An explicit imbedding of $(R\mathbf{-Mod})^{op}$ into $S\mathbf{-Mod}$

Solution 1:

This is not an answer, but a comment which should not be overlooked.

The Freyd-Mitchell embedding theorem does not apply to arbitrary abelian categories. It only applies to small abelian categories.

It is possible that $\mathsf{Mod}(R)^{op}$ has no fully-faithful exact embedding into some $\mathsf{Mod}(S)$.

The easiest example should be that of a field $R=K$; then $\mathsf{Vect}(K)^{op}$ is equivalent to the category of linear compact topological vector spaces over $K$, with continuous linear maps. This gives a faithful exact embedding $\mathsf{Vect}(K)^{op} \to \mathsf{Vect}(K)$, namely $V \mapsto V^*$. But this embedding is not full. And I really cannot imagine any fully-faithful exact embedding, because this would mean that we can encode continuous linear maps by abstract linear maps between certain modules (of course this is not a proof). Note that, however, $\mathsf{FinVect}(K)^{op}$ is an (essentially) small abelian category, which by the above functor becomes equivalent to $\mathsf{FinVect}(K)$, which is a full exact subcategory of $\mathsf{Vect}(K)$.

Now for general $R$, there is an injective cogenerator in $\mathsf{Mod}(R)$, for example $G:=\hom_{\mathbb{Z}}(R,\mathbb{Q}/\mathbb{Z})$. This means that $\hom_R(-,G) : \mathsf{Mod}(R)^{op} \to \mathsf{Mod}(R^{op})$ is faithful and exact. But again it is not full.

Perhaps one might hope for an embedding of $\mathsf{f.g.Mod}(R)^{op}$.

Solution 2:

I think that if Vopěnka's principle is true, then $\mathsf{Mod}(R)^{op}$ can't fully embed in $\mathsf{Mod}(S)$ for any non-zero $R$ and $S$ (all the references that follow are to "Locally Presentable and Accessible Categories" by Adámek and Rosický): If it did, then $\mathsf{Mod}(R)^{op}$ would be bounded (Theorem 6.6), and since it is also complete it would be locally presentable (Theorem 6.14). But if a category and its opposite are both locally presentable, then the category is equivalent to a complete lattice (Theorem 1.64).