Morphisms between biproducts in additive categories I can't understand the following description of a morphism between biproducts in an additive category, which I found in Borceux, Vol.2
If $A_1,A_2,B_1,B_2$ are four objects in an additive category $\mathscr{C}$, a morphism
$$f:A_1\oplus A_2\longrightarrow B_1\oplus B_2$$
is completely characterized by the four morphisms
$$f_{11}=p_1\circ f\circ s_1:A_1\longrightarrow B_1$$
$$f_{12}=p_1\circ f\circ s_2:A_2\longrightarrow B_1$$
$$f_{21}=p_2\circ f\circ s_1:A_1\longrightarrow B_2$$
$$f_{22}=p_2\circ f\circ s_2:A_2\longrightarrow B_2$$
$\textbf{Questions:}$
1) what does it mean $\textit{completely characterized by}$? I suppose it means that if two morphisms $f,g$ have $f_{11}=g_{11},\ldots ,f_{22}=g_{22}$ then $f=g$. Is it correct?
2) since $B_1\oplus B_2$ is a product and $f$ is a morphism into a product, isn't it true that $p_1\circ f$ and $p_2\circ f$ already characterize $f$? Why do I need to consider also compositions with the $s_i$'s?
 A: The point is that the $f_{ij}$ are the entries in a matrix:
$$\begin{pmatrix}
f_{11}& f_{12}\\
f_{21} &f_{22}
\end{pmatrix}.$$
In fact, in the case when our category is the category of (left) $R$-modules for some ring $R$, this description tells us that a map $R^n\rightarrow R^m$ (where we think of $R^n$ as $R\oplus\dots\oplus R$ with $n$ copies of $R$, and similarly for $R^m$) is completely characterised by the maps $a_{ij}:R\rightarrow R$ for $1\leq i\leq m$ and $1\leq j\leq n$. Since a linear map $R\rightarrow R$ is just multiplication by a scalar we recover the usual way of writing a linear map as a matrix:
$$\begin{pmatrix}
a_{11}&\dots& a_{1n}\\
\vdots&\ddots&\vdots\\
a_{m1} &\dots&a_{mn}
\end{pmatrix}.$$
In answer to your questions:

1) what does it mean $\textit{completely characterized by}$? I suppose it means that if two morphisms $f,g$ have $f_{11}=g_{11},\ldots ,f_{22}=g_{22}$ then $f=g$. Is it correct?

It means $f=g$ if and only if $f_{ij}=g_{ij}$ for all $i$ and $j$, and that for any components $(f_{ij})_{i,j\in\{1,2\}}$ there is some $f$ giving rise to them.

2) since $B_1\oplus B_2$ is a product and $f$ is a morphism into a product, isn't it true that $p_1\circ f$ and $p_2\circ f$ already characterize $f$? Why do I need to consider also compositions with the $s_i$'s?

Sure, to check that two matrices are the same it suffices to check that each row of one is equal to the corresponding row of the other. But the usual way to check two rows are the same is to check that each of their entries match.
In applications it's typical that we have a concrete idea of what the objects $A_1$, $A_2$, $B_1$ and $B_2$ are, but we have no way of handling $A_1\oplus A_2$ or $B_1\oplus B_2$ except by using the fact that these are the biproducts of $A_1$ and $A_2$ and $B_1$ and $B_2$. So if we have two maps $f,g:A_1\oplus A_2\rightarrow B_1\oplus B_2$, the only way to check if they are equal is to calculate their components.
A: For (1), you're right, but one can (and probably the author intended this) even strengthen this by saying that any four morphisms $f_{ij}: A_i\to B_j$ give rise to a morphism $f: A_1\oplus A_2\to B_1\oplus B_2$. For (2)... you're also right, it's just that what you say stands in no contradiction with the statement in question: namely, when you consider $p_i\circ f$, you can decompose further, since these morphisms are now morphisms from a coproduct, namely $A_1\oplus A_2$. And if you do this, you already end up with the statement you're after.
