Does the inverse of a infinitely many times differentiable transition map also infinitely many times differentiable? If a transition map between two charts is infinitely many times differentiable, then is it true that its inverse is also infinitely many times differentiable?
My intuition is that it's true, but I have no idea how to prove it. 
If it is not true, can anyone give me a counterexample?
Any help will be appreciated, thanks!
 A: Without further assumptions, it's not true. For example, here are two charts for $\mathbb R$: the maps $\phi,\psi$ both have domain $\mathbb R$ and codomain $\mathbb R$, and they are given by the formulas
$$
\phi(x) = x, \qquad \psi(x) = x^3.
$$
(Note that both are homeomorphisms from $\mathbb R$ to $\mathbb R$.)
The transition map $\psi\circ\phi^{-1}$ is $x\mapsto x^3$, which is infinitely differentiable. But the inverse is $x\mapsto x^{1/3}$, which is not even once differentiable at the origin.
What's true is that if the transition map is infinitely differentiable and has nonsingular differential everywhere, then its inverse is infinitely differentiable. (The existence of the first derivative follows from the inverse function theorem, and the existence of higher derivatives from the chain rule and induction.) 
Alternatively, you can assume that the transition map is infinitely differentiable and its inverse is once differentiable everywhere, and then the existence of higher derivatives follows as above from the chain rule.
