1
$\begingroup$

A continuous function $f$ is a homeomorphism if it is bijective, and open.

A uniformly continuous function $f$ is a uniform isomorphism if it is bijective and $f^{-1}$ is uniformly continuous.

Is it true that a homeomorphism is a uniform isomorphism? I know the converse is true since each uniformly continuous function is continuous.

I think it doesn't always hold since not every continuous function is uniformly continuous, which doesn't guarantee that a continuous function $f^{-1}$ is uniformly continuous. But I can't think of a counterexample.

Please help.

$\endgroup$
0
$\begingroup$

The map$$\begin{array}{ccc}\mathbb R&\longrightarrow&\left(-\frac\pi2,\frac\pi2\right)\\x&\mapsto&\arctan x\end{array}$$is such a counterexample. It's a homeomorphism, but the inverse is not uniformly continuous.

$\endgroup$
1
$\begingroup$

In topology we have a notion of open map, which allows us to reformulate the condition of "$f^{-1}$ is continuous" (for a bijection) as "$f$ is open". The formal definition is the category-theory one: $f$ must be an isomorphism in the category $\mathrm{Top}$, which means that there is an inverse that is also continuous and the uniform notion is defined in the same way. There is no notion (that I have seen) of "uniformly open map" that allows us to reformulate a uniform isomorphism as a bijective uniformly continuous map that is uniformly open, e.g.

$\endgroup$
0
$\begingroup$

$x \to x^{2}$ is a homeomorphsm of $(0,\infty)$ but it is not uniformly continuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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