In $(\mathbb R, d(x,y)=|x-y|)$, suppose $f:\mathbb R \to \mathbb R$ is a contraction, that is for all $x,y \in \mathbb R$, there exists a constant $A$ between 0 and 1 such that $|f(x)-f(y)| \leq A|x-y|$.
Then let $G(x)=x+f(x)$. I have already proved that $G:\mathbb R \to \mathbb R$ is a bijection and $G$ is continuous, how to show that the inverse of $G$ is also continuous?
I have tried to find connection between $| G^{-1}(x)-G^{-1}(y) |$ and $|x-y|$ but failed, any hint?
$g$ is monotone, indeed $\forall x,y \in \mathbb R, x>y$ consider
$|f(x)-f(y)|<A(x-y)<(x-y)$ $\implies -(x-y)<f(x)-f(y)<(x-y)\implies g(x)-g(y)>0$ by using left part of the inequality. Now you have a continuous bijective monotone function from $\mathbb R \to \mathbb R$ what can you say about $g^{-1}$?