Re: Monoidal product functor is strong monoidal, when?

[David Roberts <droberts.65537@gmail.com>]

Hi Takuo,

thanks for that observation, it's rather nice way to put it.

Aaron Mazel-Gee had another way to show it, which he shared with me
privately, and gave me permission to pass on to the list, copied
below.

It turns out that there are two diagrams that prove the result about
symmetric monoidal functors (if one takes an elementary approach, and
not using strictification, as Joyal–Street do in the published
'Braided tensor categories'), which are two halves of the generalised
resultoassociahedron on the middle of page 39 of
http://web.science.mq.edu.au/~street/BatanAustMSMq.pdf, originally
appearing in work of Bar-Natan in 1993 (or so, it's a little hard to
recognise). If one categorified this result, then one could have a
3-arrow that filled this polyhedral diagram of 2-arrows.

=====
On Tue, 17 Dec 2019 at 08:46, Aaron Mazel-Gee wrote:
>Hi David,

>It sounds like this is (once again) the opposite what you're looking for, but I would say that this is a special instance of a more general fact.

>Let (V,⊠) be a symmetric monoidal ∞-category, and write CAlg(V) for its ∞-category of commutative algebras (a.k.a. E_∞-algebras).

>(1) CAlg(V) admits finite coproducts, and the forgetful functor CAlg(V) --> V canonically enhances to a symmetric monoidal functor (CAlg(V),∐) --> (V,⊠).

>(1') In particular, for any pair of objects A,B∈CAlg(V), one might  write A⊠B∈CAlg(V) for their coproduct.

>(2) Using the notation (1'), for any A,B∈CAlg(V), there is a canonical enhancement of the multiplication map A⊠A --> A from a morphism in V to a morphism in CAlg(V).

>(3) For any ∞-operad O, note the existence of a forgetful functor CAlg(V) --> Alg_O(V).



>Now, take V to be the (2,1)-category Cat, equipped with the cartesian symmetric monoidal structure.  Then, an E_∞-algebra in V is nothing more or less than a symmetric monoidal category C.  By (2), the symmetric monoidal product C x C --> C admits a canonical enhancement to a symmetric monoidal functor.  And if you like, you can apply (3) with O=E_2.

======

Thanks,
David

David Roberts
Webpage: https://ncatlab.org/nlab/show/David+Roberts
Blog: https://thehighergeometer.wordpress.com

On Sun, 12 Jan 2020 at 01:15, Matsuoka Takuo <motogeomtop@gmail.com> wrote:
>
> Hi David,
>
> Your message caught my attention on my spam tray by Gmail's fault.
>
> I'm not sure what proof would be nice to you, but as far as I could see,
> the construction of a symmetric monoidality of the multiplication of a
>> _symmetric_ monoidal category is largely trivial. Let Fin denote the
> category of finite sets. Then, a symmetric monoidal category (C,@) gives
> you a symmetric monoidal functor Fin ---> Cat which associates to a finite
> set S the category C^S. The symmetric monoidal structure of C x C gives
> you a symmetric monoidal structure on the functor Fin ---> Cat
> associating C^S x C^S to S. Inspecting this symmetric monoidal functor,
> you further obtain a map of these symmetric monoidal functors which
> associates to S the multiplication functor
> C^S x C^S ---> C^S induced from the codiagonal map S + S ---> S, where
> "+" in the source denotes the coproduct operation in
> Fin. This is the desired structure.
>
> As you see, we have used the symmetric monoidality of the product
> functor Cat x Cat ---> Cat, which you have because the Cartesian product
> is a limit so preserves products. Thus, a reference you are looking for
> may be
>
> Graeme Segal, Categories and cohomology theories, Topology 13 (1974),
>
> which essentially contains a sufficient argument for this (and is indeed
> earlier than Joyal–Street).
>
> To conclude, "commutation with the braiding" comes immediately from
> the naturality of the codiagonal map, to commute with any
> automorphism of a finite set.
>
> Best regards,
> Takuo Matsuoka
>
>

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