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Let $A$ and $B$ be bialgebras, and A-Mod and B-Mod equivalent as tensor categories. Then there exists a twist $J$ for $B$ and an isomorphism of bialgebras $f\colon A\to B^J$, where $B^J$ is the twisted bialgebra. I can't find the isomorphism $f$

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Is this an exercise from some book? (Also: a little proofreading of your question and its title always helps! :) ) – Mariano Suárez-Alvarez♦ Jun 30 '11 at 17:41
No, this is not an exercise from some book. I think the proof could go as follows. For every twist $J$ for $B$ there exists a tensor equivalence $\left(id,id,\Phi\right)$ from $B$-Mod to $B^J$-Mod. Showing this is not realy difficult. If we can proof that if $A$-Mod and $B^J$-Mod are tensor aquivalnt then there exists an isomorphism $f:A\rightarrow B^J$ of bialgebra, the above hypothesis follows automatically. Because the composition of tensor functors is again a tensor functor. – anna beisi Jul 1 '11 at 7:37
Sometimes a twist is colled $2$-cocicle, $2$-cocicle twist or gauge transformation.Definition. Let $A$ be a Bialgebra. A twist for $A$ is an ivertible element $J\in A\otimes A$ which satisfies $\left(\Delta\otimes id\right)\left(J\right)\left(J\otimes 1\right)=\left(id\otimes \Delta\right)\left(J\right)\left(1\otimes J\right)$ and $\left(\epsilon\otimes id\right)\left(J\right)=\left(id\otimes\epsilon\right)\left(J\right)=1$, where $Δ$ is the comultiplication of the Bialgebra $A$. – anna beisi Jul 1 '11 at 7:44

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