Neutral element in $\hom_C(A, B)$ 
Let $C$ be an abelian category. Assuming that $\hom_C(A,B)$ has an abelian group structure, prove that the zero map $0_{AB}:A\to B$ is the neutral element of this group.

I know that the group operation can be defined as $f+g:=\nabla_B\circ(f\oplus g)\circ \Delta_A$ for every $f, g\in\hom_C(A, B)$, where $\oplus$ is the biproduct, $\Delta_X:X\to X\oplus X$ is the diagonal map and $\nabla_X:X\oplus X\to X$ is the codiagonal map. 
Defining $\varphi:=(f\oplus 0_{AB})\circ \Delta_A$, I've shown that $\pi_1\circ\varphi=f$ and that $\pi_2\circ\varphi=0_{AB}$ (where $\pi_1$, $\pi_2$ are the natural projections), which intuitively means that $f+0_{AB}=\nabla_B\circ \varphi=f$, but I don't know how to formalize this.
How do I show this formaly?
 A: An alternative argument, which might be easier.
If by having abelian group structures on $\operatorname{Hom}_\mathcal{C} (A,B)$ you actually mean that the composition of morphisms is distributive with respect to addition:
$$(g+g')\circ f = g\circ f + g'\circ f,\quad g\circ (f+f') = g\circ f + g\circ f'$$
(i.e. that your category is preadditive), then it is enough to note that


*

*the above identities imply that the neutral elements $0$ with respect to addition of morphisms satisfy $0\circ f = 0$ and $g\circ 0 = 0$ for any $f, g$ (composition of the neutral element with any morphism is the neutral element in the corresponding abelian group);

*the abelian groups $\operatorname{Hom}_\mathcal{C} (A,0) = \{ A\to 0 \}$ and $\operatorname{Hom}_\mathcal{C} (0,B) = \{ 0\to B \}$ are trivial (here $0$ is a zero object, i.e. both initial and terminal), and therefore the composition of arrows $A\to 0$ and $0\to B$ is the neutral element in $\operatorname{Hom}_\mathcal{C} (A,B)$.
So in any preadditive category with a zero object $0$, the neutral element of $\operatorname{Hom}_\mathcal{C} (A,B)$ is the morphism $A\to 0\to B$.
A: The key is to notice that $\varphi$ is actually equal to $i_1\circ f$, where $i_1:B\to B\oplus B$ is the first inclusion. 
To prove this, notice that $\pi_1\circ (i_1\circ f)=id_B\circ f=f$ and that $\pi_2\circ (i_1\circ f)=0_{BB}\circ f=0_{AB}$. As you've proved, $\pi_1\circ\varphi=f$ and $\pi_2\circ\varphi=0_{AB}$ so $\varphi$ and $i_1\circ f$ have the same components. Using the universal property of the product, we have the commutative diagram:
$$\begin{array}
BB & \stackrel{\pi_1}{\longleftarrow} & B\oplus B \\
\uparrow{f} &\,\,\nearrow{\exists!} & \downarrow{\pi_2} \\
A & \stackrel{0_{AB}}{\longrightarrow} & B  
\end{array}
$$
which says there is only one morphism $A\to B\oplus B$ whose components are $f$ and $0_{AB}$, therefore $\varphi$ and $i_1\circ f$ must be the same.
Thus $f+0_{AB}=\nabla\circ\varphi=\nabla\circ i_1\circ f=f$.
