Module of All Mappings is Module

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:

Thus $(1)$ holds.

Criterion 2

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

Let $x \in S$.

Thus $(2)$ holds.

Criterion 3

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

Thus $(3)$ holds.

Hence the result.