Canonical Injection into Cartesian Product of Modules

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Theorem

Let $G$ be the cartesian product of a sequence $\sequence {G_n}$ of $R$-modules.


Then for each $j \in \closedint 1 n$, the canonical injection $\inj_j$ 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:

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

and:

$\map {\map {\inj_a} {r x} } b = r x = r \, \map {\map {\inj_a} x} b$

$\Box$


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:

$\map {\inj_a} {r x} = r \, \map {\inj_a} x$

$\Box$


So $\inj_a$ is a homomorphism.

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

$\blacksquare$


Sources