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:
 * $\displaystyle f: A \to \bigcup_{a \mathop \in A} G_a$

Let $a \in A$.

Let $x, y \in G_a$.

Let $r \in R$.

So both $x + y \in G_a$ and $r x \in G_a$.

Let $b \in A$.

Case 1
Let $b = a$.

Then:
 * $\operatorname{in}_a \left({x + y}\right) \left({b}\right) = x + y = \operatorname{in}_a \left({x}\right) \left({b}\right) + \operatorname{in}_a(y) \left({b}\right)$

and:
 * $\operatorname{in}_a \left({r x}\right) \left({b}\right) = r x = r\operatorname{in}_a \left({x}\right) \left({b}\right)$

Case 2
Let $b \ne a$.

Then:
 * $\operatorname{in}_a \left({x + y}\right) \left({b}\right) = 0 + 0 = \operatorname{in}_a \left({x}\right) \left({b}\right) + \operatorname{in}_a \left({y}\right) \left({b}\right)$

and:
 * $\operatorname{in}_a \left({r x}\right) \left({b}\right) = 0 = r \left({0}\right) = r\operatorname{in}_a \left({x}\right) \left({b}\right)$

Therefore:
 * $\operatorname{in}_a \left({x + y}\right) = \operatorname{in}_a \left({x}\right) + \operatorname{in}_a \left({y}\right)$

and:
 * $\operatorname{in}_a \left({r x}\right) = r\operatorname{in}_a \left({x}\right)$

So $\operatorname{in}_a$ is a homomorphism.

Combined with Canonical Injection is Injection gives that $\operatorname{in}_a$ is a monomorphism.