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:

$$\forall f, g, \in G^S, \forall \lambda, \mu \in R$$:
 * VS 1: $$\lambda \circ \left({f +_G' g}\right) = \left({\lambda \circ f}\right) +_G' \left({\lambda \circ g}\right)$$;
 * VS 2: $$\left({\lambda +_R \mu}\right) \circ f = \left({\lambda \circ f}\right) +_G \left({\mu \circ f}\right)$$;
 * VS 3: $$\left({\lambda \times_R \mu}\right) \circ f = \lambda \circ \left({\mu \circ f}\right)$$.


 * VS 1:

Let $$x \in S$$.

Then:

$$ $$ $$ $$

Thus VS 1 holds.


 * VS 2:

Let $$x \in S$$.

$$ $$ $$

Thus VS 2 holds.


 * VS 3:

$$ $$ $$ $$

Thus VS 3 holds.


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

Then VS 4: $$\forall x \in G: 1_R \circ x = x$$.

Thus:

$$ $$

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