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.



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


