I am studying Hopf algebras in categories, and I hope, somebody could help me with the following.

Joost Vercruysse in his paper Hopf algebras---Variant notions and reconstruction theorems writes (Theorem 3.6) that the caracterization of Hopf algebras as bialgebras whose category of Hopf modules possesses some "good" properties is generalized to Hopf algebras in "arbitrary" braided categories:

Let $H$ be a bialgebra in a braided monoidal category $C$ with equalizers, then the following statements are equivalent:

(i) the functor $(−)^{coH}:C_H^H\to C$ is fully faithful;

(ii) the pair $(− ⊗ H,(−)^{coH})$ is an equivalence of categories between $C$ and $C_H^H$;

(iii) $(m ⊗ H)\circ (H ⊗ ∆) : H ⊗ H \to H ⊗ H$ is a $C$-isomorphism;

(iv) $H$ admits an antipode, i.e. $H$ is a Hopf algebra in $C$

(here $C_H^H$ is the category of "right-right" Hopf modules over $H$, and $(−)^{coH}$ is the right adjoint functor for $− ⊗ H$).

Vercruysse mentiones this without proof. I think, this is the effect of folklore, when something obvious for specialists is not clear for people who are not aware with the context, but I don't understand how this is proved. Could anybody enlighten me? If this is easy you can just explain this here, but if not, I hope, you could give me a reference.

  • 1
    $\begingroup$ Do you know the proof in the classical/Vect case ? I think it should extends fairly easily to this more general braided case just by writing things diagrammatically. If you're asking for a reference already for that there are many, e.g. section 3.2 in homepages.vub.ac.be/~scaenepe/Hopfalgebra.pdf . More precisely the equivalences between (i) and (ii), and between (iii) and (iv) are fairly obvious, and the reference above implies $(ii)\Leftrightarrow (iv)$. $\endgroup$ – Adrien Jul 30 at 8:39
  • $\begingroup$ Adrien, thank you! This looks very useful! $\endgroup$ – Sergei Akbarov Jul 30 at 11:32
  • $\begingroup$ ...Although I don't see how this text implies $(ii)\Rightarrow(iv)$. $\endgroup$ – Sergei Akbarov Jul 30 at 14:00
  • $\begingroup$ Regarding the braided case, maybe the following article can be of some interest to the OP: researchgate.net/publication/…. (although i am not sure if the implication $(ii)\Rightarrow(iv)$ is made explicit). $\endgroup$ – Konstantinos Kanakoglou Jul 31 at 15:55

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