Let $\mathcal C$ and $\mathcal D$ be pre-additive (enriched over the abelian groups) categories and $F : \mathcal C \Rightarrow \mathcal D$ a functor which is left-exact in the sense that it preserves kernels (or, equivalently, equalizers). Can we prove that $F$ is semi-additive in the sense that $F(f + g) = F(f) + F(g)$ without assuming that the categories have finite products or biproducts?



Does left-exactness imply semi-additivity?
So I don't know if assuming biproducts exists is enough or not but preserving kernel alone is not enough. Here is a counter-exemple. Let $C$ be the pre-additive category with only one object $\*$ and such that $Hom(\*,\*) = \mathbb{Z}$.


If we want, we can add a zero object to $C$ if we want $C$ to have all kernel.


I fix $\sigma$ a non-trivial bijection of the set of prime number. consider the following functor $F:C \to C$ defined as follow : on objects $F(\*) = \*$ (and $F(0) = 0 $). And on morphism:


$$F(0) = 0$$
$$ F(\pm p\_1^{a\_1} \dots p\_n^{a\_n} ) = \pm \sigma(p\_1)^{a\_1} \dots \sigma(p\_n)^{a\_n} $$


We extend it to the $0$ object in the only way possible. Then $F$ is clearly a functor, but it is not additive unless $\sigma$ is the identity.


And $F$ preserves all kernel : any arrow $ n: \* \to \*$ has kernel 0 if $ n \neq 0$ and kernel $\*$ (with the identity map) if $n=0$ and in both case this is preserved by $F$. This also works for the arrows $0 : \* \to 0$ and $ 0 : 0 \to \*$ and $1=0 : 0 \to 0 $.


In fact even more simply, given that $F$ come from an automorphism of the monoid $(\mathbb{Z}, \times)$ it is (part of) an equivalence of categories, and hence preserve all limits and colimits that exists.


