$\require{AMScd}$ 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 '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.


  1. 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.

  1. 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.

  1. 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.

  1. 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.

  2. 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.

  • 13
    $\begingroup$ Cool. So now we can define "cocontinuous map", "quotient cotopology", "subcospace cotopology". And maybe some coseparation axioms. And work out what colimits and limits in this category are. Oh, and think of some good examples. $\endgroup$ Mar 25, 2017 at 17:07
  • 11
    $\begingroup$ I like your idea a lot, but keep in mind that a topology can just as well be defined by closed sets, 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 maybe that would also give some interesting dual notion. $\endgroup$
    – Gro-Tsen
    Mar 25, 2017 at 18:14
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    $\begingroup$ Tom Goodwillie is pointing out something important: one should have good examples to motivate why one would introduce or explore some concept or else there is nothing to guide what is being done. That "dualizing is a natural thing to do" is not by itself a compelling reason to think about something: it should illuminate a concept of prior interest or shed light on something that is hard to understand or something like that (e.g., affine group schemes genuinely illuminate things, but not because of being a "dual" concept). Otherwise the concept has no reason to be productive vs. a dead end. $\endgroup$
    – nfdc23
    Mar 25, 2017 at 18:52
  • 32
    $\begingroup$ @nfdc23 Those considerations are appropriate if one is applying for a grant or helping a student find a thesis problem, but there are many great MO questions whose original motivation was shakier than "dualizing is a natural thing to do". Also, the OP has produced a fairly rich supply of examples, namely all metric spaces. Odds are this construction will produce something familiar - bornological spaces, perhaps? - and my curiosity is piqued, at least. $\endgroup$ Mar 25, 2017 at 19:19
  • 26
    $\begingroup$ No one made the obligatory joke that maps of cotopological spaces are "ntinuous maps"? $\endgroup$ Oct 4, 2018 at 20:16

1 Answer 1


I believe, this notion is closely related to the notion of the coarse structure which indeed had found many beautiful applications in geometric group theory and algebraic topology (including proofs of the Novikov conjecture for a lot of groups). For introduction, I'd recommend Lectures on coarse geometry by John Roe where in Chapter 2 he explains how to give a coarse structure to every metric space or a topological space (the latter upon fixing a compactification, which in your example should be Stone–Čech).

The idea of coarse geometry/topology is exactly dual to that of topology: one doesn't look at what happens at “small distances”, but rather at what happens at “large distances” in a space, and I believe, this is exactly what the definition of the cotopology also does (it defines what “large distances” mean).

However, the original definition of a coarse structure is done in a way as to define what are subsets with “not so large” distances, but I believe, it's a bit like defining topology by open subsets or closure operation. I'll try to sketch below the translation between the two notions.

Definition. A coarse structure on a set $X$ is a collection $\mathcal E$ of subsets of $X \times X$ called controlled sets, so that $\mathcal E$ contains the identity relation, is closed under taking subsets, inverses (transposes), and finite unions, and is closed under composition of relations.

The main intuitive point in the connection can be already seen in the examples: the metric cotopological structure given by $R_r(x)$ should correspond to the family of controlled sets generated by $\{(x,y)\mid d(x,y)<r\}$.

Coarse to cotopological: Given a coarse structure $\mathcal E$, define a family of equivalence relations $\sim_{E,K}$, where $E$ runs through $\mathcal E$ and $K$ through subsets of $X$ in the following way: $y \sim_{E,K} z$ iff either $y=z$ or $(x,y),(x,z)\not\in E\cup E^{-1}$ for all $x\in K$. Observe that these equivalence relations are automatically symmetric, so it will be enough for us to work only with $E=E^{-1}\in\mathcal E$.

  • This family is trivially closed under arbitrary intersections: the intersection of $\sim_{E_i,K_i}$ is just $\sim_{\bigcap E_i,\bigcup K_i}$, and $\bigcap E_i\in \mathcal E$ because $\mathcal E$ is closed under taking subsets.
  • Finally, it is closed under passing to sus of two equivalence relations again using the properties of $\mathcal E$ as such as passing to subsets, finite unions and compositions (it's by no means obvious, but I think, I have figured out at least a sketch of a bit tedious proof; to not make the post too long, I omit it, but I'm happy to discuss this in detail if needed). I can sketch the proof in the metric case: The equivalence relation generated by $R_{r}(K)$ and $R_{r}(K')$ collapses everything exactly outside the ball of radius $r$ around $K\cap K'$, so it's just $R_{r}(K\cap K')$. If the radii are different, one has to work a bit more, but the flexibility of the coarse structure allows that.

Cotopological to coarse: It easy to see that an arbitrary intersection of coarse structures is again a coarse structure. Let's say that a coarse structure $\mathcal E$ is compatible with a family $\sim_\alpha$ of equivalence relations (i.e. a cotopology) if all $\sim_\alpha$ belong to the cotopology generated by $\mathcal E$. The intersection of all compatible coarse structures should then be the coarse structure which induces the given cotopology, although I haven't checked this carefully.

In any case, the possible connection here seems worth examining in detail.

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    $\begingroup$ Ah... so "topological" is closely related to "arse". $\endgroup$
    – WhatsUp
    Nov 23, 2019 at 12:16

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