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.

But as in group theory, some of the requirements of being a homomorphism follow automatically from others, and one often sees this 'simplified' definition instead.

  • $\begingroup$ As you mentioned that the map of of composition should also should be a group homomorphism. But in the definition of pre-additive category one does not assume that the composition map is homomorphism of abelian groups. $\endgroup$
    – Babai
    May 16 '16 at 21:47

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