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I think this must be well-known, but I can't find references, so my apologies from the very beginning.

Consider first the notion of bialgebra. It is usually defined as an object $B$ in the category ${\tt Vect}_{\mathbb K}$ of vector spaces over a given field $\mathbb K$ with a system of morphisms in this category (multiplication, unit, comultiplication and counit) such that the multiplication and the unit turn $B$ into an associative algebra, the comultiplication and the counit turn $B$ into a coassociative coalgebra, and a system of commutative diagrams is provided for the compatibility. These diagrams can be understood as the requirement that the comultiplication and the couint are morphisms in the category ${\tt Alg}_{\mathbb K}$ of algebras (or, equivalently the multiplication and the unit are morphisms in the category ${\tt Coalg}_{\mathbb K}$ of coalgebras). So the whole definition can be understood as follows:

a bialgebra is exactly a comonoid (=a coassociative coalgebra) in the category ${\tt Alg}_{\mathbb K}$ of associative algebras over $\mathbb K$ (or, equivalently, a monoid (=an associative algebra) in the category ${\tt Coalg}_{\mathbb K}$ of coassociative coalgebras over $\mathbb K$).

If we try to define Hopf algebras in the same way, then we face an obvious difficulty: the diagram for antipode can't be immediately interpreted as a diagram in the category ${\tt Alg}_{\mathbb K}$ of associative algebras.

I wonder if there exists a trick for overcoming this? Is it possible that the diagram for antipode can be changed into equivalent diagrams in ${\tt Alg}_{\mathbb K}$ (so that we obtain a correct definition of Hopf algebras as objects in ${\tt Alg}_{\mathbb K}$)?

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The antipode is an anti-algebra homomorphism. Is the opposite of an algebra well-defined in the category of algebras? – darij grinberg Aug 25 '13 at 12:44
@darij grinberg: I thought, there is no problem with definition of opposite algebra. But the problem is with changing the digram for antipode so that it will be a diagram in ${\tt Alg}_{\mathbb K}$. – Sergei Akbarov Aug 25 '13 at 12:48
Ah, I see what you mean. – darij grinberg Aug 25 '13 at 12:50
The discussion in the comments at does not answer your question but it may still be useful (apologies if it just repeats things you had already considered) – Yemon Choi Aug 26 '13 at 2:34

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