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Aug 4, 2023 at 14:10 vote accept John Baez
Sep 19, 2020 at 22:06 comment added John Baez Whoops. You're right.
Sep 18, 2020 at 21:54 comment added Alexander Campbell @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.
Sep 17, 2020 at 19:48 comment added John Baez The definition of coproduct of objects in a bicategory (or 2-category) doesn't mention any non-invertible 2-morphisms, so it doesn't hurt, for this purpose, to assume all the 2-morphisms are invertible.
Sep 17, 2020 at 17:46 answer added John Baez timeline score: 6
Sep 16, 2020 at 14:31 comment added Maxime Ramzi @TimCampion : Yes, I agree, I expect so too; but I definitely don't have enough experience in the $(2,2)$-world to be sure
Sep 16, 2020 at 14:27 comment added Tim Campion @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.
Sep 16, 2020 at 9:26 comment added Maxime Ramzi @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
Sep 15, 2020 at 22:25 comment added Tim Campion 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
Sep 15, 2020 at 20:21 history asked John Baez CC BY-SA 4.0