## April 19, 2019

### Can 1+1 Have More Than Two Points?

#### Posted by John Baez

I feel I’ve asked this before… but now I really want to know. Christian Williams and I are working on cartesian closed categories, and this is a gaping hole in my knowledge.

Question 1. Is there a cartesian closed category with finite coproducts such that there exist more than two morphisms from $1$ to $1 + 1$?

Cartesian closed categories with finite coproducts are a nice context for ‘categorified arithmetic’, since they have $0$, $1$, addition, multiplication and exponentiation. The example we all know is the category of finite sets. But every cartesian closed category with finite coproducts obeys what Tarski called the ‘high school algebra’ axioms:

$x + y \cong y + x$

$(x + y) + z \cong x + (y + z)$

$x \times 1 \cong x$

$x \times y \cong y \times x$

$(x \times y) \times z \cong x \times (y \times z)$

$x \times (y + z) \cong x \times y + x \times z$

$1^x \cong 1$

$x^1 \cong x$

$x^{(y + z)} \cong x y \times x z$

$(x \times y)^z \cong x^z \times y^z$

$(x^y)^z \cong x^{(y \times z)}$

together with some axioms involving $0$ which for some reason Tarski omitted: perhaps he was scared to admit that in this game we want $0^0 = 1$.

So, one way to think about my question is: how weird can such a category be?

For all I know, the answer to Question 1 could be “no” but the answer to this one might still be “yes”:

Question 2. Let $C$ be a cartesian closed category with finite coproducts. Let $N$ be the full subcategory on the objects that are finite coproducts of the terminal object. Can $N$ be inequivalent to the category of finite sets?

In fact I’m so clueless that for all I know the answer to Question 1 could be “no” but the answer to this one might still be “yes”:

Question 3. Is there a cartesian closed category with finite coproducts such that there exist more than three morphisms from $1$ to $1 + 1 + 1$?

Or similarly for other numbers.

Question 4. Is there a category with finite coproducts and a terminal object such that there exist more than two morphisms from $1$ to $1 + 1$?

Just to stick my neck out, I’ll bet that the answer to this last one, at least, is “yes”.

Posted at April 19, 2019 4:43 AM UTC

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### Re: Can 1+1 Have More Than Two Points?

In the category Set^2 of pairs of sets, the terminal object is the pair (1,1), and “1+1” is the object (2,2). But hom((1,1),(2,2)) = 4.

Posted by: Theo Johnson-Freyd on April 19, 2019 6:11 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Ugh! I just woke up, at 4:20 a.m., and thought of this embarrassingly simple example… but you beat me to it.

Posted by: John Baez on April 19, 2019 12:24 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

The Kleisli category of the Giry monad is not a CCC. To circunvect this Chris Heunen has proposed here what he calls quasi-Borel spaces, whose category is CCC and has (countable) coproducts. There |hom(1, 1+1)| > 2.

Posted by: Jesús López on April 19, 2019 9:30 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Hi, Thanks for referring to our work, but for quasi-Borel spaces, |hom(1,1+1)|=2, and the Kleisli category of our monad isn’t a ccc.

Nonetheless this does suggest ways to get nice answers to Question 4 and John’s new question, via Kleisli categories of fancy affine monads (even on Set).

Posted by: Sam Staton on April 19, 2019 2:49 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Thanks Sam. My bad here, it’s the quasiborel spaces with the analogue of measurable maps what are a ccc, I was conflating the maps with those that in the Giry monad are Markov Kernels.

Posted by: Jesús López on April 19, 2019 5:05 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Gabriel Scherer recently proved that equality of morphisms in the free bi-cartesian closed category is decidable (this particular theory comes up in our work on categorical databases). He worked out many model-theoretic consequences that might shed light on your question.

Posted by: Ryan Wisnesky on April 19, 2019 9:33 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Okay so you can have 4 morphisms, but can you have something that isn’t a power of two, such as 3?

Posted by: Alice Ryhl on April 19, 2019 10:55 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

That’s a way to make the question more interesting! Or: can you find a category such that $hom(1,1+1)$ has two elements but $hom(1,1+1+1)$ has more than three?

Posted by: John Baez on April 19, 2019 12:57 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Consider the variety of algebras defined by a single $n$-ary operation $t(x_1,\ldots,x_n)$ satisfying equations which say that as soon as two of its arguments coincide, then the operation must return (say) its first argument.

Now by the pigeonhole principle, the free algebra on $k \lt n$ generators is just the $k$-element set of generators. But a free algebra on at least $n$ generators will contain extra elements! It’s easy to see that each one of these also defines an algebra morphism $1 \to 1 + \ldots + 1$. By including two such operations as above, you can also nest them and thereby make the hom-set from $1$ to the $n$-fold $1 + \ldots + 1$ become infinite.

More generally, perhaps it’s an interesting question which matrices $(c_{i j})_{i,j\in\mathbb{N}}$ can occur as the cardinalities of the hom-sets in a Lawvere theory. There is the obvious constraint $c_{i,j + k} = c_{i,j} \cdot c_{i,k}$. Is this condition also sufficient?

Posted by: Tobias Fritz on April 19, 2019 7:55 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

So cooking up suitable algebraic theories answers many versions of Question 4 relatively easily in the positive. However, it doesn’t seem to say much about the cartesian closed case, since these categories are rarely cartesian closed.

Posted by: Tobias Fritz on April 19, 2019 8:00 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Hey, that’s reall cool, Tobias! Too bad these categories are not generally cartesian closed.

Digressing somewhat, I thought I saw somewhere recently a result on when the category of algebras of a Lawvere theory is cartesian closed. Does anyone remember?

Posted by: John Baez on April 19, 2019 10:52 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

I don’t know about any recent papers, but Peter Johnstone’s “Collapsed toposes and cartesian closed varieties” gives one possible answer to this. The characterization is a bit too complicated to state here (at least with my mediocre expository skills). But there seem to be intriguing connections with topos theory: if a variety of algebras is locally cartesian closed, then it is already a topos!

Posted by: Tobias Fritz on April 20, 2019 12:19 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Take the category of totally ordered sets, and pairs of functions (f, g) where forall x, f x <= g x as the morphisms.

Then there are three points in 2: (F,F), (F,T), (T,T).

But is 2 equivalent to 1+1 in this category?

Posted by: chris andrew upshaw on April 19, 2019 6:03 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

I don’t understand what you mean: if $2$ is the totally ordered $2$-element set (the ordinal $2$) then it has $2$ points. Maybe you mean something else by $2$?

The ordinal $2$ is not the coproduct $1 + 1$ in the category of totally ordered sets, since for the coproduct any pair of morphisms $f, g : 1 \to X$ gives a morphism from $1 + 1$ to $X$, whereas only pairs with $f \le g$ give morphisms from $2$ to $X$.

The coproduct $1 + 1$ just doesn’t exist in the category of totally ordered sets. It exists in the category of posets, but there we have two morphisms from $1$ to $1+1$.

Posted by: John Baez on April 19, 2019 6:15 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Man I wrote that really obtusely, Wait, what I was thinking of isn’t even a category, because composition wouldn’t maintain the property I was thinking of.

I was trying to basically construct a subcategory of Set^2 that would have only 3 of the 4 morphisms from 1 to 2 in it, but still be a reasonable category.

Sorry to take up your time.

The category of sets and partial functions has 3 morphisms from 1 to 2. And I am pretty sure it has co-products, (and 2 = 1 + 1).

Posted by: christopher andrew upshaw on April 20, 2019 4:36 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Note that the terminal object in the category of sets and partial functions is the empty set, not the one-element set.

Posted by: Mike Shulman on April 20, 2019 8:46 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Take the category of meet-semilattices (not necessarily bounded). Then $1+1$ is the free semilattice on two generators, and it has three elements, or equivalently three morphisms $1 \to 1 + 1$: the two generating elements together with their meet.

However, this category is not cartesian closed, since $(1 + 1) \times -$ does not preserve coproducts: $(1 + 1) \times (1 + 1)$ has 9 elements, while $(1 + 1) + (1 + 1)$ has 15 elements.

Posted by: Tobias Fritz on April 19, 2019 7:33 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Nice. Tying it to Jesús’ comment above: A related example is the category of sets and total relations, i.e. the Kleisli category for the non-empty-powerset monad. (Since you only used free semilattices.)

Posted by: Sam Staton on April 19, 2019 8:31 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

If your category is cartesian closed (just distributive would do) then 1+1 is an internal Boolean algebra, and so Hom(1,1+1) is an ordinary Boolean algebra. So if it’s finite, it must have 2^n elements for some n.

Posted by: Richard Garner on April 20, 2019 3:19 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Very nice! Thanks—that’s just what I was missing.

Now I’m wondering this. Let $C$ be a cartesian closed category with finite coproducts, with all its objects being finite coproducts of $1$. You showed how $C$ determines a Boolean algebra. Does the Boolean algebra determine $C$ up to equivalence? Or could $hom(1,1+1+1)$ hold further mysteries, not revealed by the Boolean algebra structure of $hom(1,1+1)$?

Also: does any Boolean algebra give rise such a category $C$?

I can probably figure this stuff out, but I bet you’d be a lot faster.

Posted by: John Baez on April 20, 2019 7:00 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Good questions!

If the objects are $C$ are the finite coproducts of $1$ then it’s probably not going to be cartesian closed. But it could at least be distributive.

I can now answer your second question: yes, every Boolean algebra $B$ gives rise to such a category $C$. Take the category whose objects are the natural numbers, and where a map $n \rightarrow m$ is an $m \times n$ matrix with entries from $B$ wherein every column is an $m$-fold disjoint decomposition of $1 \in B$. Composition is matrix multiplication. This is a distributive category with $C(1,1+1) \cong B$. Compare to Example A1.4.6 in the Elephant.

As to your first question: I don’t see why in a general distributive $C$ it need be true that $C(1,1+1)$ would determine $C(1,1+1+1)$. The extra axiom you would need is that the equaliser of the first projection and the join map $(1+1)^2 \rightrightarrows 1+1$ is given by $1+1+1$. This is true in sets, and of course if it is true in $C$ then the maps $1 \rightarrow 1+1+1$ are determined by those $1 \rightarrow 1+1$. I suspect that this additional axiom is probably sufficient to determine all the homs in $C$.

I don’t currently have a counterexample where this doesn’t happen. Maybe there is one from Bart Jacobs’ theory of effectuses (where one has only a weaker form of the axiom I listed above).

Posted by: Richard Garner on April 22, 2019 7:48 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Hi. Could it be that any distributive category $D$ where all the objects are copowers of 1 is necessarily a full distributive subcategory of an extensive category? My argument for this follows. It’s a bit convoluted and there’s surely a direct route if it’s true.

Recall that since $D$ is a distributive category, the category of functors $FP(D^{op},Set)$ is a cartesian closed category with sums that contains $D$ as a full distributive subcategory. (This is true for any distributive category.) Moreover in this case $D$ is the dual of a Lawvere theory, and so $FP(D^{op},Set)$ is a cartesian closed variety.

Using Johnstone’s “Collapsed toposes…” Theorem 1.2 and Lemma 5.1, every cartesian closed variety admits a full and faithful embedding into a topos that preserves colimits and finite products. Thus $D$ is embedded into a topos.

So by your argument, $D(1,n)$ is determined by $D(1,2)$.

Posted by: Sam Staton on April 22, 2019 11:16 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Very nice! That checks out, though it seems almost too good to be true. When I get a moment I will try and unravel the construction in “Collapsed toposes” and see if it leads to a more direct argument.

Posted by: Richard Garner on April 23, 2019 3:15 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

NB if $C$ an extensive category (i.e., has pullbacks along coproduct projections, and pulling back along $A \rightarrow A+B \leftarrow B$ induces an equivalence $C/(A+B) \rightarrow C/A \times C/B$) then the equaliser axiom I describe above always holds. Unfortunately, the objects of $C$ are then not going to be just the coproducts of $1$, but in any case…

Proof: giving a map $X \rightarrow (1+1)^2 = 1+1+1+1$ is by extensivity the same as giving a fourfold coproduct decomposition of $X$ as $X = X_1+X_2+X_3+X_4$. To say that the composites $X \rightarrow (1+1)^2 \rightrightarrows 1+1$ are the same is to say that $(X_1+X_2,X_3+X_4) \cong (X_1,X_2+X_3+X_4)$. In an extensive category the only way this can happen is if $X_2\cong 0$ and because of this, $X \rightarrow (1+1)^2$ factors through $1+0+1+1 \rightarrow 1+1+1+1$.

Posted by: Richard Garner on April 22, 2019 7:52 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

A Grothendieck topos is called connected if the full subcategory on the coproducts of 1 is equivalent to the category of sets. The topos of sheaves on any non-connected space is not connected, ergo an answer to your question. Theo’s answer (sheaves on the two point discrete space) is the simplest!

If the topos is locally connected (ie has a connected components functor to Set left adjoint to the functor taking copowers of 1) then it’s connected just when the terminal object has a single connected component, and this happens just when there are exactly two maps 1->2. So in the locally connected case, if there are exactly two maps 1->2 there must be exactly three maps 1->3 and so on.

Posted by: Richard Garner on April 20, 2019 3:01 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Richard mentioned extensivity, and I’m surprised no one else did. Surely to categorify elementary algebra you don’t want just cartesian closed plus finite coproducts? Wouldn’t there be examples of such categories that are not extensive, hence the analogue of distributivity $x\times (y+z) \simeq x\times y + x\times z$ fails?

Posted by: David Roberts on April 22, 2019 10:15 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Hmm, but being cartesian closed means $x\times (-)$ has a right adjoint so preserves colimits.

Now that I think about it, I should have consulted the canonical source before starting to type :-S.

Posted by: David Roberts on April 22, 2019 10:48 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Conjecture 1. Let $C$ be a cartesian closed category where every object is a finite coproduct $1 + \cdots + 1$. Suppose $C$ has finite homsets. Then for some $k$, $C$ is equivalent to the full subcategory of $FinSet^k$ whose objects are finite coproducts of the terminal object.

Richard showed under these assumptions that $hom(1,1+1)$ has $2^k$ elements. Remember his argument. He showed that $1 + 1$ could be given the structure of a Boolean algebra object in $C$ in an obvious way. Then applying the functor

$hom(1, -) : C \to FinSet$

the set $hom(1,1+1)$ becomes a Boolean algebra. By the classification of finite Boolean algebras, it must be $\{F,T\}^k$ for some finite $k$.

Can we generalize this proof? The first trick is this: ‘Boolean algebra’ is just a name for ‘set equipped with all the operations you can do on a 2-element set, obeying all the equations those operations obey’. Then comes the hard (?) work of showing that any such set is a finite product of copies of the 2-element set, with operations defined componentwise.

So, let’s try to copy this.

Define an $T_n$-algebra to be a set equipped with all the operations you can do on an $n$-element set, obeying all the equations these operations obey.

In other words: there is a full subcategory $T_n$ of $FinSet$ where the objects are sets of cardinality $n^k$ for $k = 0, 1, \dots$. This category $T_n$ has finite products, and every object is a finite product of copies of $n$. Thus $T_n$ becomes a Lawvere theory whose $k$-ary operations are all the functions $f: n^k \to n$. A $T_n$-algebra is an algebra of this Lawvere theory.

Conjecture 2. Every $T_n$-algebra in the category of finite sets is a finite product of copies of $n$, where $n$ has been made into a $T_n$-algebra in the canonical way.

I like this conjecture. It generalizes the classification of finite Boolean algebras, which is the case $n = 2$. And I believe it implies Conjecture 1, roughly as follows.

Let $C$ be any category of the sort discussed in Conjecture 1. The $n$-fold coproduct $1 + \cdots + 1$ can be given the structure of a $T_n$-algebra in $C$. Thus, its set of points $hom(1,1 + \cdots + 1)$ becomes a $T_n$-algebra in $Set$. Using Conjecture 2 we conclude that if it’s a $T_n$-algebra in $FinSet$, it’s actually a $k$-fold product of copies of $n$. And with a little extra fussing around this should imply that $C$ is equivalent to the full subcategory of $FinSet^k$ where the objects are finite coproducts of the terminal object.

By the way, I don’t have any objection to the case where $C$ has infinite homsets! If you can handle that case, I’d love to hear about it! I just have a feeling it’s a bit more delicate, since I don’t know the full classification of infinite Boolean algebras.

Posted by: John Baez on April 23, 2019 3:58 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

I have a feeling that Richard and Sam’s work starting up here is an alternative route to tackling Conjectures 1 and 2.

Posted by: John Baez on April 23, 2019 4:01 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Maybe I’m missing something obvious in this discussion, but I thought it was the case that any $[n]$ (a coproduct of $n$ copies of the terminal object $[1]$, with $n \geq 1$) is, in a distributive category with products and coproducts, a retract of some suitable $[2]^k = [2^k]$ (we could take $k = n$ for instance).

In other words, any such $[n]$ belongs to the Cauchy completion of the category consisting of products of $[2]$.

Take for example $[3] = [1] + [1] + [1]$. The first summand $[1]$ is a retract of $[2]$ in a pretty easy way. Applying the functor $- + [1] + [1]$ preserves this retract, so $[3]$ is a retract of $[4]$.

This should mean that $\hom([1], [3])$ is determined by the Boolean algebra $B = \hom([1], [2])$. Indeed, it’s obtained by splitting the idempotent

$B \times B \stackrel{\langle \pi_1, \wedge\rangle}{\to} B \times B$

which I found simply by analyzing the previous paragraph.

Here is a theorem which might be folklore but which seems to me to be relevant:

• Let $T$ be a category with finite products, and $\bar{T}$ its Cauchy completion. Then $\bar{T}$ also has finite products, and for any Cauchy complete category with products $D$, the category $Prod(T, D)$ of product-preserving functors $T \to D$ is equivalent to $Prod(\bar{T}, D)$.

We could take $D = Set$ for instance. Now, if we take $T_2$ to be the Lawvere theory given by the full subcategory in $C$ whose objects are finite products of $[2]$, we get the same Cauchy completion $\bar{T}$ as we would if we took $T_n$, the Lawvere theory generated by $[n]$:

$Mod_{T_2} = Prod(T_2, Set) \simeq Prod(\bar{T}, Set) \simeq Prod(T_n, Set) = Mod_{T_n}.$

I have a vague memory that we were discussing this sort of thing when Tom was posting about ultrafilters and codensity monads and the like, but I’m having trouble locating it just now.

Posted by: Todd Trimble on April 24, 2019 12:50 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

I have a vague memory that we were discussing this sort of thing when Tom was posting about ultrafilters and codensity monads and the like, but I’m having trouble locating it just now.

Was it here, or maybe here?

Posted by: Tom Leinster on April 24, 2019 2:14 PM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

So in case the message of my last post didn’t transmit well: if we suppose that the objects of the form $[2^n]$ form a category equivalent to the one we get by looking at those objects in the case $C = Fin^k$ for some finite $k$, then the same is true for all the other $[m]$. That is, the category of all the $[m]$ is a full subcategory of the Cauchy completion of the category of the $[2^n]$, and since $Fin^k$ is itself Cauchy complete, it’s the same as full subcategory of $Fin^k$ consisting of the $[m]$, i.e., the finite (nonempty) coproducts of $[1]$ in $Fin^k$.

Part of my point is also that there is nothing special about $2$ here. You could replace $2$ by any $n \geq 2$ and get an analogous statement.

Posted by: Todd Trimble on April 24, 2019 11:57 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Duh, yes, of course – the equaliser defining $1+1+1$ in terms of $(1+1)^2$ is split, hence absolute. So as Todd rightly says, $C(1,1+1)$ determines $C(1,1+1+1)$ whenever $C$ is distributive. So, to summarise: $C$ is distributive with as objects are the finite copowers of $1$ if and only if it is the category of Boolean-valued matrices for the Boolean algebra $B = C(1,1+1)$.

In fact, in combination with Sam’s remarks above, this turns out to be one of the class of examples in the Johnstone paper on “Collapsed toposes”.

As Sam points out, if $C$ is distributive and its objects are the copowers of $1$, then $E = FP(C^{op},Set)$ is a cartesian closed variety. Moreover, because $C(1,1) = 1$, this variety is affine, meaning that every one-element subset is a subalgebra, or equivalently that the terminal algebra is a projective generator. Conversely, if $E$ is an affine cartesian closed variety, then it is equivalent to $FP(C^op, Set)$ where $C$ is the full subcategory of $E$ on the coproducts of $1$.

So: distributive categories $C$ whose objects are copowers of $1$ are the same as cartesian closed affine varieties. Example 8.8 of “Collapsed toposes” notes that $E$ is a cartesian closed affine (finitary) variety precisely if and only if it is the category of sheaves on a Boolean algebra $B$ which are either empty or have global support. The $B$ in question is, of course, going to $E(1,1+1)$. There is, moreover, a simple syntactic characterisations of such varieties: they are the categories of models of commutative hyperaffine theories (where hyperaffine is defined in the Johnstone paper).

At the end of Example 8.8, reference is made to Bergman’s “Actions of Boolean rings on sets” which basically works out everything that is relevant to this situation. It is good stuff and worth a read.

Posted by: Richard Garner on April 26, 2019 1:12 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Another point. If $C$ is a distributive category whose objects are the copowers of $1$, then the models of the theory $C^{op}$ are, as discussed, sheaves of global support on the Boolean algebra $C(1,1+1)$ together with the empty model.

This is related to the theory of skew Boolean algebras. A non-empty model of $C^{op}$ can also be described as a skew Boolean algebra whose Boolean algebra quotient is $C(1,1+1)$. See the paper “Boolean sets, skew Boolean algebras and a non-commutative Stone duality” by Kudryavtseva and Lawson.

Posted by: Richard Garner on April 26, 2019 1:33 AM | Permalink | Reply to this

### Re: Can 1+1 Have More Than Two Points?

Thanks, Todd and Richard!

$C$ is distributive with objects being the finite copowers of $1$ if and only if it is the category of Boolean-valued matrices for the Boolean algebra $B= C(1,1+1)$.

Great!

Boolean-valued matrices sound more like relations than functions to me; I guess you mean matrices of some special sort here? (Like those that have right adjoints in a suitable sense?)

Posted by: John Baez on April 29, 2019 12:16 AM | Permalink | Reply to this

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