Showing that a function in $\Bbb{R}^{2}$ is a diffeomorphism.

Question

Let $f:\Bbb{R}\rightarrow\Bbb{R}$ a function of class $C^{1}$ such that $|f'(t)|\leq k < 1\, \forall\, t\in\Bbb{R}$. Define $\phi:\Bbb{R}^{2}\rightarrow\Bbb{R}^2$ by

$$\phi(x,y)=(x+f(y),y+f(x)). $$

I need to show that $\phi$ is a diffeomorphism.

I want to use this theorem:

$\textbf{Theorem:}$ Let $A\subset\Bbb{R}^{n}$ an open set, and $\phi:A\rightarrow\Bbb{R}^{n}$ a function of class $C^{r}.$ If $\phi$ is injective and $\phi'(\textbf{x})$ is non-singular $\forall\, \textbf{x}\in A$, so $\phi$ is a diffeomorphism of class $C^{r}.$

In this case, $A=\Bbb{R}^{2}$ itself, and $r=1$. It was easy to show that $\phi'(x)$ is non-singular for all $\textbf{x}\in\Bbb{R}^{2},$ but I don't know how to show that $\phi$ is injetive.

What I did:

$$\phi(x_1,y_1)=\phi(x_2,y_2)\iff \begin{cases} x_1+f(y_1)=x_2+f(y_2) \\ y_1+f(x_1)=y_2+f(x_2) \end{cases} \iff \begin{cases} x_1-x_2=f(y_2)-f(y_1) \\ y_1-y_2=f(x_2)-f(x_2) \end{cases}. $$

If I proved that this two equations are both zero, I 'm done, but I don't know if this is the right way.

Answer 1

$x_1-x_2=f(y_2)-f(y_1)$ implies that $|x_1-x_2|\leq |f'(t_1)(y_2-y_1)|\leq k|y_2-y_1|$

$y_1-y_2=f(x_2)-f(x_1)$ implies that $|y_1-y_2|\leq k|x_1-x_2|$, we deduce that:

$|x_1-x_2|<k|y_1-y_2|\leq k^2|x_1-x_2|$ implies that $x_1=x_2$ since $k<1$, same idea shows that $y_1=y_2$.