continuous derivative $f(x)=2|x|+1$ but is not even $C^2$.

Anthony Carapetis
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