The Stacks project

Proof. Additive functors commute with direct sums by Lemma 12.3.7 hence (1) implies (2) and (3). If (2) holds and $0$ denotes the zero object of $\mathcal{A}$, then the map $F(0) \oplus F(0) \to F(0 \oplus 0) \cong F(0)$ being an isomorphism implies that the diagonal map $\mathop{\mathrm{Hom}}\nolimits (F(0), C) \to \mathop{\mathrm{Hom}}\nolimits (F(0), C) \oplus \mathop{\mathrm{Hom}}\nolimits (F(0), C)$ is bijective for all $C$ in $\mathcal{B}$. Thus $F(0)$ is the zero object of $\mathcal{B}$ and $F$ maps the zero morphism between any two objects to the zero morphism. Similarly holds if (3) holds. It follows that (2) and (3) are equivalent because the composition $F(A) \oplus F(B) \to F(A \oplus B) \to F(A) \oplus F(B)$ is given by the matrix

and therefore is the identity map. Assume (2) and (3) hold. Let $f, g : A \to B$ be maps. Then $f + g$ is equal to the composition

Apply the functor $F$ and consider the following diagram

We claim this is commutative. For the middle square we can verify it separately for each of the four induced maps $F(A) \to F(B)$ where it follows from the fact that $F$ is a functor and transforms zero morphisms into zero morphisms. For the triangle on the left, we use that $F(A \oplus A) \to F(A) \oplus F(A)$ is an isomorphism to see that it suffice to check after composition with this map and this check is trivial. Dually for the other triangle. Thus going around the bottom is equal to $F(f + g)$ and we conclude. $\square$