I know that there is a product in the category of small categories. I think this product is also the product in the category of pre-additive, or triangulated categories. There is also a coproduct of small categories by taking the disjoint union, so it doesn't agree with the product of small categories. I am trying to figure out what the coproduct is the category of pre-additive/abelian categories. I do not think it agrees with the product. For example, the category of rings has a full and faithful embedding into the category of pre-additive categories (take any ring $R$ to the one object category $C_R$ whose hom space is that ring). I think the coproduct $C_R \coprod C_S$ of $C_R$ and $C_S$ is $C_{R \otimes_\mathbb{Z} S}$. This does not agree with the product $C_R \prod C_S = C_{R\prod S}$. Is there a general coproduct/pushout in the category of pre-additive/additive/abelian categories? I believe that this question Direct sum of categories addresses the product, not the coproduct.

Edit: as Eric pointed out, the coproduct of $C_R, C_S$ in the category of pre-additive categories is just the disjoint union of $C_R$ and $C_S$ (not $C_{R\otimes S}$ as I wrote).

Concretely, if I have two rings $S,T$, is the coproduct of $Mod(S), Mod(T)$ in the category of abelian categories equal to $Mod(S \otimes_{\mathbb{Z}} T)$? If so, I think this has a description as the Deligne tensor product $Mod(S) \otimes_{Mod(\mathbb{Z}) } Mod(T)$. Is this Deligne tensor product the coproduct in the category of abelian categories?


Coproducts in the category of preadditive categories are just disjoint unions with a zero morphism added between each pair of objects from the different original categories. Indeed, if $C$ and $D$ are preadditive categories, then the category $E$ obtained in this way is preadditive, and a pair of additive functors out of $C$ and $D$ extends uniquely to an additive functor out of $E$ by just sending each of the new zero morphisms to zero. It follows that $E$ satisfies the universal property of the coproduct in the category of preadditive categories.

Coproducts do not exist in the category of additive categories. For instance, there is not even an initial object. The issue is that if $C$ is a category in which every object is a zero object but $C$ has many different objects, then every additive category $D$ has many different functors to $C$ (one for every function $\operatorname{Ob}(D)\to \operatorname{Ob}(C)$).

However, coproducts do exist in the 2-category of additive categories, and are the same as products. Coproducts in a 2-category are defined to only have their universal property "up to isomorphism". To be more precise, a coproduct in a 2-category of objects $C$ and $D$ is an object $E$ with morphisms $i:C\to E$ and $j:D\to E$ such that composition with $i$ and $j$ gives an equivalence of categories $\operatorname{Hom}(E,F)\simeq \operatorname{Hom}(C,F)\times \operatorname{Hom}(D,F)$ for any object $F$.

The idea behind coproducts in the 2-category of additive categories being the same as products is that an object $(c,d)$ of the product represents the formal direct sum $c\oplus d$. Explicitly, let $C$ and $D$ be additive categories. Then there is an additive inclusion functor $i:C\to C\times D$ taking $c\in C$ to $(c,0)$ for some chosen zero object $0\in D$, and similarly there is an additive inclusion functor $j:D\to C\times D$. Now given any other additive (or abelian) category $E$ and additive functors $f:C\to E$ and $g:D\to E$, we can define an additive functor $h:C\times D\to E$ by $h(c,d)=f(c)\oplus g(d)$. This satisfies $hi=f$ and $hj=g$, and it is easy to see that it is the unique such $h$ up to natural isomorphism (essentially because $(c,d)=(c,0)\oplus(0,d)$ in $C\times D$). With a bit of work we can show that this in fact gives an equivalence between the category of additive functors $C\times D\to E$ and the category of pairs of additive functors $C\to D,C\to E$ and thus $C\times D$ is a coproduct of $C$ and $D$ in the 2-category of additive categories.

The story for abelian categories is exactly the same as for additive categories.

As a final aside, you will notice that these coproducts have nothing to do with tensor products. That's because tensor products are about multiplication, but when you combine two (pre-)additive categories, there's no way in which you need to be able to "multiply" morphisms from them. Instead, you just need to be able to take direct sums of objects (to get an additive category). This is similar to how coproducts of abelian groups are direct sums, not tensor products, since there is no multiplication operation for abelian groups. (On the other hand, coproducts of commutative rings are tensor products, since they require a multiplication operation).

  • $\begingroup$ Thank you for your helpful answer! Can you clarify why there is no initial object? You seem to be saying that C (the category with many zero objects) is not a final object. Also, in what context do tensor products appear? Is it additive monoidal categoriess? $\endgroup$ – user39598 Jan 26 '19 at 18:35
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    $\begingroup$ I'm saying that because of the existence of $C$, no $D$ can be an initial object (since any $D$ has multiple maps to $C$). Taking coproducts of additive symmetric monoidal categories would seem to be closely related to tensor products, but I don't know the details off the top of my head. $\endgroup$ – Eric Wofsey Jan 26 '19 at 19:49
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    $\begingroup$ For coproducts of preadditive categories you need to adjoin zero morphisms to the disjoint union. $\endgroup$ – Jeremy Rickard Jan 27 '19 at 8:20
  • $\begingroup$ @JeremyRickard: Of course, thanks! I've corrected that. $\endgroup$ – Eric Wofsey Jan 27 '19 at 8:23

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