F(0) = 0? F: additive functor

If I define an additive functor to be a functor on abelian categories such that the action of F on Hom(A,B) is a group homomorphism, do I necessarily have that F(zero object) = zero object?

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Yes: zero object is characterized by the property 0=identity – t3suji Jul 4 '10 at 4:36
The hypotheses on $A, B$ can be weakened: we only need that $A, B$ have zero objects and that $F$ preserve zero morphisms (that is, $F$ is a $\text{Set}_{\ast}$-enriched functor). – Qiaochu Yuan Sep 3 '12 at 7:15

1 Answer

Since the OP asked for a detailed answer:

Let $A$ be an object of an abelian (or additive) category. Then $A$ is a zero object if and only if the zero endomorphism is the identity endomorphism (and then $Hom(A,A)$ is the zero ring). If $F$ is any functor, it sends the identity morphism of $A$ to the identity morphism of $F(A)$. If in addition $F$ is additive (no pun intended), it also sends the zero morphism to the zero morphism. Thus $F(0)$ is a zero object.

Another exercise in the similar spirit: show that additive functor is additive on objects: it sends finite direct sums to direct sums. (Your question is the particular case of empty direct sum.)

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