Tensor variant of Mitchell's embedding theorem Assume $\mathcal{A}$ is a small cocomplete abelian $\otimes$-category (see here for a definition). Is there a cocontinuous, full, faithful, exact $\otimes$-functor $\mathcal{A} \to \text{Mod}(R)$ for some ring $R$?
See here for a related question. The reason why I'm asking is that I want to prove that a certain (perfectly natural!) construction works in such a category $\mathcal{A}$ and it seems to me that the axioms of $\mathcal{A}$ are not enough to check that a certain sequence is exact. In module categories there are no problems.
 A: With one modification, the answer to your question is yes.  Namely, don't try to land in $\operatorname{Mod}_R$ for $R$ a commutative ring, but in $_R\operatorname{Mod}_R$, the categoy of $R$,$R$ bimodules, for $R$ an arbitrary ring.  This you can do:

Phùng Hô Hài, An Embedding theorem for abelian monoidal categories, http://arxiv.org/abs/math/0004160

Well, maybe with another modification.  A small cocomplete category is necessarily a poset, and so necessarily not abelian.  I assume you want your category to be small, abelian, and monoidal?
A: Let ${\bf 1}$ denote the unit object of ${\mathcal A}$. If $F: {\mathcal A} \rightarrow Mod(R)$ is a tensor functor, then $F( {\bf 1} ) \simeq R$. If $F$ is fully faithful, then you can recover $R$ as $End({\bf 1})$, and the functor $F$ is given by $A \mapsto Hom( {\bf 1},A)$.
This is rarely a fully faithful embedding. For example, if ${\mathcal A}$ is the category
of complex representations of a finite group, then $F$ annihilates every nontrivial irreducible representation.
