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By a tensor category, I mean here a cocomplete $k$-linear symmetric tensor category, where $k$ is a fixed ground ring. Tensor functors are assumed to be $k$-linear and cocontinuous. I will denote the unit object as $\mathcal{O}$.

Terminology: Let $C$ be a tensor category, $\mathcal{A}$ an algebra in $C$ (i.e. an object equipped with a commutative associative multiplication $\mathcal{A} \otimes \mathcal{A} \to \mathcal{A}$ and a unit $\mathcal{O}_C \to \mathcal{A}$). Then the $\mathcal{A}$-modules (i.e. objects $M$ together with a morphism $\mathcal{A} \otimes M \to M$ which is compatible with the unit and the multiplication in $\mathcal{A}$) consitute a tensor category $\text{Mod}(\mathcal{A})$.

The forgetful functor $p_* : \text{Mod}(\mathcal{A}) \to C$ is right adjoint to the tensor functor $p^* : C \to \text{Mod}(\mathcal{A}), X \mapsto \mathcal{A} \otimes X$, where here $\mathcal{A} \otimes X$ is given the obvious action. This adjunction is easily seen to be monadic. If $\mathcal{O}'$ denotes the unit object of $\text{Mod}(\mathcal{A})$, then $p_* \mathcal{O}' = \mathcal{A}$, so we have a canonical morphism $p^* \mathcal{A} \to \mathcal{O}'$.

Now let $D$ be another tensor category and $H : \text{Mod}(\mathcal{A}) \to D$ be a tensor functor. Then we get a tensor functor $F = H p^* : C \to D$ together with a morphism

$F(\mathcal{A}) = H(p^* \mathcal{A}) \to H(\mathcal{O}') = \mathcal{O}_D$

of algebras in $D$. In fact, we get a functor

$Hom(\text{Mod}(\mathcal{A}),D) \to \{(F,\sigma) : F \in Hom(C,D), \sigma : F(\mathcal{A}) \to \mathcal{O}_D \text{ alg-hom}\}$

Question Does this functor reflect isomorphisms? Or is it even an equivalence?

Remark: If $C = \text{Mod}(R)$ for some ring $R$, then the functor is an equivalence.

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up vote 6 down vote accepted

Your functor is an equivalence. You can explicitly describe an inverse as follows: given a pair $(F, \sigma)$ as you describe, define $H: Mod_{ \mathcal{A} } \rightarrow D$ by the formula $H(M) = F(M) \otimes_{ F( \mathcal{A} ) } \mathcal{O}_{D}$.

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Yes! Thanks :) – Martin Brandenburg Feb 3 '11 at 14:00

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