$\require{AMScd}$ Defining a topological space on a set $X$ is equivalent to designating certain subobjects of $X$ in ${\bf Set}$ (monomorphisms into $X$ up to equivalence) as open. The requirements on the open sets of a topological space $X$ are equivalent to requiring the following:

1. $X \rightarrow X$ and $\emptyset \rightarrow X$ are open.

2. If $X_i \rightarrow X$ are open subobjects of $X$ for finite $i \in I$ then so is their product in the category of subobjects of $X$

3. If $X_i \rightarrow X$ are open subobjects of $X$ for $i \in I$, then so is their coproduct in the category of subobjects of $X$.

My question is about what happens when one dualizes this notion:

Let $X$ be a set and consider a subset of its quotient objects $\mathcal{S}$ such that:

1. $X \rightarrow X \in \mathcal{S}$ and $X \rightarrow \{ * \} \in \mathcal{S}$.

2. If $X \rightarrow X_i \in \mathcal{S}$ for finite $i \in I$ then $\amalg_{i \in I} (X \rightarrow X_i) \in \mathcal{S}$, where the coproduct is taken in the category of quotient objects of $X$.

3. If $X \rightarrow X_i \in \mathcal{S}$ for $i \in I$, then $\prod (X \rightarrow X_i) \in \mathcal{S}$, where the product is taken in the category of quotient objects of $X$.

Does this structure arise anywhere in practice? What is known about this notion of 'co-topological spaces'? Is there a place I can learn about them?

Note: in the case of a set, the category of quotient objects is equivalent to the category of equivalence relations on the set in question. A pairwise coproduct of equivalence relations is then the equivalence relation generated by two equivalence relations. The product of a collection of equivalence relations is their intersection. Thus this notion of a 'co-topological space' is equivalent to putting a set of equivalence relations on a set $X$ closed under pairwise sum and arbitrary intersection, where the sum of two equivalence relations is the relation generated by them. We also require that collection includes the equivalence relation where all points are equivalent and the equivalence relation where no two distinct points are equivalent.

Edits:

1. Morphisms in the Category of Co-Topological Spaces.

Suppose $X$ and $Y$ are topological spaces with a set map $f: X \rightarrow Y$. We say $f$ is open if, for every open subobject $V \rightarrow Y$, the following pullback gives an open subobject $U \rightarrow X$: \begin{CD} X @>>> Y\\ @AAA @AAA\\ U @>>> V \end{CD}

Analogously, suppose there is a set map $f : X \rightarrow Y$ of co-topological spaces $X$ and $Y$. We say $f$ is co-continuous if for each open quotient object $X \rightarrow P$ the following pushout diagram forms a quotient object $Y \rightarrow Q$: \begin{CD} X @>>> Y\\ @VVV @VVV\\ P @>>> Q \end{CD}

In terms of equivalence relations this translates to requiring that if $R$ is an open equivalence relation on $X$ then the relation generated by $R'$ where $x'R'y'$ if and only if $x' = f(x)$ and $y' = f(y)$ for $x, y \in X$ such that $xRy$ is open.

2. Each metric space $(X, d)$ induces a co-topological space in the following way:

Each open ball $B_{\epsilon}(x)$ induces an equivalence relation $R_{\epsilon} (x)$ where $y R_{\epsilon} (x) z \iff (z = y$ or $z, y \notin B_{\epsilon} (x))$. Form the set $T = \{ R_{\epsilon} (x) : \epsilon \in \mathbb{R}, x \in X \}$ and close it under intersection of equivalence relations. This forms a co-topological space.

3. Each topological space $(X, T)$ induces a co-topological space in the following way:

Let $B$ be a basis for $X$. Each open set $U \in B$ induces an equivalence relation $R(U)$ where $xR(U)y$ when $x = y$ or $x, y \notin U$. Form the set $S = \{ R(U) : U \in B\}$ and close it under intersection of equivalence relations. This forms a co-topological space.