Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

Thank you in advance.

share|cite|improve this question
(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
up vote 3 down vote accepted

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.

share|cite|improve this answer
@marvinthemartian While I feel the hints I supplied make it easier to think about the problem, it turns out that one can do away with discussing two cases. I wrote an answer explaining the shorter proof. – Srivatsan Jan 6 '12 at 16:39
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
"By invariance of domain, one could conclude that a bijective continuous function h:Rn→Rn is homeomorphism." Because by invariance of domain you can conclude that its inverse in continuous. – marvinthemartian Jan 7 '12 at 14:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.