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Stream: learning: questions

Topic: Showing distinct homsets are the same?

Alex Kreitzberg (Mar 23 2025 at 02:02):

If I view the integers $Z$ as a category with "less than" as the arrows, then the Hasse diagram looks like an infinite line of dots with arrows pointing towards the right.

$Z(x, y)$ has the same number of arrows as $Z(x + c, y +c)$ for all integers c.

So in a sense, "all morphisms between adjacent integers are the same".

You can imagine this category being freely generated by the graph of integers with one edge between $n$ and $n + 1$ for all integers $n$.

I'm trying to do a bit better and clarify what it means to say "all these edges are 'really' just a translation of one edge". To my intuition, there's only one generating arrow of this category. I want to show this without collapsing all the integers into one isomorphic object.

Are there good ways to compare homsets in this way? Can I somehow articulate what it means for homsets to "be the same" even if their objects are not isomorphic?

David Egolf (Mar 23 2025 at 17:11):

If you start with a set having a single element, the free monoid generated by that set is the natural numbers, $\mathbb{N}$, together with its usual addition.

We can let this monoid act on the set of integers by addition. So, viewing $\mathbb{N}$ as a category with a single object, we get a functor $F:\mathbb{N} \to \mathsf{Set}$. This functor sends the single object of $\mathbb{N}$ to the set of integers. And it sends an element $n$ of $\mathbb{N}$ to the function $x \mapsto x + n$, the function which adds $n$ to any integer $x$.

I think that the category of elements of $F$ is the category you describe above. (Although note that we need to have identity morphisms, so intuitively the arrows correspond to "less than or equal to" instead of "less than"). Checking that this is correct (or proving me wrong!) might be a good exercise to do.

David Egolf (Mar 23 2025 at 17:15):

So this process lets us create the category you described starting from (1) a set with a single element and (2) the additive monoid of integers. I'm not sure if noticing that helps answer your question.

Patrick Nicodemus (Mar 23 2025 at 17:33):

@Alex Kreitzberg A functor from a category $C\to D$ has a function $Hom(c,c') \to Hom(F c, F c')$ for all $c,c'$. If this function is a bijection for all $c,c'$ we say that the functor $F$ is fully faithful. The function $x\mapsto x+c$ extends to a fully faithful functor from $Z$ to itself.

David Egolf (Mar 23 2025 at 17:34):

A different idea could be as follows. Consider an endofunctor $G:C \to C$. If $G$ is part of an equivalence of categories, then we might say that $C(a,b)$ is similar in some sense to $C(G(a),G(b))$.

(Edit: I see @Patrick Nicodemus already posted something roughly in this direction!)

Alex Kreitzberg (Mar 23 2025 at 20:14):

Beautiful! I believe these answer my question. Thinking for example about Patrick's $F$ and the (you're right less than or equal, sorry!) arrow from $l : 0 \rightarrow 1$ as the "generator", I can write all the generating arrows as $F(l)^{n}$ for some $n \in \mathbb Z$.

It's funny, one thought on exploring this, is "what if I have a two dimensional grid, where some of the 'roads' are broken for whatever reason" in that case I think you can have a functor to a category encoding the directions, and the arrows would map to "their direction".

So to define this "direction" functor $D$ do the following.

Define a category of directions with one object $\cdot$,
So if "up" was $u : \cdot \rightarrow \cdot$ and "right" was $r : \cdot \rightarrow \cdot$ then we'd want $r \circ u = u \circ r$.

$D$ will map from the grid $\mathbb Z \times \mathbb Z$ with some arrows removed, every integer pair maps to $\cdot$.

For homsets, if $hom((x, y), (x + 1, y))$ is nonempty, its one arrow should then map to $r$, and similarly if $hom((x, y), (x, y + 1))$ is nonempty its one arrow should map to $u$.

I think this example is related to but different than David's first example. I'm not sure it can be related to generators cleanly. I guess the functor I described above would be essentially subjective.

But anyways, I think this is all making sense, I'm getting used to manipulating these categories in the manner I want to. Thank you both for your answers.

Notification Bot (Mar 23 2025 at 20:22):

Alex Kreitzberg has marked this topic as resolved.

Alex Kreitzberg (Mar 23 2025 at 20:42):

Actually one more addendum to that last thought!

Define your arrows across the infinite grid category, omitting some arrows per your whims.

Then quotient the category freely generated on these by the relationships implied by their mapping into the direction category!

That also does what I want I think! Thanks again everybody!

John Baez (Mar 24 2025 at 02:52):

The integers with the usual $\le$ and + is a monoid
object in the category of posets, e.g. the operation + is order-preserving, so we can think of it as a monoidal category where objects are integers and there's a unique morphism from m to n iff m $\le$ n. This fact gives functions

hom(m,n) $\to$ hom(m+k,n+k)

but because the integers are also a group these functions are bijections.

So, we get this whenever we have a monoid object in the category of posets whose underlying monoid is a group. People probably call such a thing a 'partially ordered group'. There are some fun nonabelian examples in physics.

(Note our example is not a group object in the category of posets because negation is order-reversing. I bet the usual definition of 'partially ordered group' assumes or implies that.)

Notification Bot (Mar 24 2025 at 02:53):

John Baez has marked this topic as unresolved.

John Baez (Mar 24 2025 at 02:57):

(Maybe I'm being silly, but I "unresolved" this question because there's plenty more to say about such things.)

Adittya Chaudhuri (Mar 24 2025 at 13:56):

Thanks!! I was not aware of these objects, "partially ordered groups," till today. I find the fact that "group objects in the category of posets are not same as partially ordered groups" interesting.

Adittya Chaudhuri (Mar 24 2025 at 14:16):

From the point of nomenclature,

Group object in Set= Set theoretic group(Group)= Group internal to Set
Group objects in Top = Topological group = Group internal to Top
Group objects in Man= Lie group= Group internal to Man
Group objects in Cat= Categorical group(Strict 2-group)= Group internal to Cat
....... so on

but,

Group object in Pos $\neq$ Partially ordered group $\neq$ Group internal to Pos (as the inverse function is not order preserving as @John Baez mentioned.)

Question is then, why do we call these objects as "partially ordered groups" ?

Kevin Carlson (Mar 24 2025 at 14:30):

Well, the desire to talk about groups that have a partial order came up historically long before anybody realized this terminology would be mildly annoying to later category theorists. Besides, there essentially aren't any group objects in the category of partial orders, so there's little risk of confusion.

Adittya Chaudhuri (Mar 24 2025 at 14:31):

I see your point.

John Baez (Mar 24 2025 at 21:12):

I thought it was obvious that there aren't interesting abelian group objects in the category of posets, but now I'm not seeing why.

(I'm saying 'abelian' just to keep things easy: we can write the group operation as + and know x+y =y+x.)

What counts as interesting? We can take any abelian group and make it into an abelian group object in the category of posets by saying x $\ge$ y iff x=y. That counts as uninteresting!

John Baez (Mar 24 2025 at 21:15):

On the other hand suppose we have an abelian group object in the category of posets. If x $\ge$ 0 then we get -x $\ge$ 0, which sounds bad. Oh, I see: then add x to both sides and get 0 $\ge$ x. This then implies x=0.

John Baez (Mar 24 2025 at 21:18):

So: in an abelian group object in the category of posets, if x $\ge$ y then x-y $\ge$ 0 and then by our previous argument x-y =0 so x=y.

John Baez (Mar 24 2025 at 21:19):

So all distinct elements are incomparable. :face_with_symbols_on_mouth:

John Baez (Mar 24 2025 at 21:21):

(That's a QED symbol for theorems that make you angry.)

Kevin Carlson (Mar 24 2025 at 21:42):

(just to note that abelianness didn't get used in there)

John Baez (Mar 25 2025 at 04:45):

Good, so it was just a psychological crutch!

Amar Hadzihasanovic (Mar 25 2025 at 09:57):

Is there a way to axiomatise partially ordered groups with inversion being order-reversing?

I am thinking about the fact that the opposite category is a dual in the monoidal bicategory of profunctors, so presumably the opposite poset is also a dual in some appropriate bicategory of "order-preserving pro-maps".
You could have a symmetric monoidal theory with types $P$ and $P^*$, together with zig-zag equations that specify that the two types are each other's dual, and then this would allow you to specify inversion as an operation of type $P \to P^*$, which interpreted in "posets and pro-maps" would be an order-reversing map up to canonical isomorphism.

Amar Hadzihasanovic (Mar 25 2025 at 09:58):

Well, actually it would be an order-reversing pro-map.

Amar Hadzihasanovic (Mar 25 2025 at 10:00):

But perhaps it would be possible to define such a theory, so that the models whose operations are "genuine" maps and not pro-maps are exactly partially ordered groups.

James Deikun (Mar 25 2025 at 10:40):

Another way to specify them is "the pullback in $\mathsf{Cat}$ of the forgetful functor from partially ordered monoids to monoids along the forgetful functor from groups to monoids".

Amar Hadzihasanovic (Mar 25 2025 at 11:38):

Yeah, it seems to me like there are all sorts of good reasons to treat “existence of inverses” as a property of monoids rather than an operation --- iirc the most useful categorifications of Hopf algebras (those that satisfy the “most correct” categorified Tannaka reconstruction theorems) do not admit the antipode as an operation --- but I am curious about whether in this particular case one can still make things work in terms of categorical universal algebra.

Amar Hadzihasanovic (Mar 25 2025 at 12:09):

Interestingly, this article by Gunnar Fløystad starts with

This article advocates for general posets P and Q the notion of profunctor P -|→ Q as more effective than the notion of isotone (order preserving) maps P → Q between posets, especially for applications in algebra.

So it may be worth looking at generalised partially ordered groups as models of a symmetric monoidal theory in posets and profunctors (or pro-maps).

Amar Hadzihasanovic (Mar 25 2025 at 12:49):

I'd conjecture that this could be an interesting set of generators and relations for a symmetric monoidal theory (a coloured prop): we would have two types $P$ and $P^*$, and as generators

• the multiplication and unit of a monoid structure on $P$, i.e. $m: P \otimes P \to P$ and $u: I \to P$,
• the antipode $s: P^* \to P$,
• the unit and counit of a duality between $P$ and $P^*$, $\eta: I \to P \otimes P^*$ and $\varepsilon: P^* \otimes P \to I$.

We also have, structurally, symmetric bradings $\sigma_{\bullet,\bullet}$.

As equations we would have

1. equations saying that $m$ and $u$ give a monoid structure on $P$,
2. zig-zag equations saying that $\eta$ and $\varepsilon$ make $P$ and $P^*$ into each other's dual,
3. the equation $m \circ (\mathrm{id}_P \otimes s) \circ \eta = u \circ (\varepsilon \circ \sigma_{P, P^*} \circ \eta)$

Here $\varepsilon \circ \sigma_{P, P^*} \circ \eta$ is the abstract dimension of $P$ i.e. the trace of $\mathrm{id}_P$, which is a scalar.

Amar Hadzihasanovic (Mar 25 2025 at 12:54):

The idea I have in my mind is that $m \circ (\mathrm{id}_P \otimes s) \circ \eta$ should behave formally like

$\sum_{x \in P} x \cdot x^{-1}$

while $u \circ \mathrm{dim}(P)$ is like $|P| 1_P$, where $1_P$ is the group unit and $|P|$ the cardinality of $P$, so this is the equation I'd expect.

Amar Hadzihasanovic (Mar 25 2025 at 12:59):

It's interesting that this theory does not require a comonoid structure unlike the usual “Hopf algebra”-like presentation of groups.

Amar Hadzihasanovic (Mar 25 2025 at 13:00):

(Perhaps it would make sense to move this to a separate thread, sorry for diverting this one.)

John Baez (Mar 25 2025 at 14:39):

Amar Hadzihasanovic said:

Is there a way to axiomatise partially ordered groups with inversion being order-reversing?

This property, that inversion is order-reversing, follows from the usual definition of partially ordered group, which I believe is equivalent to "a monoid in posets whose underlying monoid in sets is a group". (You mentioned this approach later.)

Given such a thing we see that given

x $\ge$ y

we can add -y to both sides and get

x - y $\ge$ 0

and then add -x to both sides and get

-y $\ge$ -x

so negation is order-reversing.

(I'm using additive notation but not assuming abelianness.)

So, if the existence of inverses is treated as a mere property of a monoid in posets, they automatically obey this law.

But it's still fun to think about other approaches!

Amar Hadzihasanovic (Mar 25 2025 at 14:52):

Ok so I think I can at least show that every partially ordered group is, indeed, a model of the theory that I described.

As a preliminary, let me describe the category of posets and pro-maps more explicitly, in the style of the paper by Fløystad:

• the objects are posets,
• the morphisms $P \to Q$ are order-preserving maps $f: P \to \hat{Q}$, where $\hat{Q}$ is the poset of lower sets of $Q$ (a lower set of a poset is a subset $U$ with the property that $x \leq y$ and $y \in U$ implies $x \in U$)
• an order-preserving map $f: P \to \hat{Q}$ determines an order-preserving, union-preserving map $\hat{f}: \hat{P} \to \hat{Q}$ by $U \mapsto \bigcup_{x \in U} f(x)$; we compose two morphisms $P \to Q$ and $Q \to R$, determined by maps $f: P \to \hat{Q}$ and $g: Q \to \hat{R}$, by letting $\hat{g} \circ f: P \to \hat{R}$ be the composite;
• the identity on $P$ is determined by the map $P \to \hat{P}$ sending $x \in P$ to its lower set $\downarrow x = \{y \leq x \}$.

(This is really an instance of enriched profunctors where the basis of enrichment is the ordinal $\mathbf{2}$; the map $P \to \hat{P}$ is the $\mathbf{2}$-enriched Yoneda embedding)

Amar Hadzihasanovic (Mar 25 2025 at 15:02):

The cartesian product of posets determines a symmetric monoidal structure on this category.
With respect to this symmetric monoidal structure, the opposite poset $P^\mathrm{op}$ is a two-sided dual to $P$:

• first of all, observe that $\hat{1}$, the poset of lower sets of the terminal poset, is the ordinal $\mathbf{2}$, since the only lower sets of $1$ are $\varnothing$ and $1$;
• the counit of the duality is the pro-map $P^\mathrm{op} \times P \to \hat{1}$ sending $(x, y)$ to $1$ if and only if $x \leq y$, and to $\varnothing$ otherwise;
• the unit of the duality is the pro-map $1 \to \widehat{(P \times P^\mathrm{op})}$ which picks the set of pairs $(x, y)$ such that $x \leq y$; observe that this is indeed a lower set in $P \times P^\mathrm{op}$.

Amar Hadzihasanovic (Mar 25 2025 at 15:11):

Now I claim that every partially ordered group gives a model of the theory that I described a few posts before.

• Interpret the type $P$ as the underlying poset of the partially ordered group, and $P^*$ as the opposite poset $P^\mathrm{op}$.
• To interpret the multiplication and unit, post-compose the order-preserving maps $m: P \times P \to P$ and $u: 1 \to P$ with the Yoneda embedding $P \to \hat{P}$.
• To interpret the antipode, post-compose the order-reversing map $s: P^\mathrm{op} \to P$, $x \mapsto x^{-1}$ with the Yoneda embedding $P \to \hat{P}$.
• To interpret the unit and counit of duality, use the unit and counit pro-maps I just described.

Amar Hadzihasanovic (Mar 25 2025 at 15:15):

Now the equations relative to multiplication and unit follow immediately from the fact that a partially ordered group is in particular an internal monoid in (posets, order-preserving maps, cartesian product), and post-composing with the Yoneda embedding determines a monoidal functor from this to (posets, pro-maps, cartesian product).

The equations relative to duality follow from the fact that indeed $P$ and $P^\mathrm{op}$ are duals.

Amar Hadzihasanovic (Mar 25 2025 at 15:22):

Finally, by calculation, $m \circ (\mathrm{id}_P \times s) \circ \eta$ picks the lower set

$\bigcup \set{ \downarrow xy^{-1} \mid x, y \in P, x \leq y}$

But if $x \leq y$ then $xy^{-1} \leq 1_P$ where $1_P$ is the group unit, so $\downarrow xy^{-1} \subseteq \downarrow 1_P$, so this union is included in $\downarrow 1_P$; and conversely, since $x \leq x$ for each $x \in P$, the lower set $\downarrow 1_P$ is included in the union, so overall this is just picking the lower set of the group unit.

Amar Hadzihasanovic (Mar 25 2025 at 15:26):

Now, there are only 2 scalars in this monoidal category --- the elements of $\mathbf{2}$ --- and $\mathrm{dim}(P)$ is $\varnothing$ if $P$ is empty, and $1$ otherwise. Since every partially ordered group is nonempty, $\mathrm{dim}(P)$ is $1$ that is the unit scalar.

So we have, indeed, $m \circ (\mathrm{id}_P \times s) \circ \eta = u \circ \mathrm{dim}(P) = u$, since both sides pick the lower set of the group unit.

Amar Hadzihasanovic (Mar 25 2025 at 15:44):

To give one example of models of this theory where the presence of $\mathrm{dim}(P)$ in the final equation is non-trivial, I claim that the free $k$-vector space $kG$ on a finite group $G$ gives a model of my theory in the compact closed category of $k$-vector spaces with the tensor product:

• the dual of $kG$ is taken to be $kG$ itself, the unit is defined by $1_k \mapsto \sum_{g \in G} g \otimes g$ and counit by $g \otimes h \mapsto 1_k$ if $g = h$ and $0_k$ if $g \neq h$,
• $m$, $u$, $s$ are all linear extensions of the corresponding maps on the elements of $g$.

Amar Hadzihasanovic (Mar 25 2025 at 15:46):

Then, indeed, $m \circ (\mathrm{id}_{kG} \otimes s) \circ \eta$ is the map $1_k \mapsto \sum_{g \in G} gg^{-1} = (|G|1_k) 1_G$, which is equal to $u \circ \mathrm{dim}(kG)$ since the dimension of $kG$ is equal to $|G|1_k$.

Fernando Yamauti (Mar 26 2025 at 21:10):

Amar Hadzihasanovic said:

iirc the most useful categorifications of Hopf algebras (those that satisfy the “most correct” categorified Tannaka reconstruction theorems) do not admit the antipode as an operation

Could you clarify what's this "most correct" categorification of Hopf algebras and why it has no notion of an antipode? Are you referring perhaps to rigid monoidal cats? Or maybe trialgebras? Any references?

Alex Kreitzberg (Apr 21 2025 at 23:54):

While thinking about binary algebras, I better understood the significance behind Baez's suggestion to think of monoid objects in the category of posets.

Suppose you're given a monoid $M$, we can make this into a preorder by drawing an arrow between two elements $a$ and $c$ if there exists an element $b$ such that $ab = c$, denote this condition by $a | c$. We can even label such an arrow $b$.

This is a preorder because $a | a$ and $a | b$ with $b | c$ implies $a | c$. (proof: $a1 = 1, ax' = b, bx'' = c \implies ax'x'' = c$)

$1 | x$ for all $x \in M$, making the identity initial.

If our monoid is commutative then it's monotonic, If $a | c$ and $b | d$, then $ab | cd$.

There's a more complicated story here involving left division and right division that I wasn't able to sort out. Maybe Amar was able to get a good chunk of it sorted, but I don't need it right now.

Homomorphisms preserve divisibility, $ab =c \implies \phi(a)\phi(b) = \phi(c)$ for all homomorphisms $\phi$. So homomorphisms are also naturally monotonic maps with respect to this definition.

This gives us the structure of our functor $O : \mathsf{Comm Mon} \rightarrow \mathsf{Ord Comm Mon}$

let $f$ and $g$ be composable monoid homomorphisms. Then $O(f \circ g)(x) = f(g(x)) = O(f)(O(g)(x))$, that is $O$ preserves the composition because all it's doing is noting the presence of an extra property of functions, monotonicity. (The identity is trivially preserved)

So we have a functor on commutative monoids giving them an order! We can forget this preorder to get back our original monoid, or we could keep the preorder and forget it came from a monoid. Call these forgetful functors $U_{\mathsf{Mon}}$ and $U_{\mathsf{Pre}}$.

This gives an interesting way to frame some questions I was trying to ask about these objects. For example, the natural number preorder, ignoring its monoid structure, seems to uniquely define natural numbers. I suspect that means the fiber over the preorder $N$ under the functor $U_{\mathsf{Pre}}O$, is a contractible groupoid containing the monoid $N$.

If this is provable that would be a really fun way of answering my original question "How to show the distinct homsets are the same?" for the special cases I was exploring.

I'm struggling to get a handle on all the moving parts though so maybe there's an even more interesting and general story here.

In any case, I started thinking about this again, because while trying to understand $\mathsf{Set}^{\text{op}}$ by reading the wikipedia article Boolean algebras canonically defined (Date: 4/21/2025), I noticed the article referred to a lattice basis vs a ring basis for thinking about boolean logic.

So I suspect there's a similar sort of Functor dance between commutative rings and preorders, which in the special case of boolean logic lets us use lattices ( the usual propositional calculus way ) or algebra.

I have a bunch more ideas connected to this stuff, and I'm not convinced I didn't make a mistake above. But I still wanted to leave a note here to keep my thoughts organized.

John Baez (Apr 22 2025 at 00:07):

Interestingly this preorder you're putting on commutative monoids featured recently in my conversation with @Adittya Chaudhuri. We were using it to define the 'minimal' elements of certain commutative monoids we're studying.

You say this process gives a functor

$O : \mathsf{Comm Mon} \rightarrow \mathsf{Ord Comm Mon}$

from commutative monoids to preordered commutative monoids, which sounds right, and you also mention the (more obvious) functor

$U : \mathsf{Ord Comm Mon} \rightarrow \mathsf{Comm Mon}$

which forgets the preorder. But now you're making me wonder:

Question: is $O$ the left adjoint of $U$?

Adittya Chaudhuri (Apr 22 2025 at 03:47):

Interesting!! If $O$ is really a left adjoint to $U$, then we may compose the functor $O$ with the free monoid functor $Free \colon \mathsf{Set} \to \mathsf{CommMon}$ to obtain an adjoint pair of functors between $\mathsf{Set}$ and $\mathsf{OrdCommMon}$.

John Baez (Apr 22 2025 at 04:56):

I haven't checked whether $O$ is left adjoint to $U$, but I have a pretty good understanding of

$O \circ \text{Free} : \mathsf{Set} \to \textsf{OrdCommMon}$,

so I might as well share it with anyone who doesn't. If $X$ is a set, $O(\text{Free}(X))$ is the commutative monoid whose elements are functions

$f: X \to \mathbb{N}$

such that $f(x) = 0$ except for finitely many $x \in X$, made into a commutative monoid with the addition defined by

$(f + g)(x) = f(x) + g(x)$

and the preorder (in fact a partial order) defined by

$f \le g \iff \forall x \in X \; f(x) \le g(x)$

So it's a very nice thing! The morphisms are arbitrary commutative monoid homomorphisms, which automatically preserve the partial order.

Adittya Chaudhuri (Apr 22 2025 at 05:16):

Thanks!! It is interesting!

Alex Kreitzberg (Apr 22 2025 at 07:34):

(deleted)

Alex Kreitzberg (Apr 22 2025 at 14:45):

I think whether $O$ is left adjoint to $U$ depends on what I mean by $\mathsf{OrdCommMon}$.

If I mean "commutative monoid object in the category of preorders" I think the answer is no, because there's nothing stopping me from giving any monoid a trivial order with only reflexive comparisons. And such an ordered monoid has almost no monotonic functions into it, but lots of monoid homomorphisms into the underlying monoid.

However, if $\mathsf{OrdCommMon}$ requires for any object $M$, we have $0 \leq m$ for all $m \in M$ and $a + b = c \implies a \leq c$, then I think $O$ is a left adjoint.

I can't quite tell whether that's a natural definition for a category of ordered commutative monoids though.

James Deikun (Apr 22 2025 at 15:29):

Your second condition implies your first. Also, the second condition is basically just saying the order of $M$ is an extension of the order of $OUM$.

Alex Kreitzberg (Apr 22 2025 at 15:33):

Is it being an extension "boring" or overly restrictive? Or are you giving me a suggestion for how to say what I mean?

James Deikun (Apr 22 2025 at 15:34):

I think $O$ just gives the minimal preordering on a commutative monoid that makes addition inflationary in all arguments. But I can't think of a nice way to phrase that like "living in a certain category" because being inflationary only makes sense for endomorphisms.

James Deikun (Apr 22 2025 at 15:35):

It's up to you whether being an extension bores you or not, but it does make the fact that $O$ ends up being the left adjoint rather tautologous.

Alex Kreitzberg (Apr 22 2025 at 16:05):

Given that, I think the answer to Baez's question is "no".

While thinking about this and similar orders on monoids it did get me to wonder if I could find a functor $C : \mathsf{Pre} \rightarrow \mathsf{Mon}$ such that $\mathsf{Pre}(O(m), p) \cong \mathsf{Mon}(m, C(p))$ was an adjunction.

But I wasn't able to solidify the definitions into a form that did what I wanted.

And the more I thought about it the less I was convinced it was true as written.

But the idea was given a sufficiently nice pre-order maybe there was a monoid dual to it.

Alex Kreitzberg (Apr 22 2025 at 16:13):

I was going to come back to it after understanding adjoint functors a bit better.