# Homeomorphic spaces are uniformly isomorphic

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.

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