Let $X=\Bbb R^{n}$ be the $n$-dimensional Euclidean space. Suppose that a map $f:X \to X$ is a contraction mapping, i.e., there exists $\delta \in [0,1)$ such that $$ |f(x)-f(y)| \leq \delta |x-y| \ \ \ (x,y \in X). $$ Set $F(x) := x+f(x)$. Prove that the map $F:X \to X$ is bijective and the inverse mapping of $F$ is continuous.
The injectivity of $F$ is easy to prove. To prove $F$ is surjective, given $y \in X$, if one defines $h:=h_y:X \to X$ by $h(x)=y-f(x)$, then for $x,z \in X$, $$||h(x)-h(z)||=||y-f(x)-(y-f(z))||=||f(x)-f(z)||\leq \delta ||x-z||,$$ which means $h_y$ is a contraction on a complete metric space ($X=\Bbb R^{n}$), then $h_y$ has a unique fixed point. Let $x$ be the fixed point of $h_y$, $h(x)=x \iff y-f(x)=x \iff F(x)=y$, it follows that $F$ is onto. We conclude that $F$ is bijective.
- My first question: is there any other way to prove $F$ is surjective without using Banach fixed-point theorem?
It is easily seen that $F(x)= x+f(x)$ is Lipschitz-continuous hence uniformly continuous hence continuous. We have proved that $F$ is bijective, by Invariance of domain $f$ is homeomorphism. Hence the inverse mapping of $F$ is continuous.
- My second question: is there any other way to prove that the inverse mapping of $F$ is continuous without using Invariance of domain?
For the first question, it's likely that there is no other way. However, since we're working on a specific topological space $X=\Bbb R^{n}$, I wonder if there are another ways to prove the continuity of inverse mapping of $F$.
Any help would be much appreciated.