Arnold's proof of Liouville's Theorem on integrable systems My question happens to be almost identical to the one left unanswered/closed here, which gives a bit of background information - it may not be necessary. I hope the reason it was closed on mathoverflow is not a reason for it to be closed here, please let me know if I should reformulate the question.
To reiterate, I refer to page 278 of Arnold's book 'Mathematical Methods of classical mechanics', namely the part after Lemma 3 is proved but before section 50 on Action-angle variables. An online version can be referred to here.
My confusion first occurs when $p$ is stated as being a chart for $\mathbb{T}^k\times\mathbb{R}^{n-k}$, since it is certainly not 1-1 (all $f_i$ are mapped to 0 under it). This leads on to confusion with problem 10 - I simply cannot see a way to prove $\tilde{A}$ is a diffeomorphism, given that neither $p$ nor $g$ are diffeomorphisms. This seems to be the only gap for me at the moment - at first glance I am not sure about Problem 11 either, but will post about that separately. Any help on this would be much appreciated. 
 A: Ok, I have attempted two proofs, the first of which relies on homogeneous spaces, and the second using universal covers (neither of which I have much experience with, so take some results for granted). Any comments would be really appreciated. 
I) Since $g$ is a transitive group action and $\Gamma$ is the stabilizer of this action (as we've seen, with respect to which $x_0$ is irrelevant), we get a homeomorphism between $M_F$ and $\mathbb{R}^n/\Gamma$. Viewing $\Gamma$ is a closed subgroup of $\mathbb{R}^n$, considered as a Lie group, it follows by Cartan's Closed Subgroup Theorem that $\Gamma$ is a Lie subgroup of $\mathbb{R}^n$. Therefore $\mathbb{R}^n/\Gamma$ is a smooth manifold, and the action $g$ induces a smooth structure on $M_F$, that is, we have a diffeomorphism between $M_F$ and $\mathbb{R}^n/\Gamma$. But $\mathbb{R}^n/\Gamma$ is diffeomorphic to $\mathbb{T}^k\times\mathbb{R}^{n-k}$, since $\mathbb{T}$ is just the quotient $\mathbb{R}/\mathbb{Z}$ and $\Gamma$ is generated by $k$ copies of $\mathbb{Z}$. Since the diffeomorphism between $M_F$ and $\mathbb{T}^k\times\mathbb{R}^{n-k}$ must preserve compactness, it must be the case we have no $\mathbb{R}$ components. Therefore $k=n$ and $M_F$ is diffeomorphic to $\mathbb{T}^n$.
II) Alternatively, define the points $f_1,\dots,f_k\in\mathbb{R}^n$ by $f_i=2\pi(0,\dots,0,1,0,\dots,0)$ where the 1 occurs in the $i$'th position - then $p(f_i)=0$ for all $1\leq{i}\leq{k}$. Let $e_1,\dots,e_k\in\Gamma\subset\mathbb{R}^n$ be the generators of the stationary group $\Gamma$, which we know exist by Claim B. Let $A:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be an isomorphism mapping the vector space $\mathbb{R}^n$ with coordinates $(\phi_1,\dots,\phi_k,y_1,\dots,y_{n-k})$ to the vector space $\mathbb{R}^n$ with coordinates $\mathbf{t}=(t_1,\dots,t_n)$, such that each $f_i$ is sent to $e_i$. Note that since $k\leq{n}$, this isomorphism need not be unique. In any case, $A$ preserves the lattice structures between the two copies of $\mathbb{R}^n$, since $A(x+z_1f_1+\dots+z_kf_k)=A(x)+z_1e_1+\dots+z_ke_k$ for $x\in\mathbb{R}^n$, $z_i\in\mathbb{Z}$. Now, $p$ and $g$ are universal covering maps for $\mathbb{T}^k\times\mathbb{R}^{n-k}$ and $M_F$ respectively, and so $A$ descends to an isomorphism $\tilde{A}$ between $\mathbb{T}^k\times\mathbb{R}^{n-k}$ and $M_F$. \
But since $p$ and $g$ are both charts, they are local diffeomorphisms and hence the descension $\tilde{A}$ is in fact a diffeomorphism (I think this needs more justification). Then as above, since $M_F$ is compact and diffeomorphisms must preserve compactness, it must be the case we have no $\mathbb{R}$ components. Therefore $k=n$ and $M_F$ is diffeomorphic to $\mathbb{R}^n$.
