Problem: Let $f$ : $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ be a $C^{1}$ function such that |$f'(t)$| $\leq$ $k$ < $1$, $\forall$ $t$ $\in$ $\mathbb{R}$. Define $\phi$ : $\mathbb{R}^{2}$ $\rightarrow$ $\mathbb{R}^{2}$ by $\phi$$(x,y)$ = ($x$ + $f(y)$, $y$ + $f(x)$). Prove that $\phi$ is a diffeomorphism from $\mathbb{R}^{2}$ onto itself.
Partial solution: First we observe that $|Jac \phi|$ $\neq$ $0$. In fact, $|Jac \phi(x,y)|$ = |$1$ - $f'(x)$.$f'(y)$|, which is never zero by the hypothesis. So, the inverse function theorem gives us that $\phi$ is a local diffeomorphism.
Now, we show that $\phi$ is injective. Suppose ($x_1$,$y_1$) $\neq$ ($x_2$,$y_2$). If ($x_1$ + $f(y_1)$, $y_1$ + $f(x_1)$) = ($x_2$ + $f(y_2)$, $y_2$ + $f(x_2)$), we get:
$x_1$ - $x_2$ = $f(y_2)$ - $f(y_1)$
$y_1$ - $y_2$ = $f(x_2)$ - $f(x_1)$
But, again by the hypothesis and by the Mean Value Theorem:
|$f(y_2)$ - $f(y_1)$| $\leq$ $k$|$y_1$ - $y_2$| = $k$|$f(x_2)$ - $f(x_1)$| < |$f(x_2)$ - $f(x_1)$| < |$x_1$ - $x_2$| = |$f(y_2)$ - $f(y_1)$|
Which is a contradiction. So f is injective.
I only have to show that f is surjetive so it'll be a diffeomorphism, but I couldn't do it so far. Any hints? Thanks in advance :)