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Stream: theory: category theory

Topic: Questions on free monoidal categories

Jean-Baptiste Vienney (Aug 21 2024 at 19:17):

I'm looking for several kind of free monoidal or symmetric monoidal categories for which I didn't find references.

Free monoidal category on a single object

In Categories for the Working Mathematician (Theorem 1, p.166) he builds the free monoidal category $\mathcal{F}$ on a single object. This is a category containing an object $*$ such that the following property is satisfied:

• for all monoidal category $\mathcal{C}$ and object $A \in \mathcal{C}$, there exists a unique strict monoidal functor $J_A:\mathcal{F} \rightarrow \mathcal{C}$ such that $J_A(*)=A$.

Wonderful, that's exactly what I want.

Free monoidal category on two objects and a single morphism

I would like a monoidal category $\mathcal{G}$ which contains two objects $0,1$ and a morphism $u:0 \rightarrow 1$ such that the following property is satisfied:

• for all monoidal category $\mathcal{C}$, objects $A,B \in \mathcal{C}$ and morphism $f \in \mathcal{C}[A,B]$, there exists a unique strict monoidal functor $J_f:\mathcal{F} \rightarrow \mathcal{C}$ such that $J_f(0)=A$, $J_f(1)=B$ and $J_f(u)=f$.

There isn't anything like this in MacLane's book. This a variant of a special case Theorem 3.3 in A survey of graphical languages for monoidal categories by Peter Selinger. But this Theorem 3.3 is not exactly what I want. He builds the free monoidal category on a monoidal signature as a graphical language. But this monoidal category is strict monoidal. It follows that it is not a free monoidal category in the same sense than mine. In his definition of free monoidal category he asks for a unique strict monoidal functor up to monoidal natural isomorphism. We can't expect to have unicity of the monoidal functor in the definition of free monoidal category with such a graphical category since we could send a tensor $X \otimes Y \otimes Z$ to either $(J_f(X) \otimes J_f(Y)) \otimes J_f(Z)$ or $J_f(X) \otimes (J_f(Y) \otimes J_f(Z))$ for instance, since $(X \otimes Y) \otimes Z=X \otimes (Y \otimes Z)$ in the graphical language.

Symmetric monoidal versions

I would also like the symmetric monoidal versions of these two constructions. In the book of MacLane, there is not even a variant of the Theorem 1 p.166 for symmetric monoidal categories. In fact he doesn't even state the coherence theorem for symmetric monoidal categories. He says p.255 that he let the reader formulate it precisely himself. (He only states a version without unitors.)

Question

Are there references for:

• the free monoidal category on two objects and a single morphism, in the sense that I ask for above,
• the free symmetric monoidal category on a single object, in the sense that I explained above for monoidal categories but in symmetric monoidal version
• the free symmetric monoidal category on two objects and a single morphism, in the sense that I explained above for monoidal categories but in symmetric monoidal version?

I want in each case a proof of the appropriate version of the theorem which says that for every ... there exists a unique (and not unique up to monoidal or symmetric monoidal natural isomorphism) strict monoidal or symmetric monoidal functor such that ...

Of course constructions of the free monoidal and symmetric monoidal categories on a monoidal signature with free defined with the same unicity of the monoidal or symmetric monoidal functor as above would be fine as well as I'm interested in two of the simplest cases of monoidal signatures.

Kevin Carlson (Aug 21 2024 at 19:50):

Well, there are 2-monads for all these things, is that enough for you, or are you more interested in the detailed construction?

Jean-Baptiste Vienney (Aug 21 2024 at 19:51):

I don't worry much about the construction. I just want to have the statement: for every ... there exists a unique monoidal / symmetric monoidal functor such that ..

Jean-Baptiste Vienney (Aug 21 2024 at 19:52):

but not "unique up to monoidal or symmetric monoidal natural isomorphism"

Jean-Baptiste Vienney (Aug 21 2024 at 19:54):

This is because I want to prove that monomorphisms in the category of (symmetric) monoidal categories and (symmetric) strict monoidal functors are exactly the faithful, injective on objects (symmetric) strict monoidal functors. And I don't see how to prove it without what I'm asking for.

Kevin Carlson (Aug 21 2024 at 21:42):

Yeah, 2-monads give you isomorphisms of hom categories, not just equivalences. In particular they restrict to give a monad on the category of categories for, say, monoidal categories and strict monoidal functors.

Kevin Carlson (Aug 21 2024 at 21:42):

I’d look at Steve Lack’s 2-categories companion first if you don’t have much exposure to 2-monads.

Jean-Baptiste Vienney (Aug 21 2024 at 22:11):

Thanks, I'm going to look at this paper

Jean-Baptiste Vienney (Aug 22 2024 at 02:48):

I looked at this paper but I didn’t see an answer to my question. Can you answer my question using $2$-monads?

Jean-Baptiste Vienney (Aug 22 2024 at 02:50):

You said that there is a monad on the category of categories for monoidal categories and strict monoidal functors.

Jean-Baptiste Vienney (Aug 22 2024 at 02:53):

Let $T:\mathbf{Cat} \rightarrow \mathbf{Cat}$ denote the corresponding functor.

Jean-Baptiste Vienney (Aug 22 2024 at 02:56):

If $*$ denotes the terminal category, can we prove that for all monoidal category $\mathcal{C}$ and object $A \in \mathcal{C}$, there exists a unique monoidal functor $F:T(*) \rightarrow \mathcal{C}$ such that $F(*)=A$?

Jean-Baptiste Vienney (Aug 22 2024 at 03:01):

If $\{0,1\}$ denotes the category with two objects $0,1$ and a single morphism from $0$ to $1$ denoted $0 \le 1$, can we prove that for all monoidal category $\mathcal{C}$, objects $A,B \in \mathcal{C}$ and morphism $f:A \rightarrow B$, there exists a unique monoidal functor $F:T(\{0,1\}) \rightarrow \mathcal{C}$ a such that $F(0)=A$, $F(1)=B$ and $F(0 \le 1)=f$?

Jean-Baptiste Vienney (Aug 22 2024 at 03:02):

That’s really all I care about.

Jean-Baptiste Vienney (Aug 22 2024 at 03:03):

(In addition to the corresponding facts for symmetric monoidal categories)

Jean-Baptiste Vienney (Aug 22 2024 at 03:18):

I already know how to define these free monoidal categories. I’m going to try to prove the freeness in the sense I’m talking about myself. I’m losing my times looking for references. I feel like 2-monads will not be useful in any way to this proof.

Jean-Baptiste Vienney (Aug 22 2024 at 03:18):

(I would be happy to be proven wrong though)

Jean-Baptiste Vienney (Aug 22 2024 at 03:39):

But proving this is essentially reproving MacLane coherence theorem.

Jean-Baptiste Vienney (Aug 22 2024 at 03:39):

Damn, is there anyone here who understands monoidal categories and could help me?

Jean-Baptiste Vienney (Aug 22 2024 at 03:46):

Now I think I'm going to use another formulation of MacLane coherence theorem to derive this.

Jean-Baptiste Vienney (Aug 22 2024 at 04:16):

Forget about this topic, I think that I can prove what I need and that this is much simpler that MacLane coherence theorem.

Jean-Baptiste Vienney (Aug 22 2024 at 04:20):

I just need to prove the existence of free monoidal categories and this is much easier than MacLane coherence theorem which states that every diagram in such a monoidal category commutes.

Notification Bot (Aug 22 2024 at 04:20):

Jean-Baptiste Vienney has marked this topic as resolved.

Notification Bot (Aug 22 2024 at 05:28):

Todd Trimble has marked this topic as unresolved.

Todd Trimble (Aug 22 2024 at 05:35):

Nevertheless, even though you marked this as resolved, you might want to have a look at the nLab article [[clique]], particular the section on monoidal strictifications, to appreciate the connection between Mac Lane's theorem on the structure of the free monoidal category on one generator and the fact that every monoidal category is monoidally equivalent to a strict monoidal category. The technique has a far-reaching generalization in terms of clubs, due to Kelly.

Jean-Baptiste Vienney (Aug 22 2024 at 12:58):

Thanks, I might ask you questions on this later.

Kevin Carlson (Aug 22 2024 at 18:54):

The statements you said you need are just statements of the universal property of any free algebra for a monad, it looks like, so I don’t understand why you were unhappy at that point. I agree that this has nothing critically to do with coherence, though there are related issues nearby.

Jean-Baptiste Vienney (Aug 22 2024 at 19:26):

In fact I didn't understand what I need. In Categories for the Working Mathematician he defines a category which is a preorder and then proves that this the free monoidal category on one generating object. This is a formulation of the coherence theorem for monoidal categories. At first I wanted to define the same kind of preorder and then prove that this is the free monoidal category on two generating objects and a single generating morphism. But I don't know how to do it.

In fact I just have to define a free monoidal category like in this paper and everything works fine. You can adapt what he's doing to a generating object and a generating morphism very easily and you can add the symmetry as well. Then it is very easy to see that you have defined the free monoidal or symmetric monoidal categories.

I think I was unhappy for two reasons:

• I didn't understand my problem and it was driving me crazy.
• I didn't find any construction of free monoidal category in the paper on 2-monads.

Jean-Baptiste Vienney (Aug 22 2024 at 19:28):

I think you have two ways to prove the coherence theorem for monoidal category:
(1) start with a category which is obviously a preorder and prove that it is a free monoidal category
(2) start with a category which is obviously a free monoidal category and prove that it is a preorder

Mac Lane (at least, in the theorem 1 p.166 of Categories for the Working Mathematician) takes the approach (1). In the paper above, they take the approach (2).

Jean-Baptiste Vienney (Aug 22 2024 at 19:30):

But what I need is just the "start with a category which is obviously a free monoidal category" (but not necessarily on a single object).

Todd Trimble (Aug 22 2024 at 19:54):

Let $F(x)$ denote the free monoidal category on one object $x$, and let $C$ be a category. $F(x)$ is a coproduct (disjoint union) of codiscrete categories $F(x)_n$ where a typical object is isomorphic to an $n$-fold tensor power $x^{\otimes n}$. The free monoidal category on $C$ can be constructed as a "wreath product"

$F(x) \wr C = \sum_{n \geq 0} F(x)_n \times C^n.$

Jean-Baptiste Vienney (Aug 22 2024 at 20:10):

Nice! Can you explain what is $F(x)_{n}$ more explicitly? And what is a codiscrete category?

Jean-Baptiste Vienney (Aug 22 2024 at 20:12):

Also I would imagine that you must also add something for the monoidal unit in $F(x)_n$ if you don't want the category to be strict monoidal.

Jean-Baptiste Vienney (Aug 22 2024 at 20:15):

(And I absolutely don't want the category to be strict monoidal since it breaks the unicity in the freeness property. I need this unicity for my proof that monomorphisms in the category of monoidal categories and strict monoidal functors are injective on objects and faithful.)

Todd Trimble (Aug 22 2024 at 20:16):

"Codiscrete" means that for any pair of objects $(x, y)$, there is exactly one morphism $x \to y$. So in other words, equivalent to the terminal category. The objects of $F(x)_n$ are formal bracketed tensor products of $x$ and the monoidal unit $I$ with exactly $n$ instances of $x$. ($F(x)$ is assuredly not strict monoidal, but it is monoidally equivalent to the discrete monoidal category $\mathbb{N}$ which is strict monoidal. This is of course Mac Lane's theorem.)

Jean-Baptiste Vienney (Aug 22 2024 at 20:22):

Ok. So $F(x)_n$ is here to encode the bracketing and the instances of monoidal units. That's a nice construction. And if you want the version for symmetric monoidal categories, I guess you put $n!$ morphisms from $a$ to $b$ for each pair of objects $(a,b)$ in $F(x)_n$ instead of just one?

Jean-Baptiste Vienney (Aug 22 2024 at 20:23):

I mean, you put the symmetric group on $n$ objects between $a$ and $b$ basically.

Todd Trimble (Aug 22 2024 at 20:25):

Yes, each hom is a $S_n$-torsor, correct.

Todd Trimble (Aug 22 2024 at 20:28):

The same type of deal holds for braided monoidal categories, and for a host of other 2-categories that are monadic over $\mathrm{Cat}$. The message behind Kelly's clubs is that in a wide class of such situations, the 2-monad can be reconstructed from the free structure on one object, via this sort of wreath product.

Jean-Baptiste Vienney (Aug 22 2024 at 20:30):

This is interesting when you know that MacLane coherence theorem is all about a free structure on one object.

Jean-Baptiste Vienney (Aug 22 2024 at 20:31):

Now let me try to prove that this construction indeed gives the free monoidal category. I don't know how to prove that it is a free monoidal category. But it looks easier to prove that the universal property is verified once you know that it is a monoidal category.

Jean-Baptiste Vienney (Aug 22 2024 at 20:32):

I guess that the coproduct is also a coproduct in the category of monoidal categories (and strict monoidal functors?).

Jean-Baptiste Vienney (Aug 22 2024 at 20:33):

I'm being very vague. I just try to see how it works. A monoidal functor from $\sum_{n \geq 0} F(x)_n \times C^n$ to a monoidal category $\mathcal{D}$ should amount to a monoidal functor from each $F(x)_n \times C^n$ to $\mathcal{D}$.

Jean-Baptiste Vienney (Aug 22 2024 at 20:34):

In fact no, it doesn't make sense ahah.

Todd Trimble (Aug 22 2024 at 20:34):

I guess that the coproduct is also a coproduct in the category of monoidal categories (and strict monoidal functors?).

So the coproduct of monoidal categories is more complicated than a disjoint union (think by analogy what the coproduct of two monoids is like).

Jean-Baptiste Vienney (Aug 22 2024 at 20:35):

I no longer think that each layer is a monoidal category. But they are categories.

Todd Trimble (Aug 22 2024 at 20:35):

Ah, I see what you are saying. I am not saying that the $F(x)_n$ are monoidal categories, but together they conspire to form one.

Jean-Baptiste Vienney (Aug 22 2024 at 20:36):

Yeah, my intuition is that it pretty much like the tensor algebra.

Todd Trimble (Aug 22 2024 at 20:36):

That's not a bad intuition.

Jean-Baptiste Vienney (Aug 22 2024 at 20:36):

Todd Trimble said:

I guess that the coproduct is also a coproduct in the category of monoidal categories (and strict monoidal functors?).

So the coproduct of monoidal categories is more complicated than a disjoint union (think by analogy what the coproduct of two monoids is like).

The coproduct of monoids is this thing called the "free product" I think.

Todd Trimble (Aug 22 2024 at 20:37):

Yes, similar to the free product of groups.

Jean-Baptiste Vienney (Aug 22 2024 at 20:38):

So you define $A \star B$ to be the set of all the words on $A \sqcup B$. This is a bit more complicated for groups because in the case of groups some words must be the same so I think you define "reduced words" instead.

Jean-Baptiste Vienney (Aug 22 2024 at 20:40):

For monoidal categories you should define $A \star B$ to be the free monoidal category on the category $A+B$ maybe??

Kevin Carlson (Aug 22 2024 at 20:41):

Well, it’s not exactly free; you have to also impose relations making the maps from $A,B$ monoidal. It’s “relatively” free in that objects of $A$ only multiply freely with objects of $B$.

Jean-Baptiste Vienney (Aug 22 2024 at 20:45):

But I have the inclusion functors from $A$ to $A+B$ and from $B$ to $A+B$ so if we write $M(A)$ for the free monoidal category on a category $A$, it gives me inclusion monoidal functors from $M(A)$ to $M(A+B)$ and from $M(B)$ to $M(A+B)$.

Todd Trimble (Aug 22 2024 at 20:47):

Monoidal functors out of the free monoidal category on $A + B$ would be tantamount to a pair of mere functors out of $A, B$, not monoidal functors, is what I think Kevin is saying.

Jean-Baptiste Vienney (Aug 22 2024 at 20:50):

Let me see. First to get the inclusion monoidal functors, I should precompose by a monoidal functor $A \rightarrow M(A)$. I think such a functor could exist but I'm not sure.

Todd Trimble (Aug 22 2024 at 20:51):

I'll let you ponder all this on your own for a while, and come back later! :-)

Jean-Baptiste Vienney (Aug 22 2024 at 20:52):

No a monoidal functor from $M(A+B)$ to a monoidal category $X$ is the same as ... I don't know. But if we replace $X$ by $M(X)$ then it is the same as a functor from $A$ to $X$ plus a functor from $B$ to $X$.

Todd Trimble (Aug 23 2024 at 17:53):

Huh? Monoidal functors $M(A + B) \to X$ are equivalent to functors $A + B \to X$ (by freeness) which are equivalent to pairs of functors $A \to X, B \to X$.

Jean-Baptiste Vienney (Aug 23 2024 at 17:59):

Oh, yes of course! I was confused by $M$ being a functor so I was thinking that there is a correspondence between functor from $A$ to $B$ and monoidal functor from $M(A)$ to $M(B)$ .

Jean-Baptiste Vienney (Aug 23 2024 at 18:00):

But yeah, now I understand: you must use the adjunction.

Jean-Baptiste Vienney (Aug 23 2024 at 18:02):

Let me remember what we were saying. I was trying to show that
$M(A+B)$ is a coproduct of $A$ and $B$ in the category of monoidal categories and strict monoidal functors.

Jean-Baptiste Vienney (Aug 23 2024 at 18:04):

So, for the injections you have
$A \rightarrow M(A) \rightarrow M(A+B)$
where the second map is $M(A \rightarrow A+B)$ and the first map is the unit of the monad given by the free-forgetful adjunction.

Jean-Baptiste Vienney (Aug 23 2024 at 18:05):

Uh but $A \rightarrow M(A)$ is not necessarily a strict monoidal functor?

Jean-Baptiste Vienney (Aug 23 2024 at 18:06):

Since $M$ must be a monad on $\mathcal{Cat}$.

Jean-Baptiste Vienney (Aug 23 2024 at 18:10):

Maybe $A \rightarrow M(A)$ is strict monoidal but anyway

Jean-Baptiste Vienney (Aug 23 2024 at 18:10):

Todd Trimble said:

Huh? Monoidal functors $M(A + B) \to X$ are equivalent to functors $A + B \to X$ (by freeness) which are equivalent to pairs of functors $A \to X, B \to X$.

with that I'm convinced that $M(A+B)$ is not the coproduct of $A$ and $B$.

Todd Trimble (Aug 23 2024 at 18:47):

Jean-Baptiste Vienney said:

Uh but $A \rightarrow M(A)$ is not necessarily a strict monoidal functor?

Not at all, not even monoidal much less strict monoidal. (I assume you mean the unit of the adjunction.) If $A$ is an $M$-algebra and $u_A: A \to MA$ is an algebra map, then the algebra structure map $\xi: MA \to A$ is an isomorphism. Proof: two algebra maps $MA \rightrightarrows B$ are equal iff their compositions with $u_A$ are equal (again by freeness). Taking $B = MA$, apply this observation to the algebra maps $u_A \circ \xi$ and $1_{MA}$: their compositions with $u_A$ are both $u_A$, using the fact that $\xi \circ u_A = 1_A$ by the unit axiom for $M$-algebras. So $u_A \circ \xi = 1_{MA}$, and $\xi \circ u_A = 1_A$ as we just said.

The same type of proof applies to 2-monads (replacing equalities by isomorphisms).

Jean-Baptiste Vienney (Aug 23 2024 at 19:49):

Cool, and I guess that the structure map $MA \rightarrow A$ is rarely an isomorphism (when $A$ is already an algebra, here a monoidal category). Hmm, maybe we could an obtain a counterexample by looking at what it's doing on some tiny monoidal category $A$.

Jean-Baptiste Vienney (Aug 23 2024 at 19:51):

Honestly I don't even have much intuition for what this map is doing.

Jean-Baptiste Vienney (Aug 23 2024 at 19:52):

Finally I have some intuition.

Jean-Baptiste Vienney (Aug 23 2024 at 19:53):

I guess it's doing something like this: from a "formal" tensor product of objects in $A$, it gives their "real" tensor product in $A$.

Jean-Baptiste Vienney (Aug 23 2024 at 19:57):

This map has a section given by the functor $A \rightarrow MA$ which is the unit of the adjunction. So it is always an epimorphism.

Jean-Baptiste Vienney (Aug 23 2024 at 19:58):

Now, I think that for some monoidal categories $A$ it will not be a monomorphism.

Jean-Baptiste Vienney (Aug 23 2024 at 20:00):

Ok, I see why. This is like when you take a double tensor algebra.

Jean-Baptiste Vienney (Aug 23 2024 at 20:01):

In $MA$ you have two levels of tensors.

Jean-Baptiste Vienney (Aug 23 2024 at 20:03):

The objects of $MA$ can be written like this:
$(A_{1,1} \otimes ... \otimes A_{1,n_1}) \boxtimes ... \boxtimes (A_{p,1} \otimes ... \otimes A_{p,n_p})$ where $\otimes$ is the tensor product of $A$ and $\boxtimes$ is the tensor product of $MA$.

Jean-Baptiste Vienney (Aug 23 2024 at 20:04):

But both $A \otimes B$ and $A \boxtimes B$ will be sent to $A \otimes B$ by the structure map. Or even simpler: both the monoidal unit of $A$ and the monoidal unit of $MA$ will be sent to the monoidal unit of $A$.

Jean-Baptiste Vienney (Aug 23 2024 at 20:05):

So the structure map is never a monomorphism, whatever the monoidal category $A$ is.

Jean-Baptiste Vienney (Aug 23 2024 at 20:26):

Hmm, but we just had to write $(A,B,C,...)$ instead of $A \boxtimes B \boxtimes C ...$ here.

Jean-Baptiste Vienney (Aug 23 2024 at 20:26):

Since $MA$ is defined using the $A^n$.

Todd Trimble (Aug 23 2024 at 20:34):

In ordinary 1-category theory, if you want to form the coproduct of two $M$-algebras (or really a family of $M$-algebras), and if you assume

• The underlying category has coproducts,
• The category of algebras has reflexive coequalizers,

then what you can do is follow the construction of Theorem 2.2 here. This is a suitable quotient of $M(\sum_i A_i)$, and more or less manifests what Kevin was driving at earlier (except that we're in a 2-categorical situation, so you should use a codescent object in place of a reflexive coequalizer, but the basic idea is quite similar).

Jean-Baptiste Vienney (Aug 23 2024 at 20:57):

I didn't know what reflexive coequalizers are useful for. Now I will remember that they are useful for this. Cool. I will let you know whether I understand the construction of Theorem 2.2 or not.

Todd Trimble (Aug 23 2024 at 21:00):

Reflexive coequalizers turn out to be useful for lots of things. One of their primary virtues is that they tend to be preserved by a lot more functors than general coequalizers are, and yet they are just as expressive if there are coproducts around (because any coequalizer can be re-presented as a reflexive coequalizer by making suitable use of coproducts, as the proof of Theorem 2.2 explains).