What is the opposite category of $\operatorname{Top}$?

Question

My question is rather imprecise and open to modification. I am not entirely sure what I am looking for but the question seemed interesting enough to ask:

The opposite category of rings is the category of affine schemes. This is usually thought of as the category of spaces. Can we run the construction backwards for categories usually thought of as containing spaces?

For instance, does $\operatorname{Top}^{\operatorname{op}}$ have a nice description as some "algebraic" category?

Note that it does not seem easy to describe the opposite category of all schemes. Therefore, the above question might be asking too much. Perhaps the following is a more tractable (or not) question:

Can we find an "algebraic" category $C$ such that we can embed $C^{\operatorname{op}}$ in $\operatorname{Top}$ such that every topological space can be covered by objects in $C^{\operatorname{op}}$? Perhaps one would like to replace this criterion of being covered by objects by a more robust notion in general.

One can repeat the question for other categories of spaces like:

• Category of manifolds (perhaps closer to schemes than general topological spaces)
• Compactly generated spaces
• Simplicial Sets

and so on. A perhaps interesting example is the category of finite sets, it's opposite category is the category of finite Boolean algebras.

Answer 1

This doesn't exactly fit your criteria, but a standard answer to "what $\mathrm{Top^{op}}$ morally should be" is the category of frames. Loosely speaking, a frame is a poset that acts like the poset of open sets in a topological space. More precisely, a frame is a poset in which every finite subset has a meet, every subset has a join, and (finite) meets distribute over (possibly infinite) joins. A morphism of frames is a map which preserves finite meets and arbitrary joins. (Note that a frame actually automatically has infinite meets, but morphisms are not required to preserve them.)

What does this have to do with $\mathrm{Top^{op}}$? Well, given a topological space $X$, the poset $\Omega(X)$ of open subsets of $X$ is a frame (since open sets are closed under finite intersections and arbitrary unions). And if $X$ and $Y$ are topological spaces, a continuous map $f:X\to Y$ induces a frame homomorphism $f^*:\Omega(Y)\to\Omega(X)$ given by taking an open set to its inverse image. So this gives a functor $\Omega$ from $\mathrm{Top^{op}}$ to the category of frames.

Unfortunately, $\Omega$ is not an equivalence. However, when restricted to sober spaces, $\Omega$ is fully faithful. The objects in the image of $\Omega$ are called "spatial frames", which can be roughly thought of as "frames with enough points". The opposite category of frames is called the category of "locales", and locales can be thought of as a generalization of (sober) spaces which are in some ways more nicely behaved.

To tell this story in full would take a book, not an MSE answer, so I'll stop here for now (some more details of the story are described in sqtrat's nice answer). To learn more, some key words to look up are "pointless topology" and "Stone duality". One nice reference for these ideas (and more) is Peter Johnstone's book Stone spaces.

Answer 2

This is not a complete answer, but too long for a comment.

A frame is a complete distributive lattice $F$ in which the infinite distributive law $a\wedge \bigvee S=\bigvee\{a\wedge s\ |\ s \in S\}$ holds for all subsets $S$ and elements $a$ of $F$. A frame homomorphism is a lattice homomorphism which preserves finite meets and arbitrary joins.

Given any topological space $(X,\mathcal{T})$, $\mathcal{T}$ is a frame. Furthermore, if $f: (X,\mathcal{T}) \rightarrow (Y,\mathcal{S})$ is continuous, then $f^\ast: \mathcal{S}\rightarrow \mathcal{T}$ is a frame homomorphism, where $f^\ast(B)=f^{-1}[B]$. A completely prime filter $P$ in $F$ is a filter in $F$ such that for all $S\subset F$, if $\bigvee S \in P$, then there is an $s \in S$ such that $s \in P$.

Given any frame $F$, we can look at the collection of all completely prime filters on $F$, say $Pt(F)$. We can topologize $Pt(F)$ by defining for each $a \in F$, $U(a):= \{P \in Pt(F)\ |\ a \in P\}$. Then it's easily verified that $\{U(a)\ |\ a \in F\}$ is a topology on $Pt(F)$. Furthermore, if $g: F \rightarrow G$ is a frame homomorphism, then $\bar{g}: Pt(G)\rightarrow Pt(F)$ is a continuous map if $\bar{g}$ is defined by $\bar{g}(Q) = g^{-1}[Q]$.

Viewing these as functors from $\mathbf{Top}^{\mathrm{op}}\rightleftarrows \mathbf{Frm}$ provides us with an equivalence between the subcategories of sober spaces and spatial frames respectively.

A sober space is a topological space in which every irreducible closed subset of X is the closure of exactly one point of X. A spatial frame is a frame for which the completely prime filters separate elements in $F$, i.e.: for all $a \not\leq b \in F$, there exists a completely prime filter $P$ in $F$ such that $a \in P$ and $b \not\in P$.

In addition to Eric Wofsey's answer, another useful text is Picado J., Pultr A. Frames and locales. Topology without points (Birkhauser, 2012).

Answer 3

There is a description of $Top^{op}$ due to Barr and Pedicchio. They define a category of "grids", which are frames with additional unary operation. You can see part of the article in Google Books.

The way they motivate their construction is very straightforward, from a categorical point of view. They take a 3 element space (which they just call "3") that is a injective cogenerator for the topological spaces, and work out what properties you need to give an algebraic description of $Hom(X, 3)$ for an arbitrary topological space $X$. (If you instead take a 2 element space with one closed point, which they call "2", you get the category of frames instead.)

Answer 4

A very naive approach is to associate to a space $X$ ( locally compact and separated) the space $C(X)$ of continuous functions defined on $X$. A continuous map $f:X\rightarrow Y$ imduces a morphism $C(f): C(Y)\rightarrow C(X)$. If the space satisfies some nice conditions, maximal ideals of $C(X)$ can be identified with elements of $X$.

Now you have to identify the category whose objects are $C(X)$. You can consider the category of $R$-algebras. To a $R$-algebra $A$, you can associate the space $A(X)$ of its maximal ideals that you endow with the stronger topology for which the elements of $A$ are continuous.