Let $\phi: O_1 \subset \mathbb{R}^3 \to O_2\subset \mathbb{R}^3$ be a diffeomorphism and $S$ be a surface. Then $\phi:S \to \phi(S)$ is a diffeo. I am having some trouble with the following problem (Exercise 2.45 of the book Curves and Surfaces, by Montiel and Ros):
Let $\phi: O_1 \subset \mathbb{R}^3 \to O_2 \subset \mathbb{R}^3$ be a diffeomorphism and $S \subset O_1$ be a surface. Then $\phi:S \to \phi(S)$ is a diffeomorphism, where $O_1$ and $O_2$ are Euclidean open sets.
We know that $\phi(S)$ is another surface, so differentiability of $\phi:S \to \phi(S)$ should mean in the following sense:

Definition: Let $S$ be a surface and let $O \subset \mathbb{R}^n$ be an open set. A: A function $f:S \to \mathbb{R}^m$ is differentiable if for any parametrization $X:U \to S$ the composition is differentiable.
C: If $S_1$ is another surface, then $f:S \to S_1$ is differentiable if the map $i \circ f:S \to \mathbb{R}^3$ is differentiable in the sense of A.

I suppose we should work with the parametrizations, but I am stuck on how to write a proper proof.
Any hints or comments will be the most appreciated. Thanks in advance.
A: The formal details involve taking the usual sense of "diffeomorphism" between open sets and fitting it to the more gentle definition to manifolds. In order to prove $\phi|_S$ is a diffeomorphism, we need to show the following
1) $\phi|_S$ is differentiable: Let $\psi:U\rightarrow S$ be a paremeterzation of $S$. The composition $\phi \circ \psi: U\rightarrow O_2$ is differentiable in the "usual sense" of open sets due to the chain rule, as both $\phi,\psi$ are differentiable on open sets. This now follows from definitions A,B.
2) $\phi|_S$ is invertible with differentiable inverse: We know $\phi |_{O_1}$ is bijective, which must mean $\phi|_S$ is also bijective as a restriction. We know there exists $\phi^{-1}:\phi(O_1) \rightarrow O_1$ differentiable (between open sets). Let $\varphi$ be a parameterization of $\phi(S)$. Just like in (1), $\phi^{-1}\circ \varphi$ is differentiable and therefore from defition A, $\phi^{-1}$ is differentiable in the "surface sense"
I don't know how diffeomorphisms are defined in this book, so you might have to replace every "differentiable" in my answer with $C^k$.
Another note: I should point out $\phi(O_1)$ is open because $\text{rank}D\phi = 3$ which by the open mapping theorem, means $\phi$ is an open mapping. (This can be also proved easily by the inverse function theorem though. Let me know if you need me to elaborate and fill in the details!).
