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The category of commutative monoid objects in a symmetric monoidal category is cocartesian, with their tensor product serving as their coproduct. This sort of result seems to date back to here:

  • Thomas Fox, Coalgebras and Cartesian categories, Commun. Algebra 4 (1976), 665–667.

I'm working on a paper with Todd Trimble and Joe Moeller, and right now we need something similar one level up — that is, for symmetric pseudomonoids. (For example, a symmetric pseudomonoid in Cat is a symmetric monoidal category.)

The 2-category of symmetric pseudomonoids in a symmetric monoidal 2-category should be cocartesian, with their tensor product serving as their coproduct. I imagine the coproduct universal property will hold only up to 2-iso.

Has someone proved this already? This paper:

proves the result in the special case where the symmetric monoidal 2-category is Cat. In fact they do more, in this special case:

Theorem 2.3. The 2-category SMC of symmetric monoidal categories, strong monoidal functors, and monoidal natural transformations has 2-categorical biproducts.

Unfortunately their proof is not purely 'formal', so it doesn't instantly generalize to other symmetric monoidal 2-categories. And I believe the fact that the coproducts in SMC are biproducts must rely on the fact that Cat is a cartesian 2-category.

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    $\begingroup$ Two ideas: (1) It's surely been shown in general that if $C$ is a symmetric monoidal $\infty$-category, then in $Alg_{E_\infty}(C)$ the coproduct and tensor product coincide. This should specialize to what you want by taking $C$ to be the (2,1)-category of categories (maybe the appropriate comparisons (still!) haven't been done). (2) Perhaps this can be deduced representably from Fong-Spivak: a symm.pseudomonoid in $C$ should be a certain kind of lift of a representable 2-functor $C \to Cat$ through the forgetful 2-functor $SymPsMon \to Cat$, with coproduct and tensor product defined levelwise $\endgroup$
    – Tim Campion
    Commented Sep 15, 2020 at 22:25
  • $\begingroup$ @TimCampion : in (1), don't you mean a general (2,1)-symmetric monoidal category ? Which would not be enough if John is interested in general 2-categories with $\endgroup$
    – Maxime Ramzi
    Commented Sep 16, 2020 at 9:26
  • $\begingroup$ @MaximeRamzi Ah, good point. Although typically I should expect that coproducts in a 2-category can be detected at the (2,1) level, and similarly with the tensor product. $\endgroup$
    – Tim Campion
    Commented Sep 16, 2020 at 14:27
  • $\begingroup$ @TimCampion : Yes, I agree, I expect so too; but I definitely don't have enough experience in the $(2,2)$-world to be sure $\endgroup$
    – Maxime Ramzi
    Commented Sep 16, 2020 at 14:31
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    $\begingroup$ @JohnBaez This isn't true. The universal property of a coproduct of objects in a bicategory involves an equivalence of categories (not just groupoids): $Hom(\sum_i A_i,B) \simeq \prod_i Hom(A_i,B)$. However, if the bicategory admits cotensors by the walking arrow, then coproducts are detected at the level of the underlying (2,1)-category. $\endgroup$
    – Alexander Campbell
    Commented Sep 18, 2020 at 21:54

1 Answer 1

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The result I wanted is Theorem 5.2 here:

Daniel Schäppi, Ind-abelian categories and quasi-coherent sheaves, Mathematical Proceedings of the Cambridge Philosophical Society, 157 (2014), 391–423. doi:10.1017/S0305004114000401

The proof appears in Appendix A. He proves the result for symmetric pseudomonoids in a symmetric monoidal bicategory.

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  • $\begingroup$ Could you accept this as an answer, so that the question is marked as resolved? $\endgroup$
    – varkor
    Commented Aug 4, 2023 at 8:55
  • $\begingroup$ I tried, and StackExchange says "you can't vote on your own post". $\endgroup$
    – John Baez
    Commented Aug 4, 2023 at 11:48
  • $\begingroup$ You can't upvote it, but you should be able to click the checkmark symbol to mark it as an accepted answer. (On other people's answers, you have both options, which are independent, but if you answer your own question, you only have the option to accept.) $\endgroup$
    – varkor
    Commented Aug 4, 2023 at 11:50
  • $\begingroup$ Oh, right! Done! $\endgroup$
    – John Baez
    Commented Aug 4, 2023 at 14:10
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    $\begingroup$ Thanks! (I notice that you haven't accepted answers on many of your other questions: perhaps you had intended to accept them, but didn't realise about the checkmark. If so, it could be helpful to look through your past questions and accept the answers you intended to :) ) $\endgroup$
    – varkor
    Commented Aug 4, 2023 at 15:09

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