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Let $\mathscr{C}$ be a preadditive category. Let $A,B$ be two objects and assume that $\mathscr{C}$ has a zero object $0$ (an object both initial and terminal). Let $f$ be the unique zero morphism from $A$ to $B$, which is necessarily the composition of the unique morphism from $A$ to $0$ (viewed as terminal) with the unique morphism from $0$ to $B$ (viewed as an initial object). My question is: can we say that $f$ is the zero of the abelian group $\mathscr{C}(A,B)$? If yes, why? If not, any counterexamples?

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The answer to your question is yes $f$ is indeed the zero of the abelian group $\mathscr C(A,B)$.

Here is the proof. Since $\mathscr C(0,0)$ has only one object it is the trivial group (the group with only the identity). We know that in a pre-additive category composition is bilinear so if $r \in \mathscr C(A,0)$ and $l \in \mathscr C(0,B)$ (the only morphisms from $A$ to the zero object and from the zero object to $B$ respectively) we have that the mapping $$\mathscr C(r,l)\colon \mathscr C(0,0) \longrightarrow \mathscr C(A,B)$$ $$\mathscr C(r,l)(h)=l\circ h \circ r$$ is a group-homomorphism and so it sends $\text{id}_0$, which is the unit of the trivial group $\mathscr C(0,0)$, to the unit of $\mathscr C(A,B)$.

But, by definition, we have that $\mathscr C(r,l)(\text{id}_0)=f$ and so $f$ is the unit of $\mathscr C(A,B)$ q.e.d. .

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Yes: Deconstruct your $f:A \to B$ as $gh$, where $h:A \to 0$ and $g:0 \to B$. As there is a unique map $g:0 \to B$, we must have $g+g = g$ and hence, by bilinearity, $gh + gh = (g+g)h = gh$. But then $$f + f = gh + gh = gh = f,$$ and cancelling $f$ from both sides yields that $f$ is the zero object in the Hom-set.

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