$\require{AMScd}$ 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:

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

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 $X$

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:

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

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 $X$.

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 'cotopological 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 'cotopological 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:

- Morphisms in the Category of Cotopological Spaces.

Suppose $X$ and $Y$ are topological spaces with a set map $f: X \rightarrow Y$. We say $f$ is continuous 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 cotopological spaces $X$ and $Y$. We say $f$ is cocontinuous 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.

- Each metric space $(X, d)$ induces a cotopological 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.

- Each topological space $(X, T)$ induces a cotopological 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.

If we start instead with a topological space $(X, C)$ where $C$ is the set of closed sets on $X$, we end up with a space $(X, \mathcal{S})$ where $\mathcal{S}$ is a set of equivalence relations closed under finite intersections and arbitrary joins, where a join of equivalence relations is the smallest equivalence relation containing them.

Limits and colimits in the topology of Cotopological Spaces.

Note: the forgetful functor $F : {\bf CoTop} \rightarrow {\bf Set}$ has a left and right adjoint and therefore preserves limits and colimits. Hence if $(X_i, \mathcal{S}_i) \cong \text{ colim } \Phi$ then $X_i \cong \text{ colim } F \circ \Phi$, and the same for limits.

Take cotopological spaces $(X_i, \mathcal{S}_i)_{i \in I}$. We define a cotopology $\mathcal{S}$ on $\amalg_{i \in I} X_i$ as follows: a set $R \subset \amalg_{i \in I} X_i \times \amalg_{i \in I} X_i$ is a relation in $\mathcal{S}$ if and only if there are $\{ R_i \}_{i \in I}$, with $R_i \in \mathcal{S}_i$, such that $x_i R y_j$ for $x_i \in X_i$, $y_j \in X_j$ if and only if $i = j$ and $x_i R_i y_i$.

Defining the product cotopology, "cofinal", and "coinitial" topologies for the case of direct and inverse limits is similarly straightforward.

closedsets, and if you dualize on the closed sets, you get something presumably very different, because one place where the symmetry breaks down horribly is that a subobject has a complement whereas a quotient object does not have a canonical transversal (or whatever that would be). One can also define topologies using closure axioms and maybethatwould also give some interesting dual notion. $\endgroup$18more comments