# 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:

 $\displaystyle \lambda \circ \paren {\map {\paren {f +_G' g} } x}$ $=$ $\displaystyle \lambda \circ \paren {\map f x +_G \map g x}$ $\displaystyle$ $=$ $\displaystyle \paren {\lambda \circ \map f x} +_G \paren {\lambda \circ \map g x}$ Definition of Module $\displaystyle$ $=$ $\displaystyle \paren {\map {\paren {\lambda \circ f} } x} +_G \paren {\map {\paren {\lambda \circ g} } x}$ $\displaystyle$ $=$ $\displaystyle \map {\paren {\paren {\lambda \circ f} +_G' \paren {\lambda \circ g} } } x$

Thus $(1)$ holds.

$\Box$

### Criterion 2

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

Let $x \in S$.

 $\displaystyle \map {\paren {\paren {\lambda +_R \mu} \circ f} } x$ $=$ $\displaystyle \paren {\lambda +_R \mu} \circ \paren {\map f x}$ $\displaystyle$ $=$ $\displaystyle \paren {\lambda \circ \map f x} +_G \paren {\mu \circ \map f x}$ $\displaystyle$ $=$ $\displaystyle \map {\paren {\lambda \circ f} } x +_G \map {\paren {\mu \circ f} } x$

Thus $(2)$ holds.

$\Box$

### Criterion 3

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

 $\displaystyle \map {\paren {\paren {\lambda \times_R \mu} \circ f} } x$ $=$ $\displaystyle \paren {\lambda \times_R \mu} \circ \paren {\map f x}$ $\displaystyle$ $=$ $\displaystyle \lambda \circ \paren {\mu \circ \map f x}$ $\displaystyle$ $=$ $\displaystyle \lambda \circ \paren {\map {\paren {\mu \circ f} } x}$ $\displaystyle$ $=$ $\displaystyle \map {\paren {\lambda \circ \paren {\mu \circ f} } } x$

Thus $(3)$ holds.

$\Box$

Hence the result.

$\blacksquare$