What are abelian categories enriched over themselves?

Question

As far as I understand, an arbitrary abelian category is not enriched over itself, for example, $\mathrm{ChainComplex}(\mathrm{Ab})$ is, right? On the other hand, the categories $\mathrm{Mod}(R)$ (in particular, $\mathrm{Ab}$) are enriched over themselves. What other examples, necessary and sufficient conditions are there for abelian categories enriched over themselves? Maybe they were researched somewhere?

The question arose because the dualization of a chain complex (leading to cohomology with values ​​in the same category) is definable in $\mathrm{Mod}(R)$, but apparently undefinable in an arbitrary abelian category.

Update: (thanks to a discussion in the comments) I realized that I start initially from an abelian category $\mathcal{A}$ with a univalent functor to the category $\mathrm{Ab}$ and look for an enrichment of $\mathcal{A}$ over itself consistent with the standard enrichment of $\mathcal{A}$ over $\mathrm{Ab}$ in the sense of this functor (as far as I understand, this guarantees , that, in accordance with my motivation, the cohomology will not change, but will acquire an additional structure of objects $\mathcal{A}$). For abelian categories of the form "a subcategory of $\mathrm{Func}(I, \mathcal{A})$, where $\mathcal{A}$ is an abelian category with a standard forgetful functor in $\mathrm{Ab}$ and $I$ is small" we define a forgetful functor in $\mathrm{Ab}$ as $F \mapsto \bigoplus\limits_{i \in I} F(i)$ (and, accordingly, on morphisms). Is the category of chain complexes of $R$-modules consistently enriched over itself? And the category of sheafs?

Answer 1

To make sense of enrichment over a category $V$, you want $V$ to have a monoidal structure. Indeed, you want to be able to compose morphisms so you need a way to go from "something in $\hom(a,b)$ and something in $\hom(b,c)$ to something in $\hom(a,c)$", and this is neatly encoded by a monoidal structure where an enrichment is, among other things, a morphism $\hom(a,b)\otimes \hom(b,c)\to \hom(a,c)$ (if you're only monoidal and not symmetric monoidal you have to be careful about what kind of hom's you take, and also in what order you take this tensor, but let me not worry about it here, and pretend I'm in a symmetric monoidal category).

Now, a monoidal category $V$ has a "canonical candidate" for self-enrichment : one would want the internal hom's (which I'll denote $\mathcal{H}om$) to have the following universal property (which reflects the usual cases): $\hom(v, \mathcal Hom(v_0,v_1))\cong \hom(v\otimes v_0, v_1)$. In other words, you want $-\otimes v_0$ to have a right adjoint $\mathcal Hom(v_0,-)$, for all $v_0$. It turns out that this is enough:

Suppose $V$ is a symmetric monoidal category in which $-\otimes v_0$ has a right adjoint for each $v_0\in V$ [NB : $V$ is called 'closed (symmetric) monoidal' in this case]. Then there is a canonical enrichment of $V$ over itself which satisfies the above property.

(again, you can say similar things without "symmetric", but it gets a bit more subtle so I won't state them here)

Note that this is the "usual story" for enrichment and self-enrichment. I don't know of any other way to tell those stories, and they are the ones that tend to come up.

In your examples, $Mod(R)$ has a tensor product $\otimes_R$ when $R$ is commutative, and so it is self-enriched - note that this doesn't work if $R$ is not commutative ! If you want to make it work, you have to move to bimodules or something, and then you have choices to make to decide what the enrichment is.

Similarly, chain complexes over a (cocomplete) symmetric monoidal abelian category have a usual monoidal structure, and if they're sufficiently complete, the right adjoints from above exist. Here is one condition that is often met in practice and guarantees the existence of the right adjoints :

If $V$ is symmetric monoidal, presentable and the tensor product $-\otimes v$ preserves colimits as a functor $V\to V$, then it has a right adjoint.

Here the power is hidden in "presentable", but this is actually often met in practice, despite being a very strong assumption.

(note that this is unrelated to abelian-ness of your category : for a general category $C$ it doesn't make sense to speak about enrichment in $C$, the structure that makes this avaiable is a monoidal structure, and if it is closed, then in fact $C$ can itself be enriched over itself.

In the abelian context, you'll often want the tensor product to be at the very least nicely behaved, e.g. right exact in each variable)