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Stream: theory: category theory

Topic: Endofunctors of a cartesian monoidal category are opmonoidal

Jean-Baptiste Vienney (Jul 14 2024 at 20:26):

Let $\mathcal{C}$ be a category with finite products and $F:\mathcal{C} \rightarrow \mathcal{C}$ an endofunctor.

Consider the two projections $p:A \times B \rightarrow A$ and $q:A \times B \rightarrow B$. We then obtain $F(p):F(A \times B) \rightarrow F(A)$ and $F(q):F(A \times B) \rightarrow F(B)$. Finally, it gives us a map
$\langle F(p),F(q) \rangle:F(A \times B) \rightarrow F(A) \times F(B)$.

We also have a (unique) map $F(\top) \rightarrow \top$.

1. I think that it makes $F$ into an opmonoidal functor.
2. In the same way I'm wondering if every monad on a category with finite products is not automatically an opmonoidal monad?

Question: Are $1.$ and $2.$ true or false? Are there any references on this?

Matteo Capucci (he/him) (Jul 15 2024 at 06:26):

Every functor from a monoidal category to a Cartesian category is automatically opmonoidal in the way you describe. The usual way this fact is framed is that Cartesian cats are algebras of the product completion 2-monad on Cat, which is colax idempotent, and one of the main properties of such monads is that every functor between carrier of their algebras is automatically colax.

Vincent Moreau (Jul 15 2024 at 08:48):

I agree, $1.$ is true, where the domain and codomain of the functor may be different cartesian categories. By the way, given the fact that comonoids in a monoidal category $\mathbf{C}$ correspond to oplax/colax functors $\mathbf{1} \to \mathbf{C}$, this gives another explanation of the fact that, when $\mathbf{C}$ is cartesian, any object (i.e. functor $\mathbf{1} \to \mathbf{C}$) admits a unique structure of comonoid (i.e. the functor has a unique oplax structure) :smile:

Vincent Moreau (Jul 15 2024 at 08:50):

Matteo Capucci (he/him) said:

Every functor from a monoidal category to a Cartesian category is automatically opmonoidal in the way you describe. The usual way this fact is framed is that Cartesian cats are algebras of the product completion 2-monad on Cat, which is colax idempotent, and one of the main properties of such monads is that every functor between carrier of their algebras is automatically colax.

This looks very interesting, but I am not sure how monoidal categories are related to the product completion 2-monad. Could you explain a bit or give a reference?

Jean-Baptiste Vienney (Jul 15 2024 at 19:31):

Maybe we could obtain a “relative” version of Fox’s theorem from Vincent’s observation. I’m not sure what the precise statement would be. Maybe “a symmetric monoidal category is Cartesian iff its category of opmonoidal endofunctors is isomorphic to its category of endodunctors”.

By the way I think the statement from Fox’s paper is more complete that the one on the nLab so we should start from there.

Jean-Baptiste Vienney (Jul 16 2024 at 01:10):

Probably we should replace “opmonoidal functor” by “cocommutative opmonoidal functor” since Fox’s theorem is about cocommutative monoids. We would say that an opmonoidal functor is cocommutative when $\sout{F(A \otimes A) \rightarrow F(A) \otimes F(A)}$ equalizes the identity and the swap. EDIT: no, the structure for free of opmonoidal functor on an endofunctor of a cartesian monoidal category is not cocommutative in this sense in general. It is cocommutative in this sense when $F$ is a constant functor but this is a very special case.

Jean-Baptiste Vienney (Jul 16 2024 at 01:13):

(this is all highly speculative)

Matteo Capucci (he/him) (Jul 16 2024 at 07:36):

Vincent Moreau said:

Matteo Capucci (he/him) said:

Every functor from a monoidal category to a Cartesian category is automatically opmonoidal in the way you describe. The usual way this fact is framed is that Cartesian cats are algebras of the product completion 2-monad on Cat, which is colax idempotent, and one of the main properties of such monads is that every functor between carrier of their algebras is automatically colax.

This looks very interesting, but I am not sure how monoidal categories are related to the product completion 2-monad. Could you explain a bit or give a reference?

That theorem only says that a functor between cartesian categories has a unique colax monoidal structure, I misquoted it, I meant to write

Every functor from a Cartesian monoidal category to a Cartesian monoidal category is automatically opmonoidal in the way you describe

Vincent Moreau (Jul 16 2024 at 11:08):

Thanks!

Jean-Baptiste Vienney (Jul 16 2024 at 17:50):

Guys, I'm considering dropping a short note with a title like "A characterization of cartesian monoidal categories among monoidal categories via the category of endofunctors" on the ArXiv.

The theorem should be something close to this:

"A monoidal category $\mathcal{C}$ is cartesian if and only if the forgetful functor $U:\mathbf{OpMonEnd}(\mathcal{C}) \rightarrow \mathbf{End}(\mathcal{C})$ is an isomorphism in $\mathbf{Cat}$."

where all the terms used in this sentence will have been carefully defined.

I want to write: "The author thanks @Matteo Capucci (he/him) and @Vincent Moreau for an helpful discussion on the Zulip Category Theory Community Server".

Let me know if there is anything wrong with this.

Vincent Moreau (Jul 16 2024 at 17:54):

Sounds good to me! Don't hesitate to give us updates on this channel :smile:

Matteo Capucci (he/him) (Jul 16 2024 at 18:00):

I'm ok with that, thanks!

Vincent Moreau (Jul 16 2024 at 18:03):

This is interesting as indeed, any natural transformation between cartesian functors is automatically a monoidal natural transformation, so your $U$ indeed seems to be an iso, that's cool!

Jean-Baptiste Vienney (Jul 16 2024 at 21:18):

I think I was missing a piece concerning cocommutativity/symmetry:

The statement looks more likely if we replace "monoidal category" by "symmetric monoidal category" and "opmonoidal functor" by "symmetric opmonoidal functor". The symmetric opmonoidal functor requirement should make every object $A$ into a cocommutative comonoid.

Jean-Baptiste Vienney (Jul 16 2024 at 21:23):

(the "symmetry" diagram looks weaker than a cocommutativity diagram at first sight but when you're dealing with a constant functor one of the two symmetries in the commutative square is killed.)

Jean-Baptiste Vienney (Jul 19 2024 at 13:30):

I think I was still missing something. Here is an updated conjecture.

In the characterizations of cartesian monoidal categories as monoidal categories with extra structure, a structure of comonoid is required on every object, which must be natural and compatible with the tensor product. cf. this blog post where the compatibility with the tensor product is called "uniformity". I think I was missing this compatibility with the tensor product.

If $\mathcal{C}$ is a category, call cartesian structure on $\mathcal{C}$ the data
$(\top,(A \times B, p_{A,B}:A \times B \rightarrow A, q_{A,B}:A \times B \rightarrow B)_{A,B \in \mathcal{C}})$
of a terminal object and of a product for every pair of objects.

Every cartesian structure induces a structure of symmetric monoidal category on $\mathcal{C}$. Let's call cartesian monoidal category a symmetric monoidal category $\mathcal{C}$ whose structure of symmetric monoidal category is induced by a cartesian structure.

For every category $\mathcal{C}$, we have its category of endofunctors and natural transformations, that I note $\mathrm{End}(\mathcal{C})$. If $\mathcal{C}$ is a symmetric monoidal category, then we can consider the category $\mathrm{SymOpMonEnd}(\mathcal{C})$ of symmetric opmonoidal functors and opmonoidal natural transformations between them.

For every symmetric monoidal category $\mathcal{C}$, we have a forgetful functor
$U(\mathcal{C}):\mathrm{SymOpMonEnd}(\mathcal{C}) \rightarrow \mathrm{End}(\mathcal{C})$.

We need a bit more than this forgetful functor for our purposes. This is the fix for the missing piece I was talking about before.

If $\mathcal{C}$ is a symmetric monoidal category, then $\mathrm{End}(\mathcal{C})$ can be made into a symmetric monoidal category $\mathrm{End}_{\otimes}(\mathcal{C})$ where the monoidal structure is induced pointwisely by the one of $\mathcal{C}$. For instance, we define $(F \otimes G)(A):=F(A) \otimes G(A)$ and $(\alpha \otimes \beta)_A := \alpha_A \otimes \beta_A$ etc...

In the same way, if $\mathcal{C}$ is a symmetric monoidal category, then $\mathrm{SymOpMonEnd(\mathcal{C})}$ can be made into a symmetric monoidal category $\mathrm{SymOpMonEnd}_{\otimes}(\mathcal{C})$. We obtain accordingly a symmetric strict monoidal forgetful functor
$U_{\otimes}(\mathcal{C}):\mathrm{SymOpMonEnd}_{\otimes}(\mathcal{C}) \rightarrow \mathrm{End}_{\otimes}(\mathcal{C})$.

Note $\mathrm{StrictSymMonCat}$ the category of symmetric monoidal categories and symmetric strict monoidal functors between them.

Conjecture: A symmetric monoidal category $\mathcal{C}$ is a cartesian monoidal category if and only if $U_{\otimes}(\mathcal{C})$ is an isomorphism in $\mathrm{StrictSymMonCat}$.