$\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\Rec{Rec}\DeclareMathOperator\End{End}\DeclareMathOperator\Mod{Mod}\newcommand\fdMod[1]{#1\text-{\Mod}^\text{fd}}$In the celebrated [EGNO, Thm 5.2.3], it is shown that there is a bijection between isomorphism classes of finite-dimensional bialgebras (over a field $k$) and monoidal equivalence classes of finite ring categories over $k$ together with a fibre functor, the assignment being taking the category of finite-dimensional modules over the bialgebra.

In [EGNO, Remark 5.2.4] it is also stated that the bijection can be promoted to a equivalence of categories $$\Rep: \mathsf{FBialg} \to \mathsf{FRCF} $$ between the category of finite-dimensional bialgebras with arrows isomorphisms, and the the category of finite ring categories with a fiber functor, with morphisms being isomorphism classes of equivalences compatible with fiber functors. The quasi-inverse of this functor is the reconstruction functor $\Rec(\mathcal{C}, F)= \End(F)$.

Now, I came across these very well written lecture notes on Tensor Categories by C. Meusburger that, in Corollary 8.2.7, upgrades the previous equivalence to an equivalence

$$\Rec: \mathsf{Fib} \to \mathsf{Bial}$$

where $\mathsf{Bial}$ is the category of finite-dimensional bialgebras and bialgebra homomorphisms, and $\mathsf{Fib}$ is the category of pairs (finite ring cat, fibre functor) with arrows $k$-linear functors compatible with the fibre functors.

The functor $\Rep=\Mod: \mathsf{Bial} \to \mathsf{Fib}$ that associates the category of finite dimensional modules to any finite dimensional bialgebra $B$ is used to exhibit $\Rec$ as a equivalence of categories, in fact $\Rec \circ \Mod$ is naturally isomorphic to the identity. But I don't think that $\Mod$ is the quasi-inverse of $\Rec$, is this correct? (In fact this is never mentioned in the notes.) That is, $\Mod \circ \Rec$ is not naturally isomorphic to the identity, right? This would imply that every finite ring category is isomorphic to $\fdMod B$ for $B=\End(F)$, but I believe that it should only be (monoidally) equivalent.

But then, what is the quasi-inverse of $\Rec$?

Even if what I said above is correct, I don't see how it follows from the Corollary in the notes that any finite ring category is monoidally equivalent to $\fdMod B$ for $B= \End(F)$, as in Theorem 5.2.3. from [EGNO].

Could you please shed some light about this?

(Of course, [EGNO] = Tensor Categories by Etingof, Gelaki, Nikshych and Ostrik)

  • $\begingroup$ $\mathrm{Fib}$ seems to most naturally be a 2-category and so you should be trying to prove an equivalence of 2-categories rather than 1-categories. $\endgroup$ Commented Jul 11 at 7:19


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