# If $f: \mathbb R^n \to \mathbb R^n$ is a contraction, then $x-f(x)$ is a homeomorphism

I am stuck in following homework question.

Let $f : \mathbb R^n \to \mathbb R^n$ be a uniform contraction and $g(x) = x - f(x)$. Investigate whether $g : \mathbb R^n \to \mathbb R^n$ is a homeomorphism or not.

The definition of uniform contraction is as follows:

$(X,d)$ is metric space. $f: X \to X$ is uniform contraction if there exists $0 <\alpha <1$ such that $d(f(x),f(y)) \leq \alpha d(x,y)$.

I proved that $f$ and $g$ are continuous. But I do not have any idea about inverse of $g$.

By invariance of domain, one could conclude that a bijective continuous function $h: \mathbb R^n \to \mathbb R^n$ is homeomorphism. But I do not know how I can see $g$ is bijective or not. I do not have any other idea to prove homeomorphism. Maybe it is not a homeomorphism but I can't think of any counterexample.

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(Collecting together my comments.) HINTS: To prove injectivity of $g$, use the uniform contraction condition to conclude that $f(x) - f(y) = x-y$ is impossible unless $x = y$. To prove surjectivity of $g$, show that the map $x \mapsto f(x + v)$ is a uniform contraction for any fixed $v$, and apply the Banach fixed-point theorem. // If you are able to solve the problem based on these hints, then please post your work as an answer. – Srivatsan Jan 5 '12 at 20:33
Thank you a lot for your hints. I have to work on surjectivity a little bit but I think I will be able to solve it. – marvinthemartian Jan 5 '12 at 22:13
Yeah I got the solution thank you very much. And I will answer the question when I have time. – marvinthemartian Jan 6 '12 at 11:35

As the OP mentions, it suffices to show that $g$ is bijective:
For $v \in \mathbb R^n$, define $f_v : \mathbb R^n \to \mathbb R^n : x \mapsto f(x)+v$. It is easy to check that every fixed-point of $f_v$ satisfies $g(x) = v$; conversely, every solution of $g(x)=v$ is a fixed-point of $f_v$.
Now $f_v$ is a uniform contraction, being the composition of a uniform contraction and a translation. Therefore, the Banach fixed-point theorem guarantees that $f_v$ has a unique fixed-point, which in turn implies that $v$ has a unique pre-image under $g$. Since this is true for an arbitrary $v$, we conclude that $g$ is a bijection.
Showing that $g$ is a bijection is not sufficient, you also need to show that its inverse is continuous. This follows from the a-posteriori estimate of the Banach fixed point theorem (even better, $g$ is bi-Lipschitz). – eldering Jan 6 '12 at 18:42