Module of All Mappings is Module

Theorem
Let $\left({R, +_R, \times_R}\right)$ be a ring.

Let $\left({G, +_G: \circ}\right)_R$ be an $R$-module.

Let $S$ be a set.

Let $G^S$ be the set of all mappings from $S$ to $G$.

Then $\left({G^S, +_G', \circ}\right)_R$ is an $R$-module, where:
 * $+_G'$ is the operation induced on $G^S$ by $+_G$;
 * $\forall \lambda \in R: \forall f \in G^S: \forall x \in S: \left({\lambda \circ f}\right) \left({x}\right) = \lambda \circ f \left({x}\right)$.

This is the $R$-module $G^S$ of all mappings from $S$ to $G$.

If $\left({G, +_G, \circ}\right)_R$ is a unitary $R$-module, then $\left({G^S, +_G', \circ}\right)_R$ is also unitary.

The most important case of this example is when $\left({G^S, +_G', \circ}\right)_R$ is the $R$-module $\left({R^S, +_R', \circ}\right)_R$.

Proof
For $\left({G^S, +_G', \circ}\right)_R$ to be an $R$-module, we need to verify the following:

$\forall f, g, \in G^S, \forall \lambda, \mu \in R$:
 * $(1) \quad \lambda \circ \left({f +_G' g}\right) = \left({\lambda \circ f}\right) +_G' \left({\lambda \circ g}\right)$


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


 * $(3) \quad \left({\lambda \times_R \mu}\right) \circ f = \lambda \circ \left({\mu \circ f}\right)$

For $\left({G^S, +_G', \circ}\right)_R$ to be a unitary $R$-module, we need to verify the following:


 * $(4) \quad \forall f \in G^S: 1_R \circ f = f$

Criterion 1

 * $(1) \quad \lambda \circ \left({f +_G' g}\right) = \left({\lambda \circ f}\right) +_G' \left({\lambda \circ g}\right)$

Let $x \in S$.

Then:

Thus $(1)$ holds.

Criterion 2

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

Let $x \in S$.

Thus $(2)$ holds.

Criterion 3

 * $(3) \quad \left({\lambda \times_R \mu}\right) \circ f = \lambda \circ \left({\mu \circ f}\right)$

Thus $(3)$ holds.

Criterion 4

 * $(4) \quad \forall f \in G^S: 1_R \circ f = f$

Suppose $\left({G, +_G: \circ}\right)_R$ is a unitary $R$-module.

Then:
 * $\forall x \in G: 1_R \circ x = x$

Thus:

Thus $(4)$ holds and $\left({G^S, +_G': \circ}\right)_R$ is unitary.

Hence the result.