How to show that $\varphi(x,y)=(x+f(y),f(x)+y)$ is bijective?

Let $f:\mathbb{R}\to\mathbb{R}$ be a $C^1$ function such that $|f'(t)|\leq k<1$ for all $t\in \mathbb{R}$.

Let $\varphi:\mathbb{R}^2\to\mathbb{R}^2$ be the function given by $\varphi(x,y)=(x+f(y),f(x)+y)$.

The problem is to show that $\varphi$ is a diffeomorphism. Notice that

$$\det J_\varphi (x,y)=0\quad\Rightarrow \quad f'(x)f'(y)=1\quad \Rightarrow\quad 1=|f'(x)||f'(y)|\leq k^2<1.$$

So, $\det J_\varphi (x,y)\neq 0$ for all $(x,y)\in\mathbb{R}^2$ and thus $\varphi$ is a local diffeomorphism. Therefore, to show that $\varphi$ is a diffeomorphism it's enough to show that $\varphi$ is injective and $\varphi(\mathbb{R}^2)=\mathbb{R}^2$.

• $\varphi$ is injective.

$$\varphi(x_1,y_1)=\varphi(x_2,y_2)\quad\Rightarrow\quad x_1+f(y_1)=x_2+f(y_2)\text{ and }f(x_1)+y_1=f(x_2)+y_2$$

How to conclude that $x_1=x_2$ and $y_1=y_2$?

• $\varphi$ is surjective.

Let $(a,b)\in\mathbb{R}^2$. We need to show that there exists $(x,y)\in \mathbb{R}^2$ such that \left\{\begin{align*} x+f(y)=a\\ f(x)+y=b \end{align*}\right. How can we do it?

Thanks.

First, here is a proof of injectivity.

Let $$x_1+f(y_1)=x_2+f(y_2)$$

$$f(x_1)+y_1=f(x_2)+y_2$$

then $$x_1-x_2=f(y_2)-f(y_1)=f^{\prime}(y)(y_2-y_1)$$

further

$$y_1-y_2=f(x_2)-f(x_1)=f^{\prime}(x)(x_2-x_1)=f^{\prime}(x)f^{\prime}(y)(y_1-y_2)$$

Now since $|f^{\prime}(x)f^{\prime}(y)|< 1$ we have $y_1=y_2$ and similar for $x_1=x_2$.

For surjectivity, define a sequence $x_n$, $y_n$ via

$$x_{n+1}+f(y_n)=a$$

$$f(x_n)+y_{n+1}=b$$

Now $$|x_{n+1}-x_n|=|f(y_{n})-f(y_{n-1})|=|f^{\prime}(y)||y_{n}-y_{n-1}|\leq k|y_{n}-y_{n-1}|$$ and similarily

$$|y_{n+1}-y_n|\leq k|x_{n}-x_{n-1}|$$ This shows that $\{x_n\}$ and $\{y_n\}$ are Cauchy and their limits satisfy the requirements.