Canonical Injection into Cartesian Product of Modules

Theorem
Let $G$ be the cartesian product of a sequence $\left \langle {G_n} \right \rangle$ of $R$-modules.

Then for each $j \in \left[{1 \,.\,.\, n}\right]$, the canonical injection $\operatorname{in}_j$ from $G_j$ into $G$ is a monomorphism.

Proof
$G$ can be seen as functions $f: A \to \bigcup_{a \in A}G_{a}$.

Now, let $a \in A$ be fixed. Let $x$ and $y$ be elements of $G_{a}$, let $r \in R$. So both $x + y$ and $rx$ are elements of $G_{a}$. Let $b \in A$.

Case 1: $b = a$: Then we have that
 * $\operatorname{in}_{a}(x + y)(b) = x + y = \operatorname{in}_{a}(x)(b) + \operatorname{in}_{a}(y)(b)$, and $\operatorname{in}_{a}(rx)(b) = rx = r\operatorname{in}_{a}(x)(b)$

Case 2: $b \neq a$
 * $\operatorname{in}_{a}(x + y)(b) = 0 + 0 = \operatorname{in}_{a}(x)(b) + \operatorname{in}_{a}(y)(b)$, and $\operatorname{in}_{a}(rx)(b) = 0 = r(0) = r\operatorname{in}_{a}(x)(b)$

Therefore $\operatorname{in}_{a}(x + y) = \operatorname{in}_{a}(x) + \operatorname{in}_{a}(y)$, and $\operatorname{in}_{a}(rx) = r\operatorname{in}_{a}(x)$. So, $\operatorname{in}_{a}$ is a homomorphism. Combined with Canonical Injections are Injections gives that $\operatorname{in}_{a}$ is a monomorphism.