The following question is taken from "Arrows, Structures and Functors the categorical imperative" and "Algebraic Approaches to Program Semantics" both by: Arbib and Manes.
$\color{Green}{Background:}$
$\textbf{(1)}$ [From "Arrows, structures, and functors"]
$\textbf{Proposition:}$ 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
$\text{in}_j:A_j\rightarrow$ $\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 $\textbf{Vect}.$
$\textbf{(2)}$ There is also an exercise question similar to the above proposition in the text "Algebraic Approaches to Program Semantics" by the same two authors, and the exercise alongg with the relevant definition are included in the following screenshots:
$\color{Red}{Questions:}$
I would like to know if the following is a correct concrete description of the coproduct for category of vector spaces in $\textbf{Proposition:}$ above.
$\textbf{Coproduct}$
For $i=1,2$ and $I=\{1,2\}$ let $f:\{1,2\}\to A_1\sqcup A_2:i\mapsto f(i),$ and
For the injection map $\text{in}_j:A_j\mapsto A_1\sqcup A_2:a_j\mapsto f(i)=0$ for $i\neq j$ and $f(j)=a_j$ for $i=j,$ with $f$ being a linear map. With 'coordinatewise' addition and multiplication by a-scalar, then $f$ being linear means if $f'\in A_1\sqcup A_2,$ then $(f+f')(i)=f(i)+f'(i)$ and $(\lambda\cdot f)(i)=\lambda\cdot f(i).$ So $f+f'\in A_1\sqcup A_2$ $\lambda\cdot f\in A_1\sqcup A_2$
For the unique map $q$ that completes the commutative triangle in the theorem, $q:A_1\sqcup A_2\to C:f\mapsto q_1(f(1))+q_2(f(2)),$ $q(f)(1)=q_1(f(1))+q_2(0)=q_1(a_1)+q_2(0),$ and $q(f)(2)=q_1(0)+q_2(f(2))=q_1(0)+q_2(a_2)$
Linearity of $\text{in}_j$ and $q$ maps,
Injection map:
Let $a'_1\in A_1,a'_2\in A_2,$ and $f,f'\in A_1\sqcup A_2,$
$\text{in}_j:A_j\mapsto A_1\sqcup A_2:a_j\mapsto (f+f')(i)=0$ for $i\neq j,$ and $(f+f')(j)=a_1+a'_1$ for $i=j,$
$\text{in}_j:A_j\mapsto A_1\sqcup A_2:a_j\mapsto (\lambda\cdot f)(i)=0$ for $i\neq j,$ and $\lambda\cdot f(j)=\lambda\cdot a_j$ for $i=j,$
$q$ map:
Let $a'_1\in A_1,a'_2\in A_2,$ and $f,f'\in A_1\sqcup A_2,$ then
$q:A_1\sqcup A_2\to C:f+f'\mapsto q_1((f+f')(1))+q_2((f+f')(2)),$ $q(f+f')(1)=q_1((f+f')(1))+q_2(0)=q_1(a_1+a'_1)+q_2(0),$ and
$q((f+f'))(2)=q_1(0)+q_2((f+f')(2))=q_1(0)+q_2(a_2+a'_2)$
$q:A_1\sqcup A_2\to C:\lambda\cdot f\mapsto q_1((\lambda\cdot f)(1))+q_2((\lambda\cdot f)(2)),$ $q((\lambda\cdot f))(1)=q_1((\lambda\cdot f)(1))+q_2(0)=q_1(\lambda\cdot a_1)+q_2(0),$ and $q((\lambda\cdot f))(2)=q_1(0)+q_2((\lambda\cdot f)(2))=q_1(0)+q_2(\lambda\cdot a_2)$
I am not sure why I am getting $q_1(0), q_2(0)$ terms and what they are suppose to be equal to in the above four cases: $(f+f')(1), (f+f')(2), (\lambda\cdot f)(1), (\lambda\cdot f)(2)$.
Thank you in advance.