Module of All Mappings is Module

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Theorem

Let $\struct {R, +_R, \times_R}$ be a ring.

Let $\struct {G, +_G, \circ}_R$ be an $R$-module.

Let $S$ be a set.

Let $\struct {G^S, +_G', \circ}_R$ be the module of all mappings from $S$ to $G$.


Then $\struct {G^S, +_G', \circ}_R$ is an $R$-module.


Proof

To show that $\struct {G^S, +_G', \circ}_R$ is an $R$-module, we verify the following:

$\forall f, g, \in G^S, \forall \lambda, \mu \in R$:

$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$
$(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G \paren {\mu \circ f}$
$(3): \quad \paren {\lambda \times_R \mu} \circ f = \lambda \circ \paren {\mu \circ f}$


Criterion 1

$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$


Let $x \in S$.

Then:

\(\ds \lambda \circ \paren {\map {\paren {f +_G' g} } x}\) \(=\) \(\ds \lambda \circ \paren {\map f x +_G \map g x}\)
\(\ds \) \(=\) \(\ds \paren {\lambda \circ \map f x} +_G \paren {\lambda \circ \map g x}\) Definition of Module
\(\ds \) \(=\) \(\ds \paren {\map {\paren {\lambda \circ f} } x} +_G \paren {\map {\paren {\lambda \circ g} } x}\)
\(\ds \) \(=\) \(\ds \map {\paren {\paren {\lambda \circ f} +_G' \paren {\lambda \circ g} } } x\)

Thus $(1)$ holds.

$\Box$


Criterion 2

$(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G \paren {\mu \circ f}$


Let $x \in S$.

\(\ds \map {\paren {\paren {\lambda +_R \mu} \circ f} } x\) \(=\) \(\ds \paren {\lambda +_R \mu} \circ \paren {\map f x}\)
\(\ds \) \(=\) \(\ds \paren {\lambda \circ \map f x} +_G \paren {\mu \circ \map f x}\)
\(\ds \) \(=\) \(\ds \map {\paren {\lambda \circ f} } x +_G \map {\paren {\mu \circ f} } x\)

Thus $(2)$ holds.

$\Box$


Criterion 3

$(3): \quad \paren {\lambda \times_R \mu} \circ f = \lambda \circ \paren {\mu \circ f}$


\(\ds \map {\paren {\paren {\lambda \times_R \mu} \circ f} } x\) \(=\) \(\ds \paren {\lambda \times_R \mu} \circ \paren {\map f x}\)
\(\ds \) \(=\) \(\ds \lambda \circ \paren {\mu \circ \map f x}\)
\(\ds \) \(=\) \(\ds \lambda \circ \paren {\map {\paren {\mu \circ f} } x}\)
\(\ds \) \(=\) \(\ds \map {\paren {\lambda \circ \paren {\mu \circ f} } } x\)

Thus $(3)$ holds.

$\Box$


Hence the result.

$\blacksquare$


Sources

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