The following question is taken from Arrows, Structures and Functors the categorical imperative by Arbib and Manes.
$\color{Green}{Background:}$
$\textbf{(1)}$ $\text{Definition 1:}$ A $\textbf{product}$ in the category $\textbf{K}$ of a a family of objects $(A_i\mid i\in I)$ is an object $A$ together with a family
$$(A \xrightarrow{\normalsize{\text{in}}_i} A_i\mid i\in I)$$ of morphisms (called $\textbf{projections}$) with the property that given any other object $C$ similarly equipped with an $\textit{I}-$indexed family $(f_i:C\to A_i)$ there exists a
unique morphism $f:C\to A$ such that ${\pi}_i\circ f=f_i$ for all $i$ in $I.$ We write $A=\prod_{i\in I}A_i,$ If $I=\{1,\ldots,n\}$ we may write $A=A_1\times\ldots\times A_n.$
$\textbf{(2a)}$ $\textbf{Proposition 1:}$ Given a family $(A_i\mid i\in I$ of vector spaces, we define their $\textbf{product}$ to be the cartesian product $$\prod_{i\in I}=\{f\mid f:\to \cup_{i\in I} A_i\text{ and } f(i)\in A_i \text{ for each }i\}$$
equipped with the 'coordinatewise' addition and multiplication-by-a-scalar $$(f+f")(i)=f(i)+f'(i)\quad \text{ and } (\lambda \cdot f)(i)(i)=\lambda\cdot f(i)$$
for all $f,f'$ in $\prod_{i\in I}A_i,\lambda$ in $\textbf{R}$ and $i\in I.$ Then $\prod_{i\in I}A_i$ together with the projections $$\pi_j:\prod_{i\in I}A_i\to A_i:f\mapsto f(j)$$
is a product of $A_i)$ in the category $\textbf{Vect}$ in the sense of $\text{Definition 1:}$
Proof: Just check that $p(c)(i)=p_i(c)$ does the trick in $\text{Definition 1:}.$
$\textbf{(2b)}$ $\textbf{Proposition 2:}$ Given a family $(A_i\mid i\in I)$ of vector spaces, we define their $\textbf{weak direct sum}$ to be
$$\coprod_{i\in I}A_i=\{f\mid f:I\to \cup_{i \in I}A_i;f(i)\in A_i \text{ for each }i;\text{ and supp}(f)\text{ is finite}\}$$
considered as a subspace of $\prod_{i\in I}A_i.$ Then $\coprod_{i\in I}A_i$ together with the injections
$\text{in}_j:A_j\to \coprod_{i\in I}A_i:a_j\mapsto$ the $f$ with $f(i)=0$ for $i\neq j,$ and with $f(j)=a_j$
is a coproduct of $(A_i)$ in the category $\textbf{Vect}.$
$\textbf{(3)}$....Thus $\textbf{R}^n$ is a product in $\textbf{Vect}$ via the projections $\pi_j:\textbf{R}^n\to \textbf{R},(x_1,\dots,x_n)\mapsto x_j$ and is coproduct in $\textbf{Vect}$ via the injections $\text{in}_i:\textbf{R}\mapsto \textbf{R}^n, x\mapsto (0,\dots,x,0,\dots,0)$ (where $x$ occurs in the $i^\text{th}$ place).
As a prelude to the situation in arbitrary categories, let us observe that a linear map $f:\textbf{R}^3\to \textbf{R}^2$ corresponds to a $2\times 3$ $\textit{matrix}$ of linear maps $\textbf{R}\to \textbf{R}$
$$({f_i}^j)=\begin{pmatrix}{f_1}^1 & {f_2}^1 & {f_3}^1\\\ {f_1}^2 & {f_2}^2 & {f_3}^2\end{pmatrix}$$
as follows. Because $\textbf{R}^3=\textbf{R}+\textbf{R}+\textbf{R},$ $f$ corresponds to
$$\begin{pmatrix}f_1, & f_2, & f_3\end{pmatrix}$$
where $f_i=f\cdot \text{in}_i:\textbf{R}\to \textbf{R}^2.$ In turn, each $f_i$ corresponds to
$$\begin{pmatrix}{f_i}^1 \\\ {f_i}^2\end{pmatrix}$$
where ${f_i}^j=\pi_j\cdot f_i:\textbf{R}\to \textbf{R},$ because $\textbf{R}^2=\textbf{R}\times \textbf{R}.$ Thus $f$ corresponds to the $2\times 3$ matrix $({f_i}^j)$
As an example, using the universal properties of >>>>
$f(x,y,z)=f_1(x)+f_2(x)+f_3(x)$
$=\begin{pmatrix}{f_1}^1(x) & {f_2}^1(y) & {f_3}^1(z)\\\ {f_1}^2(x) & {f_2}^2(y) & {f_3}^2(z)\end{pmatrix}$
$\textbf{(4)}$ $\textbf{Theorem:}$ Let $\textbf{K}$ be a category in which $A+A, A\times A$ exist for every object $A.$ Then the following two conditions are equivalent.
- $\textbf{K}$ admits an $\textbf{Abm}-$ structure.
- $\textbf{K}$ is pointed (via $0_\text{AB},$ say) and for all $A,$ the map $\Delta:A+A\to A\times A$ defined by $\pi_j \cdot \Delta \cdot \text{in}_i = \delta_{ij}$ where in the matrix formulation of $\textbf{(1)}$
$$\begin{equation} \delta_{ij}= \begin{cases} \text{id}_A & \text{if } i=j\\ 0 & \text{if } i\neq j \end{cases} \end{equation}$$
is an isomorphism.
Moreover, if either of these conditions are true, the $\textbf{Abm}-$structure on $\textbf{K}$ is unique.
Note that we have "$\Delta=\begin{pmatrix}1 & 0 \\\ 0 & 1\end{pmatrix}$
$\color{Red}{Questions:}$
I understand that in the case $n=3$ in $\pi_j:\textbf{R}^n\to \textbf{R},(x_1,\dots,x_n)\mapsto x_j,$
$\pi_1:\textbf{R}^3\to \textbf{R},(x_1,x_2,x_3)\mapsto x_1$ $\pi_2:\textbf{R}^3\to \textbf{R},(x_1,x_2,x_3)\mapsto x_2$ $\pi_3:\textbf{R}^3\to \textbf{R},(x_1,x_2,x_3)\mapsto x_3$
and for $n=3$ in $\text{in}_i:\textbf{R}\mapsto \textbf{R}^n, x\mapsto (0,\dots,x,0,\dots,0),$
we have $\text{in}_1:\textbf{R}\mapsto \textbf{R}^3, x\mapsto \begin{pmatrix}x \\\ 0 \\\ 0 \end{pmatrix},$
$\text{in}_2:\textbf{R}\mapsto \textbf{R}^3, y\mapsto \begin{pmatrix}0 \\\ y \\\ 0 \end{pmatrix},$
$\text{in}_3:\textbf{R}\mapsto \textbf{R}^3, z\mapsto \begin{pmatrix}0 \\\ 0 \\\ z \end{pmatrix},$
But I don't understand how does ${f_i}^j=\pi_j\cdot f\cdot \text{in}_i$ construct the given example matrix in the explanation, $f(x,y,z)=f_1(x)+f_2(x)+f_3(x)$
$=\begin{pmatrix}{f_1}^1(x) & {f_2}^1(y) & {f_3}^1(z)\\\ {f_1}^2(x) & {f_2}^2(y) & {f_3}^2(z)\end{pmatrix}$.
The map $f:\textbf{R}^3\to \textbf{R}^2,$ it corresponds to a $2\times 3$ matrix of a linear map $\textbf{R}\to \textbf{R}$
$$({f_i}^j)=\begin{pmatrix}{f_1}^1 & {f_2}^1 & {f_3}^1\\\ {f_1}^2 & {f_2}^2 & {f_3}^2\end{pmatrix}.$$ Then it also says that it corresponds to $$\begin{pmatrix}f_1, & f_2, & f_3\end{pmatrix}$$
As a result, I am also not able to use ${f_i}^j=\pi_j\cdot f\cdot \text{in}_i$ in construct the identity matrix $\Delta$ in the $\textbf{Theorem}$ for $\textbf{(4)}.$