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

14$\begingroup$ Yes: zero object is characterized by the property 0=identity $\endgroup$– t3sujiJul 4, 2010 at 4:36

$\begingroup$ 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). $\endgroup$– Qiaochu YuanSep 3, 2012 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.)