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

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.

$$x_1-x_2=f(y_2)-f(y_1)$$ 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:
$$|x_1-x_2| implies that $$x_1=x_2$$ since $$k<1$$, same idea shows that $$y_1=y_2$$.