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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.

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    Yeah I got the solution thank you very much. And I will answer the question when I have time.2012-01-06

1 Answers 1

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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.

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    "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.2012-01-07