# Additive category and zero map

Let $A$ be an additive category. Namely

1. $A$ has a zero object,
2. $A$ has finite products and coproducts, and
3. Every Hom-set is an Abelian group such that composition of morphisms is bilinear.

Question: Is the zero morphism always the additive unit?

Since $A$ has a zero object, say $0$, for any two objects $c$ and $d$, there uniquely exists a morphism $c \to d$ called the zero morphism which can be decomposed as $c \to 0 \to d$.

For typical examples (e.g. (Ab), ($R$-Mod), (Chain complexes in $R$-Mod)), the zero map gives the additive unit of each Hom-set. But the axioms of additive categories do not seem to say that the zero maps give the additive units of Hom-sets.

Is it possible to show that the zero morphism is the additive unit of each Hom-set? Or are there any additive categories such that the zero morphism is not the additive unit of some Hom-set? How about abelian categories?

There is only one map $g : 0 \to d$. Therefore there is only one possibility for $g + g$. Thus,
$$gf + gf = (g+g)f = gf$$
for any map $f$ composable with $g$.
Strictly speaking, composition should be defined as abelian group homomorphisms $\hom(B,C) \otimes \hom(A,B) \to \hom(A,C)$ (with additional properties), so the image of $\hom(0,d) \otimes \hom(c,0) \to \hom(c,d)$ is automatically the zero map.