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Nov 9, 2021 at 18:30 vote accept user374433
Nov 8, 2021 at 0:16 comment added Simon Henry @AlecRhea Yes absolutely, it is not even monoidal.
Nov 8, 2021 at 0:14 comment added Simon Henry @მამუკაჯიბლაძე : I don't think so as these relations are also satisfied by objects with duals while the collaps of the action of (123) is only automatic for invertible objects. One way to think about it is that if X is invertible, then $X \otimes \_ $ is an equivalence of categories and so $Hom(X,X)$ is equivalent (as a monoid) to $Hom(1,1)$ where $1$ is the unit. In particular, $Hom(X,X)$ has to be commutative. It follows that as $X^{\otimes 3}$ is also invertible the action of $S_3$ on it takes values in a commutative monoids and hence has to factor through the signature map...
Nov 7, 2021 at 19:45 comment added Alec Rhea I'm confused -- the monoid of scalars on the unit in a monoidal category is always commutative, even if the category isn't symmetric (e.g. lemma 2, page 14 here); how does this mesh with your claim that the monoids of endomorphisms must all be non-commutative? Ah, the colimit you describe must not even be monoidal, got it. My bad.
Nov 7, 2021 at 16:13 comment added მამუკა ჯიბლაძე Could this (123) condition be also related to the requirement that the composites $Y\otimes X\to Y\otimes X\otimes X^{-1}\otimes X\to Y\otimes X$ and $Y\otimes X^{-1}\to Y\otimes X^{-1}\otimes X\otimes X^{-1}\to Y\otimes X^{-1}$ must be identities?
Nov 7, 2021 at 15:51 comment added Simon Henry True. I've also added a small explanation of where the (123) condition comes in.
Nov 7, 2021 at 15:51 history edited Simon Henry CC BY-SA 4.0
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Nov 7, 2021 at 15:30 comment added Maxime Ramzi The proof of Proposition 6 in Thomas Nikolaus' note on group completion also contains an explanation of this $(123)$-phenomenon
Nov 7, 2021 at 15:18 comment added user374433 Thank you very much for this wonderful answer! …I look forward to the paper :)
Nov 7, 2021 at 14:47 history answered Simon Henry CC BY-SA 4.0