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Nov 14, 2021 at 15:04 history edited Tim Campion CC BY-SA 4.0
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Nov 14, 2021 at 14:18 comment added Tim Campion @AndréHenriques I haven't carefully considered Dmitri's comment (and to be honest, I have not carefully checked the above construction I outline), but note that if C is a symmetric monoidal groupoid with duals, then those duals must be inverses for all objects, and then maybe it's a bit more believable? EDIT: Hm... maybe my formula is wrong!
Nov 14, 2021 at 13:05 comment added André Henriques If C already has duals, then formally adding duals should not change C (the operation of formally adding duals should be idempotent). Now, if C is a groupoid and C has duals then, according to Dmitri Pavlov's comment, the set of morphisms from (a,I) to (b,I) can be computed as the disjoint union of C(a⊗x,b⊗x) over all isomorphism classes of objects x. That's not C... Am I missing something?
Nov 14, 2021 at 0:52 comment added Dmitri Pavlov If C is a groupoid, this formula says that the set of morphisms from (a,I) to (b,I) can be computed as the disjoint union of C(a⊗x,b⊗x) over all isomorphism classes of objects x. In particular, the embedding of C(a,b) into this set is clearly injective. So it appears that the answer is positive in case of groupoids.
Nov 11, 2021 at 12:30 history answered Tim Campion CC BY-SA 4.0