I can't understand this sentence i the article of Majid "Tannaka-Krein theorem for quasi-Hopf algebras and other results" about the reconstruction of a quasi-algebra (in fact its dual) from a given braided category (C,c) with a forgetful functor F ; at the end it considers the action of Drinfeld-twisting upon quasi-algebra and says:
if A (quasi-algebra) corresponds to (C,F,c) then A\_twisted corresponds to (C,F,c') for a new choice of braiding c'.
i am very disturbed: the associativity constraint is modified too obviously; sorry for my very naive question.



Why does twisting quasi-Hopf algebras work (as in majid's article)
You might find these notes helpful as a second source to read fro:


<http://www-math.mit.edu/~etingof/tenscat1.pdf>


There is a careful elaboration of the various variants of Tannaka-Krein construction, and the case of quasi-bialgebras and quasi-Hopf algebras is discussed. So far as I recall, Drinfeld twisting of a quasi bi-algebra just means acting on the original category by a tensor functor which is the identity on objects (and so is only interesting as a tensor functor).


As such, it does have a fairly straightforward effect on the associator $\phi$. If you twist by an invertible element $F\in H\otimes H$, then the associator $\phi^F$ is something like $\phi^F:=B\phi B^{-1}$, where $B:=F(\Delta\otimes\operatorname{id})(F)$, or maybe $F(\operatorname{id}\otimes\Delta)(F)$, I can't remember.


It might help you to note that nothing in your question relies on the braiding; there is a reconstruction theorem for any (quasi)bi-algebra with a fiber functor, and since you're asking about the associator, it isn't too important that there's a braiding lying around.


What leads you to expect some more complicated effect on the associator?


