Let $A$ be an additive category. Denote $(A, Ab)$ the category of all additive functors from A to the category of abelian groups.
There is well known result that any functor $F \colon C \to Sets$ on a locally small category $C$ is a colimit of representable functors $F\cong \operatorname{colim}_I Hom(X_i, -)$, where indexing category $I$ can be taken to be opposite to the category of elements of $F$.
I'm interested in the similar statement for additive functors: is it true that any additive functor is a colimit of representable functors?
In the book "Homological Theory of Representations" by H. Krause I found the following
I have two questions about this proof.
- Why the map in the proof is a map of abelian groups, not just sets.
- Doesn't this lemma contradicts this discussion https://mathoverflow.net/questions/204792/is-every-abelian-group-a-colimit-of-copies-of-z for $A=\mathbb{Z}$?
There is a gap in the proof to be filled: namely, that the colimit in the category of set-valued presheaves is also the colimit in the category of additive presheaves. For this we need to use the hypothesis that $\mathcal{C}$ itself is additive.
Suppose we have a cocone in $\textbf{Add} (\mathcal{C}^\textrm{op}, \textbf{Ab})$ from $\Phi_\mathcal{C}$ to $G$. We know from the non-additive version of the lemma that there is an induced morphism $\theta : F \to G$, thinking of $F$ and $G$ as ordinary presheaves. It suffices to show that the components $\theta_C : F (C) \to G (C)$ are group homomorphisms.
Let $x$ and $y$ be elements of $F (C)$. Under the canonical identification $F (C \oplus C) \cong F (C) \times F (C)$, we have an element $\langle x, y \rangle \in F (C \oplus C)$. Explicitly, this is the unique element such that the action of the two insertions $\iota_1, \iota_2 : C \to C \oplus C$ yield $x$ and $y$ respectively. Thus, $\theta_{C \oplus C} (\langle x, y \rangle) = \langle \theta_C (x), \theta_C (y) \rangle$, by considering the following commutative diagram: $$\require{AMScd} \begin{CD} F (C) @>{\theta_C}>> G (C) \\ @A{F (\iota_1)}AA @AA{G (\iota_1)}A \\ F (C \oplus C) @>{\theta_{C \oplus C}}>> G (C \oplus C) \\ @V{F (\iota_2)}VV @VV{G (\iota_2)}V \\ F (C) @>>{\theta_C}> G (C) \end{CD}$$ But we also have the diagonal $\delta : C \to C \oplus C$, whose action on $F (C \oplus C)$ sends $\langle x, y \rangle$ to $x + y$. Hence, $\theta_C (x + y) = \theta_C (x) + \theta_C (y)$, by considering this commutative diagram: $$\begin{CD} F (C \oplus C) @>{\theta_{C \oplus C}}>> G (C \oplus C) \\ @V{F (\delta)}VV @VV{G (\delta)}V \\ F (C) @>>{\theta_C}> G (C) \end{CD}$$
A similar argument shows that $\theta_C (0) = 0$. Thus $\theta_C : F (C) \to G (C)$ is a group homomorphism, so $F$ indeed has the universal property of the colimit of $\Phi_\mathcal{C}$, in $\textbf{Add} (\mathcal{C}^\textrm{op}, \textbf{Ab})$ just as much as in $[\mathcal{C}^\textrm{op}, \textbf{Set}]$.
Note that the argument above is completely inapplicable when $\mathcal{C}$ is only $\textbf{Ab}$-enriched. This is the reason why it does not contradict the fact that not every abelian group is a colimit of a diagram of copies of $\mathbb{Z}$. There is an analogous result for the general situation where $\mathcal{C}$ is just an $\textbf{Ab}$-enriched category, but it is necessary to use weighted colimits instead of conical colimits, and ultimately it turns out to be something of a tautology.