Show $\phi$ is a diffeomorphism of $ \mathbb{R}^2 $ into itself Can you check my attempt for this problem, please?
Let $f:\mathbb{R}\to\mathbb{R}$ be a $C^{1}$ function such that $|f'(t)|\leq M<1$ for any $t\in \mathbb{R}$. Define a function $\phi:\mathbb{R}^{2}\to\mathbb{R}^{2}$ by $$\phi(x,y)=(x+f(y),y+f(x))$$ 
Show that $\phi$ is a diffeomorphism of $\mathbb{R}^{2}$ over itself.
My attempt.
Note that $\phi(x,y)=(x,y)+(f(y),f(x))$. I define $\psi:\mathbb{R}^{2}\to\mathbb{R}^{2}$ such that $$\psi(x,y)=(\psi_{1}(x,y),\psi_{2}(x,y))$$ where $\psi_{1}:\mathbb{R}^{2}\to\mathbb{R}$ and $\psi_{2}:\mathbb{R}^{2}\to\mathbb{R}$ are $$\psi_{1}(x,y)=f(y), \quad \psi_{2}(x,y)=f(x).$$
Now let $(x,y),(h,k)\in\mathbb{R}^{2}$, then
$$\psi_{1}(x+h,y+k)-\psi_{1}(x,y)=f(y+k)-f(y)=f'(y)k+r(k),$$
$r(k)=o(k)$ as $k\to 0$, since $f$ is differentiable in $\mathbb{R}$. Now, we could have written 
$$\psi_{1}(x+h,y+k)-\psi_{1}(x,y)=(0,f'(y))^{*}(h,k)+ R(h,k)$$
where $(0,f'(y))^{*}:\mathbb{R}^{2}\to\mathbb{R}$ is defined as 
$$(0,f'(y))^{*}(h,k)=\langle (0,f'(y)),(h,k) \rangle=f'(y)k$$ and $R(h,k)=r(k)...(1)$. 
Now, I prove that $R(h,k)=o(h,k)$.To do this I consider the following inequality
$$0\leq \frac{|R(h,k)|}{|(h,k)|}=\frac{|r(k)|}{|(h,k)|}$$
I consider the sum norm since two norms in $\mathbb{R}^{n}$ are to equivalent. 
$$0\leq \frac{|R(h,k)|}{|(h,k)|}=\frac{|r(k)|}{|h|+|k|}\leq\frac{|r(k)|}{|k|}$$
Now,passing the limit as $(h,k)\to (0,0)$ we have
$$0\leq\lim_{(h,k)\to (0,0)}{\frac{|R(h,k)|}{|(h,k)|}}\leq\lim_{k\to 0}{\frac{|r(k)|}{|k|}}.$$
Then
$$\lim_{(h,k)\to (0,0)}{\frac{|R(h,k)|}{|(h,k)|}}=0,$$
since $r(k)=o(k)$ as $k\to 0$. Therefore $\psi_{1}$ is differentiable in $\mathbb{R}^{2}$.
So $$d\psi_{1}(x,y)=\psi_{1}'(x,y)=(0,f'(y))^{*}...(2)$$
Analogously we have that  $\psi_{2}$ is differentiable in $\mathbb{R}^{2}$ and $d\psi_{2}(x,y)=\psi_{2}'(x,y)=(f'(x),0)^{*}...(3)$
Now, let $(x,y),(h,k)\in\mathbb{R}^{2},|(h,k)|=1$
$$|\psi'(x,y)(h,k)|=|d\psi(x,y)(h,k)|=|(d\psi_{1}(x,y)(h,k),d\psi_{2}(x,y)(h,k))|$$
$$|\psi'(x,y)(h,k)|=|d\psi(x,y)(h,k)|=|(\psi_{1}'(x,y)(h,k),\psi_{2}'(x,y)(h,k))|...(4)$$
If we consider $(1),(2),(3)$ and $(4)$, we have 
$$|\psi'(x,y)(h,k)|=|d\psi(x,y)(h,k)|=|(f'(y)k,f'(x)(h))|$$
And I consider the norm sum
$$|\psi'(x,y)(h,k)|=|d\psi(x,y)(h,k)|=|(f'(y)k|+|f'(x)h|=|(f'(y)||k|+|f'(x)||h|$$
$$|\psi'(x,y)(h,k)|=|d\psi(x,y)(h,k)|\leq M(|k|+|h|)=M(|(k,h)|)=M$$
Since $|f'(x)|\leq M$ and $|(h,k)|=1$. So
$$|\psi'(x,y)|\leq M$$
On the other hand, $\psi$ is differentiable since $\psi_{1}$and $\psi_{2}$ are differentiable. As $\mathbb{R}^{2}$ is convex. We have since the Mean Value Inequality that:
$$|\psi(x,y)-\psi(z,w)|\leq M|(x-z,y-w)| \hspace{1.0cm}\forall (x,y),(z,w)\in \mathbb{R}^{2},$$
So, $\psi$ is a M-contraction since $0\leq M<1$.
As $\phi(x,y)=(x,y)+(f(y),f(x))$ and $\psi$ is a M-contraction, for Theorem of perturbation of the identity we have $\phi$ is a homeomorphism from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$. 
Now, I write $\phi(x,y)=I(x,y)+\psi(x,y)$ where $I$ is the identity function from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$. It's clear that $I$ and $\psi$ are differentiable, for this reason we have that $\phi$ is differentiable.
 A: You have proved that $\phi$ is a differentiable homeomorphism, this is not the same as a diffeomorphism (for instance the map $\mathbb{R}\to\mathbb{R}$ given by $t\mapsto t^3$ is a counterexample of this.)
The solution here is to use Inverse function theorem:  If the Jacobian at a point is invertible, then the function is a local diffeomorphism around that point.
This is the case, since $J_\phi(x,y)=\begin{pmatrix}1&f'(y)\\ f'(x)&1\end{pmatrix}$ and we have that $\det(J_\phi(x,y))=1-f'(x)f'(y)>0$, so the Jacobian in particular is invertible. 
We have that $\phi$ is locally a diffeomorphism, to prove that it is in fact a global diffeomorphism, we just need to check that $\phi$ is injective. 
Proof of $\phi$ being injective:
Note that $\phi(x,y)=\phi(a,b)$  iff 
 $x-a=f(b)-f(y)$ and $f(x)-f(a)=b-y$.
But since  $f$ is an $M$-contraction, we have
$$|x-a|=|f(b)-f(y)|\leq M|b-y|,\hbox{ and }|b-y|= |f(x)-f(a)|\leq M|x-a|$$
Since $M<1$, this is only possible if $x=a$ and $y=b$
This proves that $\phi$ is a diffeomorphism of $\mathbb{R}^2$ onto its image. 
To prove that $\phi$ is a diffeomorphism of $\mathbb{R}^2$ onto itself, we need to prove that the image of $\phi$ is all $\mathbb{R}^2$.
Proof of $\phi$ being surjective:
Given $(z,w)$ we want to show that there is a point $(x,y)$ such that $\phi(x,y)=(z,w)$. Note that this is the same than proving that there is some $x$ such that $x+f(w-f(x))=z$.
In other words, we want to prove that (for a fix $w$) the function $g(x)=x+f(w-f(x))$ is surjective. By the intermediate value theorem, it's enough to prove that $\lim_{x\to\infty}{g(x)}=\infty$ and $\lim_{x\to-\infty}{g(x)}=-\infty$.
This is easily proved by using that $f$ is an $M$-contraction. Indeed, since $f$ is an $M$ contraction, we have that $|f(x)|\leq M|x|+c$ for some constant $c$. Using this, you can easily prove that
$|f(w-f(x))|\leq M^2|x|+C $ for certain constant $C$ ($C$ might deppend on $w$, recall $w$ is fixed here). 
Then
$$x-M^2|x|-C\leq x+f(w-f(x))\leq x+M^2|x|+C$$
i.e. 
$$x-M^2|x|-C\leq g(x)\leq x+M^2|x|+C$$
From here taking the corresponding limits, we can check that $\lim_{x\to\infty}g(x)=\infty$ and $\lim_{x\to-\infty}g(x)=-\infty$. Which is what we wanted to prove. 
