For uniformity $\mathcal{U}_1$ and $\mathcal{U}_2$, $\operatorname{id}:(X, \mathcal{U}_1)\to (X, \mathcal{U}_2)$ is a homeomorphism?

Question

A uniformity for a set $X$ is a non-empty $\mathcal{U}$ of subsets of $X\times X$ such that:

1. If $U\in\mathcal{U}$, then $U^{-1}\in\mathcal{U}$;
2. Each member of $\mathcal{U}$ contains diagonal of $X$;
3. If $U\in \mathcal{U}$, then $V\circ V\subseteq U$ for some $V\in\mathcal{U}$;
4. If $U$ and $V$ are elements of $\mathcal{U}$, then $U\cap V\in\mathcal{U}$;
5. If $U\in\mathcal{U}$ and $U\subseteq V\subseteq X\times X$, then $V\in \mathcal{U}$.

The pair $(X, \mathcal{U})$ is a uniform space.

It is known that metrics $d_i, d_2$ induce the same topology on $X$ if the identity map $id:(X, d_1)\to (X, d_2)$ is a homeomorphism.

Let $\mathcal{U_1}$ and $\mathcal{U}_2$ be uniformity on $X$ and induce the same topology on $X$.

Is it true that $id:(X, \mathcal{U}_1)\to (X, \mathcal{U}_2)$ is a homeomorphism?

Answer 1

Your notation is confused: if you write $\textrm{id}: (X, \mathcal{U}_1) \to (X,\mathcal{U}_2)$ one normally means the map in the category of uniform spaces, and consider uniform continuity the relevant property to consider. As a map between topological spaces (using the induced topology) it's just the identity from a space to itself, and thus a homeomorphism. It need not be a uniform isomorphism at all, or even uniformly continuous.

$\textrm{id}:(X, \mathcal{T}_1) \to (X, \mathcal{T}_2)$ for topologies on the same set $X$ is a homeomorphism iff $\mathcal{T}_1 = \mathcal{T}_2$, as is trivial to see. It's a boring reformulation.