Can we turn the functor "category ring" into a 2-functor in a natural way?

Question

Let $C$ be a small pre-additive category. Let $R(C)$ denote its category ring, that is, $$ R(C)=\bigoplus_{a,b\in \mathrm{Ob}(C)} C(a,b) $$ as Abelian group, where the direct sum runs over all object $a$, $b$ of $C$. The multiplication in $R(C)$ is given by composition of composable morphisms and 0 for uncomposable morphisms (extended by bilinearity).

This constuction is functorial: An additive functor $C\to D$ between small pre-additive categories induces a ring homomorphism $R(C)\to R(D)$ between the corresponding category rings in a canocical way.

Hence we have a functor $R$ from the category of small pre-additive categories (with additive functors as morhisms) to the category of rings (with ring homomorphisms as morphisms).

But the category of small pre-additive categories has a 2-categorical structure given by natural transformations.

Hence my question: Does this 2-categorical structure have a counterpart in the category of rings? More precisely, is there a natural notion of 2-morphisms between ring homomorphisms turning $R$ into a 2-functor?

Answer 1

This construction is not even a functor!

The problem is that two morphisms which are not composable in $C$ may become composable in $D$. The simplest example is to take $C$ to be the discrete category on two objects $a, b$, $D$ to be the discrete category on an object $d$, and $F : C \to D$ the unique functor between them (or rather the free preadditive category on these). Then $R(C)$ has two generators $\text{id}_a$ and $\text{id}_b$ which are not composable and hence whose product is $0$. But their image in $D$ is $\text{id}_d$ which squares to itself. So

$$R(F)(\text{id}_a \times \text{id}_b) = R(F)(0) = 0 \neq R(F)(\text{id}_a) \times R(F)(\text{id}_b) = \text{id}_d.$$

So $R(F)$ is not a ring homomorphism.

This construction is not a functor, but it can be understood in terms of a functor as follows. I'll restrict my attention to the case that $C$ has finitely many objects. Given such a preadditive category you can talk about the category obtained by formally adjoining finite direct sums (this is a functor, even a 2-functor), and in this new category I claim that the category ring is just $\text{End}(\oplus_{c \in C} c)$. The significance of this ring is that it is Morita equivalent to $C$, meaning that right modules over it are naturally equivalent to presheaves on $C$ valued in abelian groups.

Answer 2

Rings are precisely those pre-additive categories with exactly one object. Thus the category of rings (noncommutative here) is a $2$-category. If $f,g : R \to S$ are $1$-morphisms, then a $2$-morphism $f \to g$ is an element $s \in S$ such that for all $r \in R$, we have $s f(r) = g(r) s$.

So take a natural transformation $\eta : F \to G$ between additive functors $F,G : C \to D$. Then we may take $s = \sum_{x \in C} \eta(x)$ as an element of $R(D)$. But wait, this sum does not have to be finite. Then let's define $R(-)$ simply as the product and not the direct sum. This causes other problems, see the answer of Agusti Roig.

Now the desired equation $s F(r) = G(r) s$ for some morphism $r : x \to y$ in $C$ is equivalent to

$\sum_{u : F(u)=F(x)} \eta(u) F(r) = \sum_{v : G(v)=G(x)} G(r) \eta(v)$.

This seems to be true only if $F$ and $G$ are injective on objects.

So my answer would be: No, unfortunately $R$ cannot be made into a $2$-functor.

But somehow, $R$ should be a $2$-functor, and perhaps we can modify the whole setting a but, so that it works.

Answer 3

This is not an answer, but an elementay question about this category ring $R({\cal C})$. Because of Martin's remark, I'll use the product notation: elements of $R({\cal C})$ are tuples

$$ f= (f_{ab}) \in \prod_{a, b \in {\mathrm ob} {\cal C}} {\cal C}(a,b) \ , $$

in which every $f_{ab} : a \longrightarrow b$ is a morphism of ${\cal C}$. Right?

(By the way: are we still in our universe when we make such a product? I mean: this won't be a "ring" at all, that is: a small ring.)

So, the problem that Martin has pointed out comes from the fact that, when you multiply two of these tuples $f=(f_{ab})$ and $g= (g_{ab})$ a sum appears in the $ab$-component of $g\cdot f$:

$$ \sum_x g_{xb}\circ f_{ax} \ . $$

Is it so?