$g(x)=x+f(x)$ defined on $\mathbb{R}^n$ where $f$ is 1/2-lipschitz. Prove that g is an homeomorphism

Question

Let $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ an application such that for every $x, y \in \mathbb{R}^n$, $|f(x)-f(y)|\leq\frac{1}{2}|x-y|$. Let $g : \mathbb{R}^n \rightarrow \mathbb{R}^n$ defined by $g(x)=x+f(x)$

Prove that $g$ is an homeomorphism.

Note : $g$ is continuous, because it's the sum of two continuous function. For all $x,y, \frac{1}{2}|x−y|≤|g(x)−g(y)|≤\frac{3}{2}|x−y|$. This implies $g$ is injection, and the image by $g$ of an open set is an open set. It remains to prove $g$ surjection

Source : quiz in topology course

Answer 1

$g$ is continuous, because it's the sum of two continuous function. For all $x,y, \frac{1}{2}|x−y|≤|g(x)−g(y)|≤\frac{3}{2}|x−y|$. This implies $g$ is injection, and the image by $g$ of an open set is an open set.

It remains to prove $g$ surjection :

Let $y \in \mathbb{R}^n$. $t \in \mathbb{R}^n \rightarrow f(y-t) \mathbb{R}^n$ is $\frac{1}{2}$-lipschitz function. By Banach fixed point theorem, there is $m \in \mathbb{R}^n$ such that $m = f(y-m)$. Then $g(y-m) = y-m + f(y-m) = y$. So $g$ is surjective.