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Stream: learning: questions

Topic: Abelianisation Adjoint

Fawzi Hreiki (Nov 02 2020 at 15:05):

The inclusion $\text{CMon} \hookrightarrow \text{Mon}$ of commutative monoids into monoids has a left adjoint (often called abelianisation). Likewise, we have such adjoints for groups, rings, etc.. It is known that, for a topos $E$, $E$ has free monoids iff it has a NNO. But what structure in general is needed on a category $C$ with finite products (or finite limits) to have such abelianisation adjoints $\text{Mon}(C) \rightarrow \text{CMon}(C)$?

Jacques Carette (Nov 02 2020 at 15:44):

The construction at the nlab requires quite a bit, i.e. quotients and forming sub-structures. So which meta-theory you are in will come in quite strongly. I don't quite know what minimal vocabulary you'll need to state all those things, but it feels like quite a lot, i.e. quite a bit more than just having finite products.

Jade Master (Nov 02 2020 at 16:11):

Fawzi Hreiki said:

The inclusion $\text{CMon} \hookrightarrow \text{Mon}$ of commutative monoids into monoids has a left adjoint (often called abelianisation). Likewise, we have such adjoints for groups, rings, etc.. It is known that, for a topos $E$, $E$ has free monoids iff it has a NNO. But what structure in general is needed on a category $C$ with finite products (or finite limits) to have such abelianisation adjoints $\text{Mon}(C) \rightarrow \text{CMon}(C)$?

I think that coequalizers would do the trick? I'm pretty sure that belianization can be thought of as coequalizing the symmetry composed with multiplication with the multiplication.

John Baez (Nov 02 2020 at 16:13):

Coequalizers sound right, but perhaps we need products to "get along" with coequalizers in some way for the coequalizer to actually inherit a monoid structure.

John Baez (Nov 02 2020 at 16:57):

To be a bit less mysterious, I'm suggesting we'll need binary products to distribute over coequalizers. I haven't checked this.

Jacques Carette (Nov 02 2020 at 16:59):

But the question remains: what vocabulary do you need to write down 'coequalizer' ? And do you need additional vocabulary to start they they distribute?

Morgan Rogers (he/him) (Nov 02 2020 at 17:04):

A coequalizer is a finite colimit; what "vocabulary" are you talking about?

John Baez (Nov 02 2020 at 17:11):

Jacques is talking more about the metatheory; I think first we should settle this question in the plain old framework of mathematics, because it's so simple: in what sort of category C with finite products can we abelianize a commutative monoid object, getting a left adjoint to Mon(C) $\to$ CMon(C)?

My conjecture: it's enough for C to have coequalizers and for binary products to distribute over these.

fosco (Nov 03 2020 at 16:11):

John Baez said:

Jacques is talking more about the metatheory; I think first we should settle this question in the plain old framework of mathematics, because it's so simple: in what sort of category C with finite products can we abelianize a commutative monoid object, getting a left adjoint to Mon(C) $\to$ CMon(C)?

My conjecture: it's enough for C to have coequalizers and for binary products to distribute over these.

The monoids are not already commutative; and the left adjoint is to the embedding CMon(C) → Mon(C); in fact the abelianization functor G ↦ Gᵃ is left adjoint to the inclusion (and there is also a right adjoint, that sends a monoid to its center). Of course, both were typos :smile:

It seems very natural to ask the same question in terms of the PROP of monoids, $\mathbb{T}_M$ and the theory of commutative monoids $\mathbb{T}_{CM}$ (this is just the category of finite sets and functions, and a commutative monoid in $\mathcal C$ is just a monoidal functor $\mathbb{T}_{CM} \to \mathcal C$...); there is a morphism of PROPs $i$ from $\mathbb{T}_{M}$ to $\mathbb{T}_{CM}$, because the latter is just the former with an additional equation (commutativity). $i^*$, precomposition with $i$, gives the functor that forgets commutativity; adjoints to this functor are governed by the request that left/right Kan extensions sends a monoidal functor $G : \mathbb{T}_{M} \to \mathcal C$ into another monoidal functor $\mathbb{T}_{CM} \to \mathcal C$.

However, this is just a high-level approach, that might yield the same conjecture as John's.

John Baez (Nov 03 2020 at 16:49):

Yeah - I didn't mean to write "abelianize a commutative monoid object" - I meant "abelianize a monoid object".

John Baez (Nov 03 2020 at 16:54):

There's definitely a nice morphism of props here, as you say. But to do the abelianization using props - that is, to abelianize a monoid in $\mathcal{C}$ and get a commutative monoid in $\mathcal{C}$, for example by constructing the left Kan extension you want, we need $\mathcal{C}$ to be "nice enough".

John Baez (Nov 03 2020 at 17:01):

I know more about this for Lawvere theories than for props. Let's use $\mathrm{Mod}_{\mathcal{C}}(\mathbb{T})$ to denote the category of models of a Lawvere theory $\mathbb{T}$ in a category with finite products $\mathcal{C}$. Given a map of Lawvere theories $i : \mathbb{T} \to \mathbb{T}'$, we always get a precomposition map $i^\ast: \mathrm{Mod}_{\mathcal{C}}(\mathbb{T}') \to \mathrm{Mod}_{\mathcal{C}}(\mathbb{T})$. And this always has a left adjoint when $\mathcal{C}$ is cocomplete, with finite products distributing over colimits.

John Baez (Nov 03 2020 at 17:01):

So for example it always works for $\mathcal{C} = \mathsf{Set}$.

John Baez (Nov 03 2020 at 17:03):

This is a consequence of a special case of Theorem 3.3 here:

• Todd Trimble, Multi-sorted Lawvere theories.

In fact Todd shows it more generally for multi-sorted Lawvere theories, and he shows that the precomposition map $i^*$ is not only a right adjoint, but monadic.

John Baez (Nov 03 2020 at 17:06):

But in the case where $\mathbb{T}$ is the theory of monoids and $\mathbb{T}'$ is the theory of commutative monoids, I'm conjecturing that it's enough for $\mathbb{C}$ to have finite products and coequalizers, with the finite products distributing over the coequalizers. In fact I'll conjecture this whenever $\mathbb{T}'$ is obtained from $\mathbb{T}$ merely by adding extra laws - that is, equations between morphisms.

John Baez (Nov 03 2020 at 17:08):

And we don't need a Lawvere theory to describe monoids or commutative monoids: a prop is enough.

John Baez (Nov 03 2020 at 17:10):

So we can ask for any symmetric monoidal category $\mathcal{C}$ if the forgetful functor $\mathsf{ComMon}(\mathcal{C}) \to \mathsf{Mon}(\mathcal{C})$ has a left adjoint. And I will conjecture that this happens whenever $\mathcal{C}$ has coequalizers, with the tensor product distributing over coequalizers.

John Baez (Nov 03 2020 at 17:10):

And while I'm at it, I'll conjecture that this happens whenever we have a pair of props, with the second obtained from the first merely by adding extra laws.

John Baez (Nov 03 2020 at 17:11):

There are lots of obvious-sounding - or at least plausible - results about Lawvere theories and props that I haven't seen written up. There should really be a book that analyzes these things in detail.

Fawzi Hreiki (Nov 03 2020 at 17:20):

I had a quick look in the book Algebraic Theories by Adamek, Rosicky, and Vitale and I couldn’t find anything on this

Jade Master (Nov 03 2020 at 18:51):

A maybe sufficient but not necessary condition is that $C$ is a locally presentable category. Then because the forgetful functor is accessible and preserves finite limits (right?), it has a left adjoint by the adjoint functor theorem. Check out Theorem 2.2 here: https://ncatlab.org/nlab/show/adjoint+functor+theorem

John Baez (Nov 03 2020 at 19:03):

Sufficient but not necessary for what, exactly?

John Baez (Nov 03 2020 at 19:03):

Let me stick to the concrete question of whether we can abelianize monoids in C.

John Baez (Nov 03 2020 at 19:07):

C being locally presentable is certainly not necessary: for example neither FinSet nor Top are locally presentable, but we can abelianize monoids in these categories.

John Baez (Nov 03 2020 at 19:22):

I'm not quite sure how you're getting the sufficiency, Jade, but I'll take a guess. You want the forgetful functor U: ComMon(C) $\to$ Com(C) to have a left adjoint, so you're planning to check that if C is locally presentable, then:

1) ComMon(C) and Com(C) are locally presentable
2) U is accessible and it preserves all small limits

(You said "preserves finite limits", but that's not enough for Theorem 2.2 to kick in.)

John Baez (Nov 03 2020 at 19:24):

I bet this will all be true, but it seems like a massive hassle compared to just constructing the left adjoint "by hand", which seems to require only that C be symmetric monoidal, with equalizers, and that the tensor product distribute over equalizers.

John Baez (Nov 03 2020 at 19:30):

I have a dumb question for the experts in the room: does a locally presentable category automatically have all small limits?

John Baez (Nov 03 2020 at 19:34):

My intuition says yes, but I'm not finding that spelled out anywhere.

Todd Trimble (Nov 03 2020 at 19:40):

John Baez said:

I have a dumb question for the experts in the room: does a locally presentable category automatically have all small limits?

Yes, definitely. Not a dumb question. It's in "the Bible" (Adamek-Rosicky, Locally Presentable and Accessible Categories), Corollary 1.28.

John Baez (Nov 03 2020 at 19:43):

Thanks! I was looking through the bible and not finding it!

John Baez (Nov 03 2020 at 19:44):

I think I'm gonna stick this fact in the nLab. The nLab treatment is very heavily biased towards colimits, which makes sense of course, but one wants limits too.

John Baez (Nov 03 2020 at 19:47):

So it comes from the fact that a reflective subcategory of a complete (resp. cocomplete) category is again complete (resp. cocomplete).

John Baez (Nov 03 2020 at 19:49):

So why is a reflective subcategory of a complete category again complete? Yet another hole in my education.

John Baez (Nov 03 2020 at 19:50):

The nLab says it's because the inclusion of the reflective subcategory in the larger category is monadic.

Todd Trimble (Nov 03 2020 at 19:53):

John Baez said:

The nLab says it's because the inclusion of the reflective subcategory in the larger category is monadic.

Yes, it's true. The subcategory can be identified with the objects in the super-category for which the unit of the induced monad is an isomorphism (we were discussing that a little yesterday). You might find more information under "idempotent monad" (monads for which the counit is an isomorphism) -- all idempotent monads arise this way (via a reflective subcategory).

John Baez (Nov 03 2020 at 19:54):

The stuff you're saying, I kinda know. The hole in my education is something like "a monadic functor reflects limits"... is that it?

Todd Trimble (Nov 03 2020 at 19:54):

Yes, that's right.

John Baez (Nov 03 2020 at 19:56):

Now that I think about it, I've probably been told that a dozen times... for me, facts of that sort tend to go in one eye and out the other. Which is something I'm trying to cure.

Todd Trimble (Nov 03 2020 at 19:57):

It's not too hard to see. Limits are universal objects that are targets of cones. For example, if $X \times Y$ is a product (of underlying objects of algebras), then you can produce an algebra structure $T(X \times Y) \to X \times Y$ just by exploiting the universal property of a product.

John Baez (Nov 03 2020 at 20:05):

Thanks. I think my problem is that I got into category theory through knot theory, TQFTs and quantum mechanics, so I quickly got to like braided monoidal categories, symmetric monoidal categories, 2-categories and then higher categories, but none of this stuff focused my attention on the behavior of limits and colimits... which is half of category theory. So when I saw people saying something like "monadic functors reflect limits" I'd just say "yeah, yeah, another fact".

John Baez (Nov 03 2020 at 20:06):

But now the stuff I'm doing is, more and more, forcing me to learn such facts.

Todd Trimble (Nov 03 2020 at 20:07):

Yeah, for me it was the other way around. First I had to have universal properties really well under control.

John Baez (Nov 03 2020 at 20:14):

To me things like braided monoidal categories and such are intensely visual and thus very easy to understand.

Todd Trimble (Nov 03 2020 at 20:15):

But your conjecture above is interesting. The proposed abelianization (I'm American; it's with a 'z') I guess is a reflexive coequalizer of a pair of maps of the form $X \otimes X \to X$, one being the multiplication on the monoid $X$, the other being that twisted by a symmetry. That's a reflexive pair because both are left inverses of the evident map $X \to X \otimes X$ formed with the help of the unit. And you need that the monoidal product preserves reflexive coequalizers in each argument. I imagine your conjecture is correct, although I've never really thought about it before.

John Baez (Nov 03 2020 at 20:16):

Okay, trimming it down to reflexive coequalizers - a very Trimble-esque move in my opinion!

John Baez (Nov 03 2020 at 20:18):

If we need coequalizers to "impose extra properties" - e.g. to force an algebraic structure to obey extra laws, as I was discussing - what sort of special extra properties require only reflexive coequalizers?

Todd Trimble (Nov 03 2020 at 20:18):

It's actually an important move (and not my own). There is a strong tendency for monoidal products to distribute over reflexive coequalizers even if they don't distribute over arbitrary coequalizers. This is true for example for locally presentable categories.

John Baez (Nov 03 2020 at 20:19):

Oh, interesting.

Todd Trimble (Nov 03 2020 at 20:20):

Oh wait, I'm not sure about that last sentence. But the general point stands.

John Baez (Nov 03 2020 at 20:21):

I was imagining we could analyze maps of multi-sorted Lawvere theories $f: \mathbb{T} \to \mathbb{T}'$ into various kinds, corresponding to how much the forgetful functor $f^\ast : \mathrm{Mod}_C(\mathbb{T'}) \to \mathrm{Mod}_C(\mathbb{T}))$ forgets.

$f$ can add "extra stuff" (sorts), "extra structure" (operations), or "extra properties" (equations).

John Baez (Nov 03 2020 at 20:21):

Meaning that $f^\ast$ forgets stuff, structure or properties.

John Baez (Nov 03 2020 at 20:23):

If $f^*$ only forgets properties, the left adjoint of $f^\ast$, maybe I'll call it $f_!$ (???), seems to involve only coequalizers.

John Baez (Nov 03 2020 at 20:24):

That is, we can define it whenever $C$ has coequalizers, and I guess products distribute over them.

John Baez (Nov 03 2020 at 20:25):

But if $f^*$ forgets structure, we seem to need quite general (countable) colimits in $C$ to get a left adjoint for it.

Todd Trimble (Nov 03 2020 at 20:25):

I'm not sure about "stuff", but if the algebra categories $\text{Mod}_C(T)$ are monadic over $C$ (which is not guaranteed without nice properties of $C$), then you need only reflexive coequalizers. I tell ya, the reflexive coequalizer stuff is really important.

John Baez (Nov 03 2020 at 20:26):

What?

Todd Trimble (Nov 03 2020 at 20:27):

Sorry, what what?

John Baez (Nov 03 2020 at 20:28):

So if I'm working in $C = \mathsf{Set}$, say, and I form the free monoid on a set, I'm only using reflexive coequalizers?

John Baez (Nov 03 2020 at 20:28):

Or the free ring on a (multiplicative) monoid?

John Baez (Nov 03 2020 at 20:29):

In the first case I'm surely using countable coproducts (and products).

Todd Trimble (Nov 03 2020 at 20:31):

We're probably talking past each other. If you know already that the categories $\text{Mod}_C(T) = \text{Prod}(T, C)$ are monadic over $C$, then the forgetful functors $f^\ast$ have left adjoints under existence of reflexive coequalizers. Sorry for the miscommunication.

John Baez (Nov 03 2020 at 20:32):

Okay, I'm thinking about that in the case where $f: T \to T'$ is the map from theory of "plain objects" to the theory of monoids.

John Baez (Nov 03 2020 at 20:33):

Then $f^*$ is the forgetful functor from the category of monoids in $C$ to $C$.

John Baez (Nov 03 2020 at 20:33):

And I want to know how nice $C$ must be, for $f^*$ to have a left adjoint.

John Baez (Nov 03 2020 at 20:34):

Of course $C$ must have finite products for us to even play this game.

But you seem to be saying that if $C$ also has reflexive coequalizers, $f^* : \mathrm{Mon}(C) \to C$ has a left adjoint.

Todd Trimble (Nov 03 2020 at 20:34):

So anyway, yes, if your Lawvere theory $T$ is presentable by finitely many operations and finitely many equations, and if the functor $c \times -: C \to C$ preserves countable colimits for each object $c$ of $C$, then $\text{Prod}(T, C)$ is monadic over $C$.

Todd Trimble (Nov 03 2020 at 20:35):

John Baez said:

Of course $C$ must have finite products for us to even play this game.

But you seem to be saying that if $C$ also has reflexive coequalizers, $f^* : \mathrm{Mon}(C) \to C$ has a left adjoint.

No, I wasn't saying that.

John Baez (Nov 03 2020 at 20:36):

Okay, good.

John Baez (Nov 03 2020 at 20:37):

The question I've been asking about is when it does. Or more generally, under which conditions $f^* : \mathrm{Mod}_C (\mathbb{T}') \to \mathrm{Mod}_C (\mathbb{T})$ has a left adjoint, for some map of Lawvere theories $f$.

Todd Trimble (Nov 03 2020 at 20:39):

And I'm trying to answer that. If $C$ has sufficiently many colimits and if the functor $c \times -: C \to C$ preserves such colimits for each object $c$ of $C$, then all those forgetful functors have left adjoints. It doesn't have much to do with stuff, structure, properties unless I'm confused.

Todd Trimble (Nov 03 2020 at 20:42):

I was about to say: in the case of the forgetful functor $\text{Prod}(T, C) \to C$, the (underlying object of the) free algebra on an object $c$ is given by a coend

$\int^{n: \text{FinSet}} T(n, 1) \cdot c^n$

(as calculated in $C$) where $T$ is your Lawvere theory. This is a countable colimit if $T$ can be presented with only finitely many operations and equations.

Todd Trimble (Nov 03 2020 at 20:45):

Oh, but you're speaking of multisorted Lawvere theories. I think my answer is pretty much the same, but I'd have to think to be sure. There's a lot of stuff in that note on my nLab web page that's relevant.

John Baez (Nov 03 2020 at 20:46):

Todd Trimble said:

And I'm trying to answer that. If $C$ has sufficiently many colimits and if the functor $c \times -: C \to C$ preserves such colimits for each object $c$ of $C$, then all those forgetful functors have left adjoints. It doesn't have much to do with stuff, structure, properties unless I'm confused.

My question is about what kinds of colimits count as "sufficiently many colimits". For the forgetful functor from commutative monoids in C to monoids in C to have a left adjoint, I was conjecturing it's enough for C to have coequalizers (and $c \times -$ to preserve them). Or, for example, the forgetful functor from "groups where every element is 5-torsion" to "groups". These are cases of "forgetting properties". In these cases the left adjoint to the forgetful functor just "imposes extra equations", which we can do with coequalizers.

But for the forgetful functor from "commutative monoids" to "plain objects", we are forgetting structure. In this case the left adjoint to the forgetful functor involves coproducts as well as coequalizers.

Todd Trimble (Nov 03 2020 at 20:48):

Oh, gotcha. Haven't thought about it much.

John Baez (Nov 03 2020 at 20:48):

The only reason I brought in multisorted Lawvere theories is that then we can investigate "forgetting stuff", as well. But never mind.

Todd Trimble (Nov 03 2020 at 20:49):

This started off being about Jade's question, and I suspect your conjecture is correct there.

John Baez (Nov 03 2020 at 20:49):

So now you're adding extra twist of the knife: reflexive coequalizers. And this makes me wonder if these are sufficient for "extra properties of some particularly mild sort".... and if so, what does "mild sort" mean!

Todd Trimble (Nov 03 2020 at 20:50):

I don't know!

John Baez (Nov 03 2020 at 20:52):

Yeah, it's a weird question: I'd need to get a catalog of examples before I could get any intuition for this.

John Baez (Nov 03 2020 at 20:54):

I was imagining something a little less refined, like a 3-level factorization system on the category of multisorted Lawvere theories, where any morphism $f: \mathbb{T} \to \mathbb{T}'$ can be factored as one that only adds extra sorts, followed by one that only adds extra operations, followed by one that only adds extra equations... or something like that...

... and then maybe this will match up with some 3-level factorization system on the categories of models of these Lawvere theories, like the usual "forgetting just stuff, forgetting just structure, forgetting just properties" system.

Dan Doel (Nov 03 2020 at 20:54):

nlab says every congruence is a reflexive pair. So it might be that everything you'd actually want to say is mild enough.

John Baez (Nov 03 2020 at 20:55):

Interesting. Where does it say that?

Dan Doel (Nov 03 2020 at 20:55):

On the reflexive coequalizer page.

Todd Trimble (Nov 03 2020 at 20:57):

I mean, that much is true: a congruence is simply an internal equivalence relation $E \hookrightarrow X \times X$, and the induced pair of maps $E \rightrightarrows X$ are both left inverse to the map $X \to E$ coming from the reflexivity axiom.

John Baez (Nov 03 2020 at 20:59):

Okay, thanks. I'm not at all sure of the implications of this for my wonderings.... if any.

Todd Trimble (Nov 03 2020 at 21:01):

One thing is, your response to Jade was very general, looking at symmetric monoidal categories rather than just cartesian monoidal ones. Whereas the remark above about congruences refers to cartesian monoidal structure.

John Baez (Nov 03 2020 at 21:02):

Right.

John Baez (Nov 03 2020 at 21:03):

This is one reason I started dreaming of a big fat book that settled all these questions. There should be a big fat chapter on props, and a big fat chapter on Lawvere theories.

John Baez (Nov 03 2020 at 21:04):

And I guess I'm thinking of it in terms of a "big fat book" because I'd like to read the answers, but not actually do the work to figure them out. :upside_down:

Dan Doel (Nov 03 2020 at 21:05):

I guess I was thinking that all the equations you're adding in an algebraic theory would somehow be equivalence relations. But I'm not sure if that actually makes sense.

Todd Trimble (Nov 03 2020 at 21:06):

Depends on the doctrine where "algebraic theory" is being interpreted, I guess. I think that's right for the doctrine of cartesian monoidal categories. But I'm not sure how to interpret this for say the doctrine of symmetric monoidal categories (where "algebraic theory" means essentially "operad").

Todd Trimble (Nov 03 2020 at 21:10):

I think I'm on too little sleep, and typing so hurriedly, that I'm not thinking as clearly as I would like. (That's the trouble with me and this Zulip thing, which creates a kind of hectic sense in me that I don't really enjoy).

John Baez (Nov 03 2020 at 21:24):

There's no rush. The text just sits there, for years if necessary.

John Baez (Nov 03 2020 at 21:27):

I would be very happy to learn for which maps $f: \mathbb{T} \to \mathbb{T}'$ between Lawvere theories the functor $f^* : \mathrm{Mod}_C (\mathbb{T}') \to \mathrm{Mod}_C (\mathbb{T}')$ will have a left adjoint whenever $C$ has finite products and coequalizers, with the former distributing over the latter. And I will still be happy to learn this next year.

I conjecture it happens whenever $\mathbb{T}'$ is formed from $\mathbb{T}$ solely by adding extra equations: i.e., $f: \mathbb{T} \to \mathbb{T}'$ is full.

Todd Trimble (Nov 04 2020 at 12:05):

Todd Trimble said:

It's actually an important move (and not my own). There is a strong tendency for monoidal products to distribute over reflexive coequalizers even if they don't distribute over arbitrary coequalizers.

One of my favorite examples of that is the substitution product (say on graded sets), where $(X \circ Y)_n = \sum_k X_k \times Y^{\otimes k}(n)$ (here $\otimes$ is the Day convolution induced by thinking of $\mathbb{N}$ as a discrete monoidal category). Here $- \circ Y$ preserves all coequalizers and indeed all colimits, whereas $X \circ -$ preserves reflexive coequalizers but not much else.

David Michael Roberts (Nov 05 2020 at 03:07):

I was wondering, @Todd Trimble do reflexive coequalisers have better properties than arbitrary coequalisers?

sarahzrf (Nov 05 2020 at 03:27):

not to answer for todd, but i can give at least one example of a better property they have: reflexive coequalizers are sifted colimits (i.e., they commute with finite products in Set)

sarahzrf (Nov 05 2020 at 03:28):

so, for example, the category of models in Set (or, i think, in any grothendieck topos?) of a lawvere theory is gonna have reflexive coequalizers computed on the underlying sets (im pretty sure)

David Michael Roberts (Nov 05 2020 at 05:33):

Oh, nice. That's halfway to being pullback-stable, which is important for lots of things I do.

sarahzrf (Nov 05 2020 at 06:00):

hmmmm, not quite >.< image.png

David Michael Roberts (Nov 05 2020 at 08:42):

Well, I'm not an $\infty$-category theorist at present, but I'll file that point for later.