I am currently reading Arnold's book "Mathematical Methods of classical mechanics" on page 278 and I don't see through his arguments there at a point.
Especially, I am talking about the part that begins with "It is now easy to prove Lemma 2".
Lemma 2 says: Let $M^n$ be a compact connected differentiable $n-$ dimensional manifold on which we are given n pairwise commutative and linearly independent vector fields at each point. Then $M^n$ is diffeomorphic to an n-dimensional torus.
In his previous discussion, he established a surjective map $g: \mathbb{R}^n \rightarrow M^n.$ This map is smooth and surjective, but not injective. In particular, it is the restriction of a group action $g(t_1,...,t_n) := \Phi_{t_1,...,t_n}(x_0)$ where $x_0 \in M^n$ is fixed and $\Phi: \mathbb{R}^n \times M^n \rightarrow M^n.$ You find the definition of this action in terms of flows on 274, in case it is important to answer the question.
Now, we can define the group $\Gamma$ as the stabilizer of this action, so $t \in \Gamma$ if and only if $g(t) = x_0.$ Arnold shows then that this group is a discrete subgroup of $\mathbb{R}^n$ and hence $\Gamma:= \mathbb{Z}e_1+...+\mathbb{Z}e_k$ for linearly independent $e_1,...,e_k$ and $k \le n$.
Then, on page 278, he takes $f_i=2 \pi (0,...,0,1,0,...0)$ where the $1$ is at position i and $i \in \{1,...,n\}.$ From this he defined a map $A : \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $A(f_i)=e_i$ with the $e_i$ taken from the representation of $\Gamma$ above. This does -since $k \le n$- not uniquely determine $A$, but he says that $A$ is an isomorphism, so I guess the rest of $A$ can be defined arbitrary, as long as we get an isomorphism.
To get the connection to the torus, he defines the natural projection $p: \mathbb{R}^{2n} \rightarrow T^k \times \mathbb{R}^{n-k}$ ( I guess the $2n$ is a typo and it should just be $n$) such that $p(\phi_1,...,\phi_k; y_1,...,y_{n-k}):= ( \phi \mod 2 \pi ; y).$ This map is smooth, surjective and well-defined but clearly not injective, i.e. $p(f_i) = 0.$ for all $i \in \{1,..,n\}.$
Now, problem 10 states that this give us a map $\tilde{A}: T^k \times \mathbb{R}^{n-k} \rightarrow M^n$ that is diffeomorphic. But clearly, $p$ and $g$ are no diffeomorphisms, so this is at least not obvious to me where this diffeomorphic property should come from. Does anybody have an idea? I would really like to close this gap in the proof.
Now he finishs the proof by saying that if we know that $\tilde{A}$ is a diffeo, then $\tilde{A}^{-1}(M^n)= T^k \times \mathbb{R}^{n-k}$ is clearly compact, from which we get $k=n$ and the claim follows.
By the way: I got the hint that covering space theory might help. Clearly, $p$ is a smooth covering map and $g$ might be one, too (at least the surjectivity holds). So as $A$ is a diffeomorphism does this define the diffeomorphism $\tilde{A}$, too?