Using connectedness to prove surjectivity...

Question

$\textbf{My problem:}$ Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a contraction and $\phi :\mathbb{R^2} \rightarrow \mathbb{R^2}$ defined by $\phi (x,y)=(x+f(y),y+f(x))$. Prove that $\phi (\mathbb{R^2} )=\mathbb{R^2}.$

$\textbf{My attempt:}$ I though to prove that $\phi (\mathbb{R^2} )$ is an open and closed subset of $\mathbb{R^2} $. Then, because $\mathbb{R^2} $ is connected, we can conclude the proof. However, I don't achieved it...

Could anyone help me please?...

Answer 1

Let $(x,y)\in\mathbb R^2$ and let

$$g: u \mapsto x - f\left(y-f(u)\right)$$

\begin{align} \left|g(u) - g(v)\right|& = \left|x-f(y-f(u)) - x + f(y-f(v))\right|\\ &= \left|f(y-f(u))-f(y-f(v))\right|\\ &\le k\left|y-f(u) - y + f(v)\right|\\ &=k\left|f(u)-f(v)\right|\\ &\le k^2|u-v|. \end{align}

This proves that $g$ is a contraction of $\mathbb R$. Let $s$ be a fixed point of $g$ and $t=y-f(s)$ then \begin{align}\phi(s,t) &= (s+f(t), t+f(s))\\ &= (g(s) + f(y-f(s)), y-f(s) + f(s))\\ &= (x - f(y-f(s)) + f(y-f(s)), y)\\ &= (x,y)\end{align}.