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Take regular surfaces $M_1, M_2$ in $\mathbb{R}^3$ and suppose that we have the charts $$x: U \rightarrow M_1 \qquad y: U \rightarrow M_2.$$

Define the function $F=y\circ x^{-1}: x(U) \rightarrow y(U) $.

Why is $F$ a diffeomorphism between the surfaces $x(U)$ and $y(U)$?
Clearly $F$ is bijective, but I cannot see why $F$ is differentiable.

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Why do the charts have the same domain? That seems odd. – Dylan Moreland Feb 2 '12 at 20:12
It is an assumption. In my notes they say that $F$ is a diffeomorphism by definition. This should be trivial. We have that $x^{-1}$ is continuous bij definition and $y$ is differentiable by definition. – Nadori Feb 2 '12 at 20:20
A standard way to make your argument is to locally extend $x$ and $y$ to be diffeomorphisms between open subsets of $\mathbb R^3$ -- so you'll have to think of $U \equiv U \times \{0\} \subset \mathbb R^3$. You need the inverse function theorem to do this. Once you have the local extensions, your function $F$ is (locally) the composites of two diffeomorphisms. – Ryan Budney Feb 2 '12 at 20:53
up vote 3 down vote accepted

It's obvious: It is the exact purpose of charts $x: {\mathbb R}^2\to M_1$ and $y: {\mathbb R}^2\to M_2$ to enable the definition of differentiability: A map $f:\ M_1\to M_2$ is differentiable (regular, etc.) if the composition $$\psi:=y^{-1}\circ f\circ x:\quad {\mathbb R}^2\to{\mathbb R}^2\ ,$$ obtained by expressing $f$ in the local coordinates of $M_1$ and $M_2$, is differentiable (regular, etc.) as a map between ordinary cartesian spaces.

In your case $f$ is given as $f:=y\circ x^{-1}$ which implies $\psi={\rm id}_{{\mathbb R}^2}$, and this is certainly differentiable and regular.

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