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Nov 29, 2021 at 8:01 comment added Bugs Bunny No, it is not. Take ${\mathcal C}$ to be vector space. Then $Nat(U_A,U_A)$ is typically not a vector space.
Nov 26, 2021 at 11:47 vote accept yohei ohta
Nov 26, 2021 at 9:55 comment added yohei ohta Thx! I would like to ask another related question. I think Tannaka reconstruction theorem holds for an arbitary complete closed monoidal category $\mathcal{C}$. Let A be a monoid in $\mathcal{C}$ ,$A\text{-}\mathrm{Mod}$ be a category of left A-modules in $\mathcal{C}$ and $U_A:A\text{-}\mathrm{Mod} \to \mathcal{C}$ be a forgetful functor. Is $\mathrm{Nat}(U_A,U_A)$ a object of $\mathcal{C}$?
Nov 26, 2021 at 8:25 comment added Bugs Bunny It is a set because your category is "presentable". It has a generator ${}_AA$, a value on which determines a natural endomorphism.
Nov 26, 2021 at 0:41 comment added yohei ohta Is $\mathrm{End}(U_A)$ a set? If F and G are functors, I think $\mathrm{Nat}(F,G)$ is not a set in general.
Nov 25, 2021 at 12:11 history answered Bugs Bunny CC BY-SA 4.0