In topology, a function is continuous if the inverse image of an open set is open. A function is open if the image of an open set is open.
Uniformity continuity can be defined in a similar way as continuity. A map $f$ between two uniform spaces is uniformly continuous if the inverse image (under $f \times f$) of an entourage is an entourage.
Now consider the following "uniform analog" of openness: the image (under $f \times f$) of an entourage is an entourage.
Question. Was this property studied? And if yes, is there a well-established name for it?
Motivation. I need this property in a very special case and I am looking for references.
The maps in question appear to be uniformly open surjections. I believe that uniformly open maps were introduced in E. Michael, ‘Topologies on spaces of subsets’, Trans. Amer. Math. Soc. $\mathbf{71}$ ($1951$), $152$-$182$. If $\langle X,\mathscr{U}\rangle$ and $\langle Y,\mathscr{V}\rangle$ are uniform spaces, a map $f:X\to Y$ is uniformly open if for each $U\in\mathscr{U}$ there is a $V\in\mathscr{V}$ such that $f[U[x]]\supseteq V[f(x)]$ for each $x\in X$.
Suppose that $f$ is a uniformly open surjection, $U\in\mathscr{U}$, and $V\in\mathscr{V}$ is such that $f[U[x]]\supseteq V[f(x)]$ for each $x\in X$. Let $\langle y_0,y_1\rangle\in V$; then there are $x_0,x_1\in X$ such that $f(x_i)=y_i$ for $i=0,1$ and $\langle x_0,x_1\rangle\in U$, so $V\subseteq(f\times f)[U]$, and hence $(f\times f)[U]\in\mathscr{V}$.
I’ve not read the paper, but Vitaly V. Fedorchuk & Hans-Peter A. Künzi, ‘Uniformly open mappings and uniform embeddings of functions spaces’, Topology and its Applications $\mathbf{61}$ ($1995$), $61$-$84$ appears to discuss them in considerable detail.