Relation between Ind-completion and "additive"-ind-completion

Question

Suppose that $\mathcal{C}$ is a skeletally small additive category.

To enlarge $\mathcal{C}$ and produce a bigger category whose "small" objects can be identified with those in $\mathcal{C}$, one may consider the ind-completion $\operatorname{Ind}\mathcal{C}$ of $\mathcal{C}$ in the sense of Grothendieck and Verdier. Denote by $\operatorname{Fun}(\mathcal{C}^{op},\operatorname{Set})$ the category of all functors from $\mathcal{C}^{op}$ to $\operatorname{Set}$. Then $\operatorname{Ind}\mathcal{C}$ is defined as the full subcategory of $\operatorname{Fun}(\mathcal{C}^{op},\operatorname{Set})$ whose objects are the filtered colimits (in $\operatorname{Fun}(\mathcal{C}^{op},\operatorname{Set})$) of representable functors.

We may also consider the category $\operatorname{AddFun}(\mathcal{C}^{op},\operatorname{Ab})$ of additive functors from $\mathcal{C}^{op}$ to the category of abelian groups and define $\operatorname{AddInd}\mathcal{C}$ as the full subcategory of $\operatorname{AddFun}(\mathcal{C}^{op},\operatorname{Ab})$ whose objects are the filtered colimits (in $\operatorname{AddFun}(\mathcal{C}^{op},\operatorname{Ab})$) of representable functors. This construction is thoroughly studied, for instance, in the paper "Locally finitely presented additive categories", by W. Crawley-Boevey.

I am trying to understand the relation between $\operatorname{Ind}\mathcal{C}$ and $\operatorname{AddInd}\mathcal{C}$. I have seen in two papers statements claiming that $\operatorname{Ind}\mathcal{C}$ and $\operatorname{AddInd}\mathcal{C}$ are equivalent, but do not yet understand why this should be true. Any references or hints would be very welcome.

Answer 1

The point is what Ivan hints at in his last paragraph, that additivity is a property rather than an extra structure.

In fact, we have:

Suppose $C$ is an additive category. Then the forgetful functor $Fun^\times(C,\mathbf{Ab})\to Fun^\times(C,\mathbf{Set})$ is an equivalence of categories.

(if you only assume $C$ to be preadditive, then the same holds with "commutative monoids" in place of $\mathbf{Ab}$). Here $Fun^\times$ denotes the category of finite-product preserving functors.

Let me skip the proof of this fact for now.

The second point is that of course, a filtered colimit of representables preserves finite products products. Furthermore, $\mathbf{Ab}\to \mathbf{Set}$ preserves filtered colimits.

So the forgetful functor $\mathrm{AddInd}(C)\to \mathrm{Ind}(C)$ is fully faithful (there is a map from this arrow to the arrow described above, with $C^{op}$ in place of $C$, where both vertical arrows are fully faithful, and the lower horizontal arrow is an equivalence), preserves filtered colimits, and hits the representables : it is therefore an equivalence.

Let me spend a second on the claim I mentioned above. It is relatively easy to prove using Lawvere theories, namely : let $L$ denote the Lawvere theory of abelian groups. Then

i) If $D$ is any additive category, the forgetful functor $Fun^\times(L,D) \to D$ is an equivalence

and

ii) $Fun^\times(L,\mathbf{Set} ) = \mathbf{Ab}$ (the image of the Yoneda embedding of $L^{op}$ under this equivalence is exactly the full subcategory of finitely generated free abelian groups).

Both points are special cases of the fact that $Fun^\times(L,D) =$ abelian groups in $D$, whenever $D$ is a category with finite products.

With these two points in mind, we can write the forgetful functor as the following composite : $$Fun^\times(C,\mathbf{Ab}) = Fun^\times(C,Fun^\times(L,\mathbf{Set})) = Fun^\times(L,Fun^\times(C,\mathbf{Set}))\to Fun^\times(C,\mathbf{Set})$$

where the last functor is an equivalence because $Fun^\times(C,\mathbf{Set})$ is additive (which I will leave as an exercise)

Answer 2

The equivalence you mention holds more generally whenever your base of enrichment has a finitely presentable unit. This certainly includes $Ab$ but also many other examples: $Cat$, $sSet$, $GAb$, $DGAb$, etc.

Assume that $\mathcal V=(\mathcal V_0,\otimes,I)$ is a symmetric monoidal closed complete and cocomplete category. Then you can define the Ind-completion of a $\mathcal V$-category $\mathcal C$ as the closure of the representables in $[\mathcal C^{op},\mathcal V]$ under (conical) filtered colimits; let's call it simply $Ind(\mathcal C)$, and denote by $J:\mathcal C\to Ind(\mathcal C)$ the inclusion.

This free cocompletion can be recognised as follows (see e.g. 4.2 from "Notes on enriched categories with colimits of some class" by Kelly and Schmitt): A $\mathcal V$-functor $F:\mathcal C\to\mathcal K$ exhibits $\mathcal K$ as the Ind-completion of $\mathcal C$ if and only if:

1. $F$ is fully faithful;
2. $\mathcal K$ has filtered colimits;
3. $Fc$ is finitely presentable in $\mathcal K$ for any $c\in\mathcal C$ (i.e. $\mathcal K(Fc,-):\mathcal K\to \mathcal V$ preserves filtered colimits);
4. The closure of $\mathcal C$ in $\mathcal K$ under filtered colimits is $\mathcal K$ itself.

Now assume that the unit $I$ of $\mathcal V$ is finitely presentable, that is $\mathcal V_0(I,-):\mathcal V_0\to Set$ preserves filtered colimits. Let $\mathcal C$ be a $\mathcal V$-category and $J:\mathcal C\to Ind(\mathcal C)$ be the inclusion in its enriched Ind-completion (then $J$ satisfies (1)-(4) above). We can then consider the underlying ordinary functor $J_0:\mathcal C_0\to Ind(\mathcal C)_0$ of $J$; then the properties (1), (2), and (4) still hold for $J_0$ (independently of the unit) and (3) holds since $$ Ind(\mathcal C)_0(J_0c,-)\cong \mathcal V_0(I,Ind(\mathcal C)(Jc,-)_0) $$ is a composite of filtered-colimit-preserving functors.

It follows by the characterization above of Ind-completions (for $\mathcal V=Set$) that $Ind(\mathcal C)_0$ is the Ind-completion of $\mathcal C_0$; in other words $$ Ind(\mathcal C)_0\simeq Ind(\mathcal C_0).$$

Answer 3

Let $\mathcal{V}$ be a cocomplete monoidal category which can be presented by a limit theory, so that $\mathcal{V} = \mathsf{Lex}(\mathbb{T},\text{Set})$, of course this is the case of your question.

Now, let me give the answer under the assumption that $C$ has finite $\mathcal{V}$-enriched colimits. In this case the Ind-completion is described by the formula $$\mathsf{Ind}(C) = \mathcal{V}\mathsf{Lex}(C^\circ, \mathcal{V}).$$

Now, there is a forgetful functor $$\mathcal{V}\mathsf{Lex}(C^\circ, \mathcal{V}) \to \mathsf{Lex}(C^\circ_0, \mathcal{V}), $$ where by $C^\circ_0$ we intend the underlying category of $C^\circ$. This forgetful functor is clearly faithful and conservative. I always thought it should also be monadic and recently in a private conversation Adrian Miranda sketched me a convincing argument hinting that it is both monadic and comonadic, which I will skip.

Indeed it is evident though, that having a good representation of $\mathsf{Lex}(C^\circ_0, \mathcal{V})$ could help us in understanding the situation. Now, follow the isos.

$$\mathsf{Lex}(C^\circ_0, \mathcal{V}) \cong \mathsf{Lex}(C^\circ_0, \mathsf{Lex}(\mathbb{T}, \text{Set})) \cong \mathsf{Lex}(\mathbb{T}, \mathsf{Lex}(C^\circ_0, \text{Set})) \cong \mathsf{Lex}(\mathbb{T}, \mathsf{Ind}(C_0)).$$

This means that, in the special case of additive categories, we get a representation of the additive Ind-completion of $C$ in the category of internal abelian group objects in the (Set) Ind-completion of $C_0$.

$$\mathsf{AddInd}(C) \to \mathsf{Ab}(\mathsf{Ind}(C_0)). $$

In the general case, we get $\mathbb{T}$-models in the $\mathsf{Ind}$-completion of $C_0$, of course.

In the special case of additivity, maybe (?) the reason for which the connection is tighter is that being additive is very close to be a property, more than a structure of the category $C_0$, so that already taking the Ind-completion does most of the job.