I have a question about the action-angle theorem on p. 283 in Arnold's textbook on classical mechanics.

The theorem says that the transformation $(p,q) \mapsto (I,\phi)$ is symplectic, where $I$ are the action variables and $\phi$ the action angles.

He says that he only sketches the proof which might be the source of confusion.

I will state the proof and explain what exactly causes the troubles.

So first we consider the $1$-form $pdq$ on the manifold $M_f:=\{(p_1,..,p_n,q_1,..,q_n)=:(p,q) \in M; F_1(p,q)=f_1,...,F_n(p,q)=f_n\}$ where $F_1,..,F_n$ have linearly independent derivatives and $M$ is a symplectic manifold of dimension $2n$.

addendum: It can be shown that $\omega|_{M_f} = 0$ and he also assumed that $\frac{\partial I}{\partial f}|_{M_f}$ is invertible in a previous proof.

Therefore, $S(x)= \int_{x_0}^{x} pdq|_{M_f}$ is invariant under deformations of paths $(x_0 \rightarrow x)$ (by Stokes' theorem).

addendum: It can be shown that if $M_f$ is connected and compact it is diffeomorphic to a torus.

Still, $S$ is multiple-valued as when we integrate around one circle $\gamma_i$ of this torus, we get a period $\Delta_i (S)= \int_{\gamma_i} dS = 2 \pi I_i. $

Now he continues by saying: Let $x_0$ be a point on $M_f$, in a neighbourhood of which the $n$ variables $q$ are coordinates of $M_f$ such that the submanifold $M_f \subset \mathbb{R}^{2n}$ is given by $n$-equations of the form $p= p(I,q)$, $q(x_0)=q.$

In a simply connected neighborhood of the point $q_0$ a single-valued function is defined

$S(I,q) = \int_{q}^{q} p(I,q) dq.$

$\textbf{Question:}$ Now my question is: Why is it possible to take $(I,q)$ as coordinates, i.e. what is the argument that explains why the coordinates $q$ can be taken as local coordinates (and Arnold actually says that it is easy to conclude that this also holds true in the "large " around $M_f$) in a nbh of $M_f$ with center $x_0$?