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The question:

Let $F:\mbox{Ab} \to \mbox{Ab}$ be an additive functor; if $f$ is a zero homomorphism, then so is $F(f)$; if $A$ is the zero group, then so is $F(A)$.

This boils down to showing that, for an additive functor $F(0)=0$. Obviously this is true, but my category theory is very basic, and I can't quite yet get a handle on what is allowed.

Is it simply just the case that

$$F(f+g)=F(f)+f(g) \implies F(0) = F(f+(-f))=F(f)+F(-f)=F(f)-F(f)=0?$$

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You would need to show that $F(-f) = -F(f)$. – Arturo Magidin Apr 8 '11 at 13:15
up vote 2 down vote accepted

By definition an additive functor induces homomorphisms on the hom-groups, that is $$ \begin{array}{c} \mbox{Hom}_{\mathbb Z}(A,B) \to \mbox{Hom}_{\mathbb Z}(FA,FB)\\ f \mapsto F(f) \end{array} $$ is an homomorphism. And well, homomorphisms preserve the neutral element.

Notice that $\mbox{Hom}_{\mathbb Z}(A,B)$ is a group with the point-wise operation inherited from $\mathbb Z$. Therefore the neutral element is the $0$ morphism

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