How do I prove the Frobenius Theorem?

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I am studying the basics of differential geometry, and in partiular I am focusing on the Frobenius theorem.

I think I have understood the concept, but I have not clear its demonstration.

I have seen that it is needed to prove both sufficiency and necessity, but I cannot understand necessity.

If it can hell, I am studying from the folloing book: Nonlinear control systems, from Alberto Isidori.

It is a really good book, but as I said, I am studying only the basics of differential geometry, so it is hard to understand some concepts.

In the proof of the Frobenius theorem, I am completely lost when I have to prove the sufficiency, where it is used the concept of flow of a vector field which I cannot understand.

To make my question complete I will try to expose what are my doubts:

in the proof of the book it starts by considering a distribution:

$\Delta (x)=span{f_1(x)...f_d(x)}$

and it is added an additional set of vectors:

$\Delta (x)=span{f_1(x)...f_d(x),f_{k+1}(x)....f_{n}(x)}=\mathbb{R}^{n}$

Now, it is considered the concept of flux of $\dot{x}=f(x)$, so $\Phi _{t}^{f}(x)$ solves $\dot{x}=f(x)$ with $x(0)=0$, and it is smooth

Since it is smooth, we can say:

$\frac{d}{dt}\Phi _{t}^{f}(x) =f(\Phi _{t}^{f}(x))$

with $\Phi _{0}^{f}(x)=x$ and $t$ small enough.

Now, if $x_0$ small enough, we have the following mapping:

$\Phi _{t}^{f}:x\mapsto\Phi _{t}^{f}(x)$

which is a local diffeomorphism.

At this point, it is considerend a composition of fluxes:

$\Phi _{t_1}^{f_1}(x)\circ ..... \circ \Phi _{t_n}^{f_n}(x)$

and I can consider a mapping:

$\Psi : U_\varepsilon \rightarrow \mathbb{R}^{n}$

$(z_1....z_n)\mapsto \Phi _{t_1}^{f_1}(x)\circ ..... \circ \Phi _{t_n}^{f_n}(x)$

and if $\varepsilon$ small enough:

1. List item $\Psi$ is a local diffeomorphism 2. any $z \in U_\varepsilon$ , the first $k$ columns of $\frac{d\Psi }{dz}$ are vecors of which are linearly independent $\Delta (\Psi (z))$

After that, it is said that if we consider $U^{0}$ is an image of $\Psi$ around $x_0$.

Now, from $**1.**$ it is said that we can deduce that it exists $\Psi^{-1}$ which is a smooth mapping on $x_0$. So, I consider:

$\begin{pmatrix} \Phi _1(x)\\ ....\\ \Phi _n(x) \end{pmatrix}=\Psi^{-1}$

which are the last $(n-k)$ functions which are the independent functions we were looking in $**2.**$ and also:

$\frac{d\Psi ^{-1}}{dx}|_{x=\Psi (x)} \cdot \frac{d\Psi }{dz}=I$

this is end of the demonstration.

Also, if you have the book, you can notice that it is a little different, since I have written a bit if what I think I have understood and a bit if what my professor wrote in his notes.

Hnestly I really don't understand much of what I have to do to prove the sufficiency of the Frobenius theorem, and so what I have written above is really unclear.

Can someone please explain to me in a form which is more clear and intuitive the proof for the sufficiency of the Frobenius theorem?