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I asked this initially in math.stackexchange:

The group algebra $k(G)$ of any group $G$ satisfies as a Hopf algebra the following identities: $$ S\otimes S\circ \Delta=\sigma\circ\Delta\circ S $$ $$ \nabla\circ S\otimes S=S\circ\nabla\circ\sigma $$ where $S$ is the antipode, $\Delta$, the comultiplication, $\nabla$, the multiplication, and $\sigma:x\otimes y\mapsto y\otimes x$.

Is this valid for all Hopf algebras (in any braided monoidal category) or only for some special class?

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Those identities are usually described by saying that the antipode map is an antihomomorphism of coalgebras and of algebras, and should be proved in most textbooks on the subject. – Mariano Suárez-Alvarez Aug 12 '14 at 8:42
up vote 3 down vote accepted

This also holds in an arbitrary braided monoidal category and is not hard to see. See for example Majid: Foundations of quantum group theory. He gives a graphical calculus proof of this in Figure 9.14. Strictly speaking, you need to turn this proof upside down (i.e. dualize) to get the corresponding identity for the comultiplication. There are two notions of anti-(co)algebra homomorphism in a braided monoidal category. One involving the braiding, one its inverse. For example, $S^{-1}$ is also an anti-(co)algebra morphism but involves the inverse braiding.

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This is what I need, thank you! – Sergei Akbarov Aug 17 '14 at 1:32

That's true in any Hopf algebra. See Sweedler, Hopf algebras, Prop. 4.0.1.

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As far as I understand, Sweedler considers Hopf algebras in the category of vector spaces. Is the same true for all (braded?) monoidal categories? – Sergei Akbarov Aug 12 '14 at 13:57

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