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If $B$ is a bialgebra in the category $\tt{Vect}$ of vector spaces (over $\mathbb C$, for example) then $B$ can't have two different antipodes.

Is this true for bialgebras in an arbitrary symmetric monoidal category?

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This is really obvious if you look at the standard proof of the uniqueness of the antipode. The antipode is defined as the $*$-inverse of the identity in the convolution algebra. The convolution algebra is a honest algebra, not an algebra-over-a-category; so inverses are unique if they exist. – darij grinberg Aug 4 '14 at 18:05
@darij grinberg: This becomes obvious only if the assocoativity of convolution in $Hom(B^c,B^a)$ is already proved. This fact is not obvious -- in the book by Dascalescu, Nastasescu and Raianu, for example, they use the Sweedler notations for proving it. But the Sweedler notations, I would say, is a risky way in arbitrary monoidal categories. Chari and Pressley mention this fact, but without references. These were two sources for me. I saw before the pictures like those that Evan Jenkins gave in his answer, but I did not understand them and I did not know that they can be used here. – Sergei Akbarov Aug 5 '14 at 1:15
Oh -- but it is very easy to prove the associativity of convolution without Sweedler's notation, just using the axioms of a coalgebra and of an algebra. – darij grinberg Aug 5 '14 at 1:18
I did not see such a proof. Do you mean this trick with string diagrams? – Sergei Akbarov Aug 5 '14 at 1:20
Denoting the multiplication of the algebra by $m$ and the comultiplication of the coalgebra by $\Delta$, we have ... – darij grinberg Aug 5 '14 at 1:29
up vote 8 down vote accepted

Yes, as the following string diagram proof shows. String diagram proof of uniqueness of antipodes

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Evan, I don't understand these pictures. Do you know a text where they are explained? – Sergei Akbarov Aug 4 '14 at 7:59
Yes, I guessed. However, a text with explanations would be highly appreciated. – Sergei Akbarov Aug 4 '14 at 10:01
@UwF: I guessed also. :) But this style of explanation - "I drop a hint, you guess!" - is a bit annoying when you need a reference for a paper. Are there texts where this technique is described more or less intelligible? :) – Sergei Akbarov Aug 4 '14 at 14:53
@SergeiAkbarov A good start would be the seminal paper by Joyal and Street, The Geometry of Tensor Calculus (in Adv. Math.). The technique is scattered all across the literature, so much so that there is an embarrassment of riches, but the nLab page referenced in Dmitri's answer might also be a good place to look. – Todd Trimble Aug 4 '14 at 19:30
@SergeiAkbarov I learned this diagrammatic technique from papers by Shahn Majid, I think he described it quite well in his book "Foundations of quantum group theory". But I understand what you mean, and I agree, especially since there are so many variations. One should take the time to explain the symbols if one uses diagrammatic notation. – UwF Aug 4 '14 at 19:40

Nice answer by Evan Jenkins. (Here are references for string diagrams:

The bottom line is that being a Hopf algebra is a property of a bialgebra rather than an extra structure. A Hopf algebra is usaully defined as a bialgebra with an antipode. Then the "propertiness" means that the antipode is unique. This indeed holds in any braided monoidal category.

However, there is another definition of a Hopf algebra which emphasizes the "propertiness". It is based on the generalization of the following simple group-theoretic fact: A monoid is a group if and only if the maps $$(a, b) \mapsto (ab, b), \qquad (a, b) \mapsto (a, ab)$$ are invertbile. Indeed, the inverses are given by $$(a, b) \mapsto (ab^{-1}, b), \qquad (a, b) \mapsto (a, a^{-1}b).$$

Generalizing this to bialgebras (within any braided monoidal category), a Hopf algebra is a bialgebra $B$ for which the maps $$B\otimes B \xrightarrow{1\otimes d} B\otimes B\otimes B \xrightarrow{m \otimes 1} B\otimes B$$ $$B\otimes B \xrightarrow{d\otimes 1} B\otimes B\otimes B \xrightarrow{1\otimes m} B\otimes B$$ are invertible. These maps are called left and right fusion maps.

This definition is equivalent to the definition via an antipode. The anipode can be recovered as $$B \xrightarrow{1\otimes \eta} B\otimes B \xrightarrow{l^{-1}} B\otimes B \xrightarrow{\epsilon\otimes 1} B$$

where $l$ is the left fusion map. Or it can be recovered by a similar composite involving the inverse of the right fusion map.

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Dima, thank you! – Sergei Akbarov Aug 5 '14 at 0:57

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