# When are zero morphisms preserved?

I came across a problem where I want to say the functor I am working with preserves zero morphisms and I did some quick internet searching to try and figure out when this is actually true and came up empty. I know additive functors preserve zero objects, is there a weaker condition for zero morphisms? I know that the categories I am working in are both additive but not the functor.

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What kind of functor do you have in mind where it's hard to determine whether it preserves zero morphisms? The only useful condition I know is preserving zero objects. –  Qiaochu Yuan Mar 5 '13 at 22:22
@QiaochuYuan The general kind, I know nothing about my functor really. But if there was a condition that would help that I could prove holds that would help. –  Sean Ballentine Mar 5 '13 at 22:28
If you have a functor between two categories with a zero object, then the functor preserves zero morphisms if and only if it preserves the zero object. But if you can't even prove that in your example then you probably can't say much... –  Zhen Lin Mar 5 '13 at 22:30
@CliveNewstead I am not looking for a snappy name just an alternate condition. –  Sean Ballentine Mar 5 '13 at 22:30
Just a remark: The notion of zero morphisms also makes sense in categories without zero objects. This is the same as a category enriched in pointed sets. –  Martin Brandenburg Mar 5 '13 at 22:40

Let $\mathcal{A}$ and $\mathcal{B}$ be categories with a zero object. Then, whatever definition of zero morphism you are using, it is the case that a morphism is zero if and only if it factors through the zero object (and any such factorisation is unique).
1. Suppose $F : \mathcal{A} \to \mathcal{B}$ preserves zero morphisms. Note that the zero object $Z$ is characterised by the property that $0_Z = \textrm{id}_Z$; but $F$ preserves $0$ and $\textrm{id}$, so $F Z$ must also be a zero object.
2. Suppose $F : \mathcal{A} \to \mathcal{B}$ preserves the zero object. Then it preserves the factorisations through the zero object, because $F$ preserves composition, so $F$ must also preserve zero morphisms.
This is the same. First, the functor preserves morphims $a\to 0$ and $0\to b$, since they are uniqe. Further, every zero morphism $a\to b$ decomposes in product former morphims: $a\to 0\to b$.