Although $\mathbf{Top}$, the category of topological spaces, is not regular because products do not preserve quotient maps in general, its opposite category is regular. However, the category $\mathbf{Top}^{op}$ is not exact.
My question is the following:
What does the ex/reg completion of $\mathbf{Top}^{op}$ look like?
Of course, since we started with a dual category, we have to dualize back to get the "coex/coreg" or "co-(ex/reg)" completion of $\mathbf{Top}$.
Recall that the ex/reg completion of a regular category $C$ is formed by adding in the "missing" quotients. The objects in $C_{ex/reg}$ are the equivalence relations in $C$ and the morphisms from $(X, E_X)$ to $(Y, E_Y)$ in $C_{ex/reg}$ are relations $F \subseteq X \times Y$ such that $F \circ E_X = E_Y \circ F = F$, $E_X \subseteq F^{\circ} \circ F$, and $F \circ F^{\circ} \subseteq E_Y$, where $F^{\circ}$ is the inverse relation of $F$.
It is known that equivalence relations in $\mathbf{Set}^{op}$ have a very easy form. Namely, any coreflexive corelation on a set must actually also be cosymmetric and cotransitive, and hence a coequivalence corelation, because $\mathbf{Set}^{op}$ is a Mal'cev category. Given any subset $Y$ of a set $X$, the corresponding quotient of $X \coprod X$ is the one that identifies $(y, 1)$ with $(y, 2)$ for $y \in Y$, but nothing else. Conversely, given any equivalence relation $E$ on $X \coprod X$ that only identifies pairs with the same first coordinate, the corresponding subset of $X$ is $\{x \in X \vert ((x, 1), (x, 2)) \in E\}$.
Also, since the forgetful functor from topological spaces to sets preserves (finite) colimits and regular monos, and $\mathbf{Set}^{op}$ is already exact, we get an induced $RegMono$-preserving right exact functor from $((\mathbf{Top}^{op})_{ex/reg})^{op}$ to $\mathbf{Set}$. I suspect that this functor will probably not be faithful anymore, because the original forgetful functor is not conservative.