Does left-exactness imply semi-additivity? 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?
 A: 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.
