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?
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? 


Yes, as the following string diagram proof shows. 


Nice answer by Evan Jenkins. (Here are references for string diagrams: http://ncatlab.org/nlab/show/string+diagram#details) 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 grouptheoretic 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. 

