Show that if $f$ is a $C^2$ map such that $|f'(t)| \leq k < 1$ for all $t \in \mathbb{R}$ then $\varphi(x, y) = (x + f(y), y + f(x))$ is a diffeomorphism of $\mathbb{R}^2$ onto itself.
Note that the determinant of the jacobian matrix at $(x, y)$ is $1 - f'(x)f'(y)$. By the hypothesis on $f'$, we know that this is always positive, and thus, by the Inverse Function Theorem $\varphi$ is a local diffeomorphism.
Now, it remains to prove that $\varphi$ is injective, from what follows that it is a global diffeomorphism. Any hints on this will be the most appreciated.
$\phi(x_1,y_1)=\phi(x_2,y_2) \Longleftrightarrow (x_1+f(y_1),y_1+f(x_1))=(x_2+f(y_2),y_2+f(x_2))$
Then $$x_1+f(y_1)=x_2+f(y_2)$$ $$y_1+f(x_1)=y_2+f(x_2)$$
so $$|x_1-x_2|=|f(y_2)-f(y_1)|$$ $$|y_1-y_2|=|f(x_2)-f(x_1)|$$
Since $f$ has a bounded derivative,it is Lipschitz cntinuous,thus $$|f(x)-g(y)| \leq k|x-y|, \forall x,y \in \Bbb{R}$$
Thus $|x_1-x_2| \leq k^2|x_1-x_2|$ and since $k^2<k<1$ then $x_1=x_2$
Similarly $y_1=y_2$ thus $(x_1,y_1)=(x_2,y_2)$