How algebraic can the dual of a topological category be?

Question

(I'm going to try to use definitions from Abstract and Concrete Categories: The Joy of Cats by Adámek, Herrlich, and Strecker, since both of the adjectives in the title of my question seem to have at least three definitions in the context of category theory.)

Recall that given a category $X$, a concrete category over $X$ is a pair $(A,G)$ where $A$ is a category and $G : A \to X$ is a faithful functor (which we often call the forgetful functor). There is a notion of a concrete category being topological (AHS Definition 21.1, nLab) which abstracts the nice property of the category of topological spaces that limits and colimits are computed in a very uniform way relative to the forgetful functor to sets. In particular, limits and colimits are computed in $\mathrm{Set}$ and then lifted to $\mathrm{Top}$ by computing the initial or final topology relative to the diagram in question. In an arbitrary topological category $(A,G)$, one has access to a similar procedure. There are a lot of examples of this kind of category, such as uniform spaces with uniformly continuous maps, (extended) pseudometric spaces with $1$-Lipschitz maps, preorders with monotone maps, and measurable spaces with measurable maps (with this standard forgetful functors). Moreover, there are extensions of $\mathrm{Top}$ that are still topological but are also nicer as categories (such as the categories of convergence spaces and pseudotopological spaces, which are quasitoposes).

There is also a notion of a concrete category (or a functor more generally) being algebraic (AHS Def. 23.19, nLab), which abstracts the nice properties of categories of models of algebraic theories (such as groups, rings, vector spaces over a fixed field, etc.). There's also several related weaker and stronger conditions (e.g., monadicity and essential algebraicity). One of the nice aspects of the category of locales, $\mathrm{Loc}$, is that its dual (the category of frames) is monadic (over $\mathrm{Set}$) and therefore algebraic in this sense.

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My question is about the extent to which you can have a category that resembles both these $\mathrm{Top}$-like categories and $\mathrm{Loc}$. These two conditions are strong and having both of them would be nice, but there seems to be a degree of incompatibility between them. Specifically, as mentioned in AHS Example 23.6, if a topological category $(A,G)$ is algebraic (or just essentially algebraic), then $U$ is an equivalence of categories. Since I'm asking about the dual, this might not be an immediate problem, but the property of being a topological category is self-dual in the sense that $(A,G)$ is topological (over $X$) if and only if $(A^{\mathrm{op}},G^{\mathrm{op}})$ is topological (over $X^{\mathrm{op}}$). That said, $\mathrm{Top}^{\mathrm{op}}$ is a quasi-variety, which makes it surprisingly close to being algebraic over $\mathrm{Set}$. Note also that $\mathrm{Set}^{\mathrm{op}}$ is actually monadic over $\mathrm{Set}$, since it is equivalent to the category of completely distributive complete Boolean algebras.

All of the concrete categories in my questions are over $\mathrm{Set}$ (although a particularly interesting example where this is not the case would also be welcome).

Question 1. Is there a topological category $(A,G)$ such that $G$ is not an equivalence of categories and $A^{\mathrm{op}}$ is monadic? Algebraic? Essentially algebraic?

Given that topological categories don't need to resemble $\mathrm{Top}$ particularly, it's also natural to ask the following more specific question.

Question 2. Is there a topological category $(A,G)$ with $A^{\mathrm{op}}$ monadic or algebraic such that any of the common categories of spaces (e.g., compact Hausdorff spaces, compactly generated weak Hausdorff spaces, topological spaces, locales, etc.) is a full subcategory of $A$? In particular, are either of $\mathrm{Conv}^{\mathrm{op}}$ or $\mathrm{PsTop}^{\mathrm{op}}$ monadic or algebraic (where $\mathrm{Conv}$ is the category of convergence spaces and $\mathrm{PsTop}$ is the category of pseudotopological spaces)?

Answer 1

I have not properly digested your framework of definitions; the following example may or may not fit within it. (If not, it would be helpful to explain why not.)

Let $\operatorname{CH}$ be the category of compact Hausdorff spaces, and for $X\in\operatorname{CH}$ let $C(X)$ be the ring of continuous functions $X\to\mathbb{R}$. It is well-known that continuous maps $X\to Y$ biject with morphisms $\alpha\colon C(Y)\to C(X)$ of topological $\mathbb{R}$-algebras. It is slightly less well-known, but still classical, that any ring map $\alpha\colon C(Y)\to C(X)$ is a morphism of topological $\mathbb{R}$-algebras. Indeed, it is easy to see that $u\leq v$ in $C(X)$ iff $v-u$ is a square, and $\|u\|\leq 1/n$ iff $1-nu$ and $nu-1$ are both squares, so the ring structure determines the order and the topology, and the $\mathbb{R}$-algebra structure is determined by the $\mathbb{Q}$-algebra structure together with the topology, and the claim follows easily from this. Thus, we find that $\operatorname{CH}^{\operatorname{op}}$ is equivalent to a full subcategory of the category of commutative rings.

Answer 2

Check out the following papers.

• Barr and Pedicchio, $\text{Top}^\circ$ is a quasi-variety
• Barr and Pedicchio. Topological spaces and quasi-varieties.
• Dimov, Pedicchio, and Tironi. Frames and Grids.
• Adamek and Pedicchio. A remark on topological spaces, grids, and topological systems.

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I was asked to summarise the content of these papers in the comments. Let's do that. That said, the introduction of Topological spaces and quasi-varieties is simply great.

The main result of Topological spaces and quasi-varieties. is that $\text{Top}^\text{op}$ is a quasivariety. Let me recall a couple of theorems.

Let me stress that in this answer when I speak of varieties, I mean infinitary ones, no bounds whatsoever on the arity of the operations. So that even the category of frames is a variety. On this specific topic, I recommend reading the (appendix of the) beautiful paper by Brandeburg, Large limit sketches and topological space objects.

Thm. (Vitale, On the characterization of monadic categories over Set) A category $\mathcal{K}$ is monadic over $Set$ iff it satisfies the following properties.

• it is Barr-exact.
• there exists a regular projective object $P$, and all its tensors.
• $P$ is a regular generator.

Thm. (Pedicchio, On k-Permutability for categories of T-algebras) A category $\mathcal{K}$ is a (single sorted) quasi-variety iff it satisfies the following properties.

• it has finite limits and coequalizers of equivalence relations.
• there exists a regular projective object $P$, and all its tensors.
• $P$ is a regular generator.

Cool, so it's not exactly monadic over sets, but it is epi-reflective in a monadic category. That sounds great, but also a bit too general. Q. What kind of quasi-variety is it? It's easy to describe the variety in which it's reflective. That's the variety of Grids (Barr and Pedicchio, $\text{Top}^\circ$ is a quasi-variety). A Grid $\mathbb{G} = (G,c)$ is a couple where $G$ is a frame and $c$ is an additional unary operator sotsfying a bunch of additional properties.

Answer 3

First of all, I was somewhat confused about terminologly when I originally wrote the question. Quasivarieties are algebraic (in the sense of Adámek-Herrlich-Strecker), so $\mathrm{Top}^{\mathrm{op}}$ is algebraic over $\mathrm{Set}$. This only leaves the strongest form of the question, whether $A^{\mathrm{op}}$ can be monadic over $\mathrm{Set}$ for a topological category $A$ (in other words whether $A$ can be a covariety).

It turns out that this can happen; there is a non-trivial topological covariety. Since topological categories are always complete and cocomplete, the result of Vitale mentioned in Ivan's answer implies that a topological category $A$ is a coquasivariety if and only if it satisfies the following:

1. there is a regular injective object $I$ that is a regular cogenerator and
2. (coregularity) pushouts of regular monomorphisms are regular monomorphisms.

And $A$ is a covariety if it additionally satisfies the following:

3. (Barr-coexactness) cocongruences are cokernel pairs.

In the context of topological categories, 1 implies 2 (which seems like something which would be a known fact, but I couldn't find it anywhere): Fix $G : A \to \mathrm{Set}$, a topological functor, and $I$, a regular injective regular cogenerator of $A$.

Lemma. For any objects $B$ and $C$ of $A$, a function $f : GB \to GC$ lifts to a morphism in $A$ if and only if for every $g : C \to I$, $(Gg) \circ f$ lifts to a morphism in $A$.

Proof. Obviously if $f = Gh$ for some morphism $h : B \to C$, then for every $g : C \to I$, $(Gg) \circ f = G(g \circ h)$. Conversely, assume the lifting condition in the statement of the lemma. Let $X = \prod_{g : C \to I} I$. Let $i : C \to I$ be the canonical map satisfying $\pi_g \circ i = g$ for each projection $\pi_g : \prod_{g : C \to I} I \to I$. Note that since $I$ is a regular cogenerator, $i$ is a regular monomorphism (i.e., an isomorphism onto its image). By our assumption and the universal property of products, we get a morphism $h : B \to X$ satisfying that for each projection $\pi_g : X \to I$, $G(\pi_G \circ h) = (Gg) \circ f$. The set-theoretic image of $h$ is contained in the set-theoretic image of $i$. Therefore we have a morphism $i^{-1} \circ h : B \to C$, but this is precisely the required lift of $f$. $\square$

The lemma implies that the isomorphism type of any object $B$ in $A$ is uniquely determined by its underlying set $GB$ and the set of 'continuous' maps from $B$ to $I$. Moreover, we get that a map $f : B \to C$ is a regular monomorphism if and only if every $g : B \to I$ extends to an $h : C \to I$ (i.e., an $h$ such that $g = h \circ f$).

Proposition. $A$ is coregular (and so in particular $A^{\mathrm{op}}$ is a coquasivariety).

Proof. Since $G$ is topological, $A$ is complete and cocomplete, so we only need to verify that pushouts of regular monomorphisms are regular monomorphisms. Let $f : B \to C$ be a regular monomorphism, let $g : B \to D$ be an arbitrary morphism, and let $E$ be the pushout in the following diagram: $\require{AMScd}$ \begin{CD} B @>g>> D\\ @V f V V @VV i_D V\\ C @>>i_C> E \end{CD}

We need to show that $i_D$ is a regular monomorphism. $G$ is a continuous functor and preserves monomorphisms, so we have that the underlying set of $E$ is the pushout of $Gf : GB \to GC$ and $Gg : GB \to GD$. In other words we can identify it with $D \sqcup (C \setminus f[B])$ (abusing notation a little bit). By the lemma, all we need to show is that every 'continuous' map from $D$ to $I$ extends to a 'continuous' map from $E$ to $I$ (by the comment before the statement of the proposition). Fix $h : D \to I$. Since $I$ is a regular injective object, we can find $i : B \to I$ such that $i \circ f = h \circ g$. Let $j: E \to I$ be the map given by the universal property of $E$ applied to $h$ and $i$. This is the required extension of $h$. Since we can do this for every $h : D \to I$, $g : D \to I$ is a regular monomorphism. $\square$.

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So this means that all we need to do is find a topological category with a regular injective regular cogenerator in which cocongruences are cokernel pairs. I originally worked out the following example as a subcategory of the category of pseudotopological spaces, but then I realized it was just equivalent to the category of pretopological spaces. Recall that a pretopological space is a set $X$ together with a function $N$ from $X$ to filters on $X$ satisfying that for each $a \in X$, $X \in N(a)$ and $a \in B$ for every $B \in N(a)$. The elements of $N(a)$ are called neighborhoods of $a$. A function $f : A \to B$ is continuous if for each $a \in A$, the preimage of any neighborhood of $f(a)$ is a neighborhood of $a$. Let $\mathrm{PrTop}$ denote the category of pretopological spaces with continuous maps and let $G : \mathrm{PrTop} \to \mathrm{Set}$ be the obvious forgetful functor.

Given pretopological spaces $A$ and $B$, a map $f : A \to B$ is a regular monomorphism if and only if it is an injection and for each $a \in A$, $N(a) = \{f^{-1}[U] : U \in N(f(a))\}$ (which is a filter since $f$ is an injection). Given a pretopological space $B$ and a subset $X$ of $GB$, this also describes the canoncial subspace pretopology on $X$ induced by the pretopology on $B$.

Let $I$ be the pretopological space with underlying set $\{0,1,2\}$ with $N(0)$ the filter generated by $\{0,1\}$ and $N(1) = N(2) = \{\{0,1,2\}\}$.

Proposition. $I$ is a regular injective object in $\mathrm{PrTop}$.

Proof. Let $B$ be a pretopological space and let $A$ be a subspace of $B$. Let $f : A \to I$ be a continuous map. Let $g : B \to I$ be defined by $g(x) = f(x)$ if $x \in A$ and $g(x) = 1$ if $x \notin A$. We need to verify that $g$ is continuous. Fix $b \in B$. If $b \notin A$ or if $b \in A$ but $f(b) \neq 0$, then $N(g(b)) = \{I\}$, so the preimage of any neighborhood of $g(b)$ is a neighborhood of $b$. If $b \in A$ and $f(b) = 0$, then we have that $\{0,1\}$ is the only non-trivial neighborhood of $f(b)$. Since $f$ is continuous, $f^{-1}[\{0,1\}]$ is a neighborhood of $b$ in $A$. Since $A$ is a subspace of $B$, there is some neighborhood $U$ of $b$ in $B$ such that $A \cap U = f^{-1}[\{0,1\}]$, but $g^{-1}[\{0,1\}] = f^{-1}[\{0,1\}] \cup (B \setminus A)$, so we have that the $g$-preimage of the only non-trivial neighborhood of $g(b) = f(b)$ is a neighborhood of $b$ in $B$. Since we can do this for any such point, $g$ is continuous. $\square$

Proposition. $I$ is a regular cogenerator of $\mathrm{PrTop}$.

Proof. Fix a pretopological space $A$. Let $X = \prod_{f : A \to I} I$ and let $g : A \to X$ be the canonical map satisfying $\pi_f \circ g = f$ for each $f : A \to I$.

First note that for any set $B \subseteq A$, the indicator function of $B$ (thought of as a $\{0,1\}$-valued function to $I$) is continuous, so continuous maps from $A$ to $I$ separate points and $g$ is an injection. Finally, it is immediate that for any $a \in A$ and $U \in N(a)$, the map $f : A$ defined by $f(a) = 0$, $f(x) = 1$ for $x \in U \setminus \{a\}$, and $f(x) = 2$ otherwise is continuous. Therefore for any $a \in A$ and $U \in N(A)$, we have that $g[U]$ is a neighborhood of $g(a)$ in the subspace $g[A]$ and so $g$ is an isomorphism onto its image, which is a regular subobject of a power of $I$. Since we can do this for any $A$, $I$ is a regular cogenerator of $\mathrm{PrTop}$. $\square$

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Now we just need to show that cocongruences are cokernel pairs. A cocongruence on an object $A$ is a map $r : A \oplus A \to R$ satisfying that for any object $B$, the set $$ \{(f,g) \in \mathrm{hom}(A,B)^2 : (\exists h : R \to B)f \oplus g = h \circ r\} $$ is an equivalence relation on $\mathrm{hom}(A,B)$. This is also witnessed by maps $e : R \to A$, $s : R \to R$, and $t : R \to R \oplus_A R$ corresponding to reflexivity, symmetry, and transitivity of the equivalence relation). Morally speaking, a cocongruence on $A$ is something that looks like it should be the pushout $A \oplus_B A$ for some regular subobject $B$ of $A$. What it means for this to be a cokernel pair is that it actually is. $R$ is a cokernel pair if when we let $i:E \to A$ be the equalizer of the component maps $r_0,r_1 : A \to R$, then $R$ is the pushout in the following diagram: $\require{AMScd}$ \begin{CD} E @>i>> A\\ @V i V V @VV r_1 V\\ A @>>r_0> R \end{CD} This of course fails in $\mathrm{Top}$ and the key is that pushouts in $\mathrm{PrTop}$ are 'floppier' than pushouts in $\mathrm{Top}$, so the only way to have a cocongruence on $A$ is to actually specify a subspace $B$ of $A$ and let $R = A \oplus_B A$.

It will be useful in a minute to have a precise characterization of pushout pretopologies over subspaces. Fix a pretopological space $A$ and a subspace $B \subseteq A$. The pretopology on $A \oplus_B A$ is the finest (i.e., biggest neighborhood filters) pretopology that makes the two inclusion maps $i_0,i_1 : A \to A \oplus_B A$ continuous. This is accomplished by letting neighborhoods of $a \in i_0[B]=i_1[B]$ be sets $U \subseteq A \oplus_B A$ satisfying that $i_0^{-1}[U]$ and $i_1^{-1}[U]$ are neighborhoods of $i_0^{-1}(a) = i_1^{-1}(a) \in B$ and neighborhoods of $a \in i_0[A \setminus B]$ be sets $U \subseteq A \oplus_B A$ satisfying that $i_0^{-1}[U]$ is a neighborhood of $i_0^{-1}(a)$.

(Note that you can already see the floppiness of pretopological pushouts relative to topological pushouts in the example of gluing two two-point indiscrete spaces at a point. Topologically this gives you the three-point indiscrete space, but pretopologically this gives you $\{0,1,2\}$ where $N(0)$ is generated by $\{0,1\}$, $N(1)$ is generated by $\{0,1,2\}$, and $N(2)$ is generated by $\{1,2\}$. As wrote up in an MSE answer recently, this is related to the easiest example of a cocongruence in $\mathrm{Top}$ that is not a cokernel pair..)

One thing to note for the following proof is that any topological functor $G : A \to \mathrm{Set}$ maps cocongruences to cocongruences. Cocongruences in $\mathrm{Set}$ are precisely pushouts of subsets.

Proposition. Cocongruences in $\mathrm{PrTop}$ are cokernel pairs.

Proof. Let $r : A \oplus A \to R$ be a cocongruence with component maps $r_0,r_1 : A \to R$. Since $R$ is a cocongruence, $r_0$ and $r_1$ are both regular monomorphisms (i.e., isomorphisms onto their image subspaces). By the comment just before the proposition, $Gr : G(A \oplus A) \cong GA \sqcup GA \to GR$ is a cocongruence in $\mathrm{Set}$. Let $Gi : GB \to GA$ be the equalizer of $Gr_0$ and $Gr_1$ and let $B$ be the corresponding pretopological subspace (with inclusion map $i : B \to A$). By the universal property of pushouts, we get a map $f : A \oplus_B A \to R$ which is a bijection of underlying sets (and more specifically is an isomorphism of $G(A \oplus_B A)$ and $GR$ as cocongruences in $\mathrm{Set}$). Therefore we know that the pretopology on $R$ is at least as coarse as the pushout pretopology and we need to show that it is no coarser. In particular this means that we need to show that for any point $a$ in $R$ (whose underlying set we can think of as $GA \sqcup_{GB} GA$), any neighborhood of $a$ in the pretopology $A \oplus_B A$ is a neighborhood of $a$ in the pretopology $R$.

Let $C$ be the pushout in the following diagram: $\require{AMScd}$ \begin{CD} A @>r_0>> R\\ @V r_1 V V @VV j_0 V\\ R @>> j_1 > C \end{CD} Since $R$ is a cocongruence, there must be a map $t : R \to C$ satisfying that $j_0 \circ r_1 = t \circ r_1$ and $j_1 \circ r_0 = t \circ r_0$. (This is the map witnessing transitivity of $R$ in general.)

Identify the underlying set of $C$ with $B \sqcup ((A \setminus B) \times \{0,1,2\})$, where $B \sqcup ((A \setminus B) \times \{0,1\})$ is the image of $j_0$ and $B \sqcup ((A \setminus B) \times \{1,2\})$ is the image of $j_1$. Let $C_k = B \sqcup ((A \setminus B) \times \{k\})$ for each $k < 3$. Note that there is only one relevant pretopology on each $C_k$ (the pretopology corresponding to that on $A$), and this is the subspace pretopology of $C_k$ relative to any of the other pretopologies considered in this argument. Note also that $C_0 \cup C_2$ is the image of $t$.

Fix a point $a \in R$ and consider $t(a) \in C_0 \cup C_2$, and let $U$ be a neighborhood of $a$ in the pushout pretopology on $C_0 \cup C_2 = B \sqcup ((A \setminus B) \times \{0,2\})$ (i.e., the pretopology on $A \oplus_B A$). Let $V = U \cup ((A \setminus B)\times \{1\})$. We need to argue that $V$ is a neighborhood of $t(a)$ in the pretopology on $C$. We have that for both $k \in \{0,2\}$, if $t(a) \in C_k$, then $U \cap C_k$ is a neighborhood of $t(a)$ in the pretopology on $C_k$. Therefore, if $t(a) \in C_0$, then $(U \cap C_0) \cup ((A \setminus B) \times \{1\})$ is a neighborhood of $t(a)$ in the pretopology coming from $R$ (since $C_0$ is a subspace), and likewise if $t(a) \in C_2$, then $(U \cap C_2) \cup ((A \setminus B) \times \{1\})$ is a neighborhood of $t(a)$ in the pretopology coming from $R$. Therefore, by our characterization of pushout pretopologies, $V$ is a neighborhood of $t(a)$ in the pushout pretopology on $C = C_0 \cup C_1 \cup C_2$. Since $t$ is continuous, this implies that $t^{-1}[V] = t^{-1}[U]$ is a neighborhood of $a$ in $R$. Since we can do this for any $a \in R$ and any neighborhood of $t(a)$ in the pushout pretopology on $C_0 \cup C_2$, we have that the map $f$ is an isomorphism and so in particular $R$ is the cokernel pair of the inclusion map of $B$ into $A$. $\square$

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So now we can conclude that $A^{\mathrm{op}}$ is monadic over $\mathrm{Set}$ by Vitale's charcterization. In particular, the monad sends a set $X$ to the pretopological space $I^X$. It should be possible to give an explicit algebraic description of the dual category, analogously to the grids of Barr and Pedicchio, but I don't know a clean presentation. Also, since $\mathrm{Top}$ is a full subcategory of $\mathrm{PrTop}$, this gives a positive answer for several of the categories in my second question.

I'm curious about the extent to which this can be pushed further. For instance, can a category satisfying these conditions be a quasitopos (which $\mathrm{PrTop}$ is not)? Bigger generalizations of $\mathrm{Top}$ (such as $\mathrm{Conv}$ and $\mathrm{PsTop}$) also seem like they should have the requisite 'floppiness' to have all cocongruences be cokernel pairs (which is why I was trying to find an example among subcategories of $\mathrm{PsTop}$), but the existence of a regular injective regular cogenerator is far less clear. This answer to an older question of mine makes me suspect there cannot be any such thing for $\mathrm{PsTop}$.