Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: M x X viewed as an (M -> M)-set

Harrison Grodin (Apr 13 2026 at 17:36):

Let $(M,1,\cdot)$ be a monoid, and let $E = (M^M, \text{id}, \circ)$ be the monoid of endomorphisms on the carrier $M$ (not necessarily monoid homomorphisms). I'm interested in the following $E$-set:

$G : \mathbf{Set} \to E\text{-}\mathbf{Set}$
$U(GX) = M \times X$
$e \cdot (m, x) = (e(m), x)$

Notice, $G$ behaves kind of like the free $M$-set functor $F : \mathbf{Set} \to M\text{-}\mathbf{Set}$, but it extends the action to all endomorphisms $e$, not just those of the form $m' \cdot (-)$ (which, of course, forms forms a monoid homomorphism $\varphi : (M,1,\cdot) \to (M^M, \text{id}, \circ)$ (defined by $\varphi(m) = m \cdot (-)$).

Using this $\varphi$, we get a functor $\varphi^* : E\text{-}\mathbf{Set} \to M\text{-}\mathbf{Set}$. If we consider $G$ as a functor mapping into the [[full image]] of $\varphi^*$, then $U : \text{FullImage}(\varphi^*) \to \mathbf{Set}$ is the right adjoint to $G : \mathbf{Set} \to \text{FullImage}(\varphi^*)$! (The induced monad is the free $M$-set monad.)

My question is as follows: is this $G$ "well-known", or easy to construct out of $F$ (or the free $E$-set) somehow? I know there's lots of structure available (e.g., $\varphi_!$ and $\varphi_*$), but it's not clear to me how to assemble this into $G$.

Harrison Grodin (Apr 13 2026 at 17:37):

(I came up with this construction "manually", so maybe there's something cleaner! Really, all I want is a way to decompose the monad $M \times (-) : \mathbf{Set} \to \mathbf{Set}$ in such a way that the left adjoint lands in a category where the objects come equipped with $(M \to M)$-actions.)

Harrison Grodin (Apr 13 2026 at 17:44):

(Another remark: I proved that this works given

• any monoid $E = (N, 1, \cdot)$
• a monoid homomorphism $\varphi : (M, 1, \cdot) \to (N, 1, \cdot)$
• an $N$-action structure on $M$, $\alpha : N \times M \to M$
• such that $\alpha (\varphi(c), 1) = c$.
  My construction/proof uses all of this structure, although maybe there's another approach that uses less structure. It would even be helpful to me if the above structure or something similar shows up/has a common name...)

John Baez (Apr 13 2026 at 17:59):

Btw, a minor notational thing: mathematicians use +, 0 for commutative monoids and $\cdot$, 1 for not-necessarily-commutative monoids, so seeing all these +'s and 0's gets the wrong part of my brain to turn on.

Harrison Grodin (Apr 13 2026 at 18:01):

(Sorry, good point - edited! The construction I'm presenting works for any monoid on $M$; although, I'm mainly interested in commutative monoids $M$ anyway, so if there's some way to answer my question that uses commutativity somehow, I'd be perfectly happy.)

John Baez (Apr 13 2026 at 18:03):

So let's see if I understand this. You're taking a monoid $M$ and a set $X$ and getting $M^M$ (by which you mean the monoid of endomorphisms of $M$, not the monoid of all functions from $M$ to $M$) to act on $M \times X$ in the obvious way. And you're wanting a slick way to show that there's a functor from $\mathsf{Set}$ to $M^M$-$\textsf{Set}$ that sends any set $X$ to $M \times X$, made into an $M^M$-set this way.

I'm just trying to restate your question in a more informal way.

Harrison Grodin (Apr 13 2026 at 18:04):

Yes, exactly! (And, I'd like that functor to have a right adjoint.)

Harrison Grodin (Apr 13 2026 at 18:05):

Oops, race condition - I do care about all functions $M$ to $M$ (not just monoid homomorphisms)! (Although if restricting to monoid homomorphisms helps somehow, that would be interesting to know.)

Harrison Grodin (Apr 13 2026 at 18:08):

(For a concrete example, let let $(M,1,\cdot) = (\mathbb{N},0,+)$; I'm interested in applying $e = (n \mapsto \lceil n \log n \rceil)$.)

John Baez (Apr 13 2026 at 18:10):

Okay, all functions. In that case you're really just using the fact that $M^M$ is a monoid that acts on $M$. So we might change your question to this more general one where we've got a monoid $A$ that acts on a set $S$.

I've got a monoid $A$ that acts on a set $S$. Then for any set $X$, $A$ acts on $S \times X$ in the obvious way. Can we show that there's a functor from $\mathsf{Set}$ to $A$-$\textsf{Set}$ that sends any set $X$ to $S \times X$, made into an $A$-set in this way, and does this functor have a right adjoint?

John Baez (Apr 13 2026 at 18:11):

This is more general so a priori less likely to be true, but I don't see how taking $A = M^M$, $S = M$ helps, so I'd be more inclined to think about this more general version.

Harrison Grodin (Apr 13 2026 at 18:20):

Yes, this is a great framing, thanks. Although, I'm willing to land in a category with $A$-sets as objects but possibly more restrictive homomorphisms (this [[full image]] idea).

Given just the action, I can define the functor $\mathbf{Set} \to A\text{-}\mathbf{Set}$.

To define a right adjoint, my idea was to restrict the functor to $\mathbf{Set} \to \text{FullImage}(\varphi^*)$ (the objects are $A$-sets but the morphisms are morphisms of $S$-sets), where $\varphi$ is a monoid homomorphism from $S$ (which now needs a monoid structure) to $A$.

Then, I'm able to define a right adjoint when I also assume that $\varphi(s) \cdot 1 = s$ (which, admittedly, feels like it comes out of nowhere).

John Baez (Apr 13 2026 at 18:24):

This $\text{FullImage}$ thing feels wrong to me; I have to go to work now, and let someone else pick up the baton now and carry it further, but given the functor from $\mathsf{Set}$ to $A$-$\mathsf{Set}$ that we're talking about, if it has a right adjoint it should be fairly simple to describe this right adjoint directly and check that it's a right adjoint straight from the definition of "adjoint". If that works, there are also people here who might be able to solve the problem in a slicker way.

Harrison Grodin (Apr 13 2026 at 18:55):

Sounds good, thanks for the advice! I'd be happy to avoid the full image as well, although so far that's the only solution that I've found that gives me a right adjoint inducing the $S \times (-)$ monad*.

*I do need to assume a monoid structure on $S$ for this to even be a monad...

David Egolf (he/him) (Apr 13 2026 at 19:29):

I'm reminded of this result, which might be relevant:

This is from "Monoid Properties as Invariants of Toposes of Monoid Actions" by Jens Hemelaer and Morgan Rogers.

John Baez (Apr 13 2026 at 19:44):

Yes, that could imply my guess if we specialize by taking $N$ to be the trivial monoid.

In case it helps Harrison: $\mathsf{Psh}(M)$ is a funky name for the category of $M^{\text{op}}$-sets, where $M^{\text{op}}$ is the monoid $M$ with the reversed multiplication

$x \cdot_{\text{op}} y := y x$

An $M^{\text{op}}$-set is also known as a right $M$-set. If you don't like this you can say the $A$ in my comment was secretly $M^{\text{op}}$. (We are now using $M$ in a different way than in your original question.)

John Baez (Apr 13 2026 at 19:46):

So, the key idea is that we get the right adjoint by using $\text{Hom}_N$. If you don't know what that means, I'm sure @David Egolf (he/him) would be happy to explain. :grinning_face_with_smiling_eyes:

Harrison Grodin (Apr 13 2026 at 19:54):

Yes, this sounds quite reasonable... although maybe with $N = M^M$ and $B = M$ or something? Wondering if I can then horizontally compose the adjunctions, where my $G$ is $F$ (free $M$-set) followed by $(-) \otimes_M M$ or something... (pondering on scratch paper now)

Harrison Grodin (Apr 13 2026 at 20:07):

Oh, yes, I think this is exactly the right thing!!

Harrison Grodin (Apr 13 2026 at 20:13):

Letting my $N = M^M$, I get that

• $FX \otimes_M M = {(FX \times M)/\sim} \cong FX$ (which makes sense - $M = F1$)
• $U(\text{Hom}_N(M, A)) = \{f : M \to UA \mid f(m' \cdot m) = m' \cdot f(m)\}$, which is the same as $UA$ when $M = \mathbb{N}$ (the only nontrivial choice is $f(0)$)

Harrison Grodin (Apr 13 2026 at 20:14):

This is fantastic, thanks so much!!! What a great paper (link), thanks @David Egolf (he/him)!

Harrison Grodin (Apr 13 2026 at 20:25):

Ohh, although, I suppose it's simpler to do what @John Baez suggests ($M \triangleq 1$, $N \triangleq M^M$, and $B \triangleq M$), indeed, and the same argument applies. :)

Harrison Grodin (Apr 13 2026 at 20:26):

(oh, duh, because $M = F1$!)

John Baez (Apr 13 2026 at 22:22):

Yes, this is looking good. And this theorem should be fairly easy to prove, or at least convince yourself of, since there are explicit "formulas" for both functors in the adjunction.

Harrison Grodin (Apr 13 2026 at 22:28):

Indeed - in retrospect, it feels like it was staring me in the face!

Harrison Grodin (Apr 13 2026 at 22:53):

(After equipping a set $S$ with an $A$-action, just define the left adjoint $FX = X \rtimes S$, the copower of $X$ with $S$; and then $UA = \text{Hom}(S, A)$, which is a set. And they're obviously adjoint, because this is the definition of copower! :man_facepalming:)

John Baez (Apr 13 2026 at 23:00):

Well, at some point you have to check that these functors exist, and that they're adjoint.

Morgan Rogers (he/him) (Apr 14 2026 at 06:17):

Thanks for dusting off our paper @David Egolf (he/him) :smile_cat:

Harrison Grodin (Apr 14 2026 at 19:59):

Upon further thought, this may not be exactly what I'm looking for... the right adjoint is $R(A) = \text{Hom}(S, A) = \{ f : S \to UA \mid \forall a : M, f(m \cdot s) = m \cdot f(s) \}$, not what I wrote above. This seems rather bad - with $L(X) = X \rtimes S$, I get $R(L(X)) = \{ (f_1, f_2) : S \to X \times S \mid \forall m : M, f_1(m \cdot s) = f_1(s) \land f_2(m \cdot s) = m \cdot f_2(s) \}$. When $S = \mathbb{N}$ and $M = \mathbb{N}^\mathbb{N}$, this $R(L(X))$ is just $X$, since $f_1(s) = f_1(s + 0) = f_1(0)$ and $f_2$ has to be the identity function. But I wanted $R(L(X)) = S \times X$...

Harrison Grodin (Apr 14 2026 at 20:02):

Seemingly relatedly, for my application, I do want a map $M \times RA \to RA$ that gives me an $M$-action for everything in the image of $R$. When I chose $R$ to be $U$ (using my "full image" idea), this was immediate, but for $R(A) = \text{Hom}(S, A)$, this isn't the case...

Harrison Grodin (Apr 14 2026 at 20:06):

Since the full image idea does work but just feels a bit ad-hoc, maybe I should frame the question differently. The full image category on $\varphi^*$ (given a monoid homomorphism $\varphi : M \to N$) is a full subcategory of $\mathbf{Psh}(N)$. Is it possible that the inclusion has a left adjoint $L$? If so, then my original adjunction will just be the following composite of adjunctions:

John Baez (Apr 14 2026 at 20:07):

My eyes go blurry when I see a wall of symbols like that; for me at least it's better if you instead, or also, say in words what you're thinking. For example why did you "want" R(L(X) = S $\times$ X?

Harrison Grodin (Apr 14 2026 at 20:08):

Sorry, right - I was writing the symbols to record my thoughts, but maybe it's not so human-readable. :) Maybe the following message is the better one for reading - I want a map $M \times RA \to RA$ for whatever the right adjoint $R$ is.

Harrison Grodin (Apr 14 2026 at 20:10):

With my full image idea, this is immediate, since $RA$ just gets the carrier of $A$ (what I've been writing $UA$). But with $RA = \text{Hom}(S, A)$, I no longer have this (unless the monoid is commutative to begin with or something, which I certainly don't want to assume).

Harrison Grodin (Apr 14 2026 at 20:11):

The full-subcategory idea "smells right" to me for my application - I'm really working in type theory, and it would make lots of sense if my construction was just a reflector/modality applied to the free $N$-set...

John Baez (Apr 14 2026 at 20:13):

Yesterday I thought I had pierced the veil and figured out what you were trying to do. Now I no longer know what you're trying to do, and I'd probably only understand it if you gave a complete new description, not just corrections like "I used to want this, and now I want this".

Other readers, younger and more focused on what you're doing, may not have my problem!

Harrison Grodin (Apr 14 2026 at 21:40):

I also thought what you suggested was the right solution, but realized upon studying this adjunction more that it's not quite what I'm looking for. Let me try to give a fresh description of the problem:

I have two monoids, $M$ and $N$, along with a monoid homomorphism $\varphi : M \to N$. (Typically for me, $M = \mathbb{N}$ with addition and $N$ is arbitrary functions $\mathbb{N} \to \mathbb{N}$ with composition.)

I'm interested in viewing the free $M$-set as an $N$-set, which I can do since (in my cases of interest) I have an $N$-action structure on $M$, written ${\cdot} : N \times M \to M$. I'm working in a type theory whose semantics lets me choose a pair of adjoint functors $F \dashv U : \mathcal{C} \to \mathbf{Set}$, and I want it to be the case that objects of $\mathcal{C}$ come equipped with an $N$-action, viewable through $U$ as maps $N \times UA \to UA$ for all $A : \mathcal{C}$.

Because of this last condition (which I was so used to when I posed my earlier question that I didn't even think to mention it, sorry!), it seems like I should choose my category $\mathcal{C}$ to have objects being $N$-sets and let $U$ genuinely extract the carrier of an $N$-set.

If I let $\mathcal{C}$ be the category of $N$-sets, though, the left adjoint to $U$ will be the free $N$-set, but I only care about the free $M$-set! And for type theory-esque reasons, I can only pick a single adjunction. So, my idea is to let $\mathcal{C}$ have objects be $N$-sets, but have a morphism from $A \to B$ be a morphism of $M$-sets from $\varphi^*(A) \to \varphi^*(B)$ (this is the [[full image]] category of $\varphi^*$).

With this in mind, I'm set. The free $M$-set functor $\mathbf{Set} \to \mathcal{C}$ is genuinely left adjoint to $U$: since the maps of $\mathcal{C}$ only preserve $M$-set structure and thus the "free-ness" is successfully only with respect to the $M$-set structure, the notion of free is what I want, $M \times X$. To prove this, I only need the mildly odd condition connecting the monoid homomorphism $\varphi$ and the action $\cdot$ that $\varphi(m) \cdot 1 = m$.

The question is, though, whether this can be recovered out of more elementary parts. $\mathbf{Set}$, $\mathbf{Psh}(M)$, and $\mathbf{Psh}(N)$ all have tons of structure: there are

• a free-forgetful adjunction $F \dashv U : \mathbf{Psh}(M) \to \mathbf{Set}$,
• similarly $F \dashv U : \mathbf{Psh}(N) \to \mathbf{Set}$,
• an adjoint triple $\varphi_! \dashv \varphi^* \dashv \varphi_* : \mathbf{Psh}(M) \to \mathbf{Psh}(N)$,
• the adjunction we discussed yesterday $(-) \rtimes M \dashv \text{Hom}(M, -) : \mathbf{Psh}(N) \to \mathbf{Set}$,
• and more!
  So, I'd kind of expect my adjunction to decompose into the composite of some known adjunctions or something.

Harrison Grodin (Apr 14 2026 at 21:46):

My best guess (based on domain-specific intuition) for what might happen is that the inclusion functor $\iota$ from the full image category might have a left adjoint $L$:

Harrison Grodin said:

Is it possible that the inclusion has a left adjoint $L$? If so, then my original adjunction will just be the following composite of adjunctions:

That would be great, since it would render this "full image" business as just the category of algebras for the idempotent monad $\iota \circ L : \mathbf{Psh}(M) \to \mathbf{Psh}(M)$. And then my construction would just be $L \circ F$, applying this elusive $L$ to the free $M$-set in such a way that its carrier remains $M \times X$.

So: I'd be very happy if I could find this left adjoint $L$! If that doesn't exist, maybe this construction I'm interested in can come out of some other composite of adjoints or something, but I feel less certain about where to look...

Harrison Grodin (Apr 14 2026 at 21:47):

(Thanks for going through this, btw - I appreciate your time and thoughts!)