Canonical Injection into Cartesian Product of Modules/Proof 2

Proof
$G$ can be seen as functions:
 * $\ds f: A \to \bigcup_{a \mathop \in A} G_a$

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:
 * $\map {\map {\inj_a} {x + y} } b = x + y = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$

and:
 * $\map {\map {\inj_a} {r x} } b = r x = r \, \map {\map {\inj_a} x} b$

Case 2
Let $b \ne a$.

Then:
 * $\map {\map {\inj_a} {x + y} } b = 0 + 0 = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$

and:
 * $\map {\map {\inj_a} {r x} } b = 0 = \map r 0 = r \, \map {\map {\inj_a} x} b$

Therefore:
 * $\map {\inj_a} {x + y} = \map {\inj_a} x + \map {\inj_a} y$

and:
 * $\map {\inj_a} {r x} = r \, \map {\inj_a} x$

So $\inj_a$ is a homomorphism.

Combined with Canonical Injection is Injection gives that $\inj_a$ is a monomorphism.