# Unit of Ring of Mappings iff Image is Subset of Ring Units

## Contents

## Theorem

Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.

Let $U_R$ be the set of units in $R$.

Let $S$ be a set.

Let $\struct {R^S, +', \circ'}$ be the ring of mappings on the set of mappings $R^S$.

Then:

- $f \in R^S$ is a unit of $R^S$ if and only if $\Img f \subseteq U_R$

where $\Img f$ is the image of $f$.

In this case, the inverse of $f$ is the mapping $f^{-1} : S \to U_R$ defined by:

- $\forall x \in S : \map {f^{-1} } x = \map f x^{-1}$

## Proof

From Structure Induced by Ring with Unity Operations is Ring with Unity, $\struct {R^S, +', \circ'}$ has a unity $f_{1_R}$ defined by:

- $\forall x \in S: \map {f_{1_R}} x = 1_R$

### Unit of Ring of Mappings implies Image is Subset of Ring Units

Let $f^{-1}$ be the product inverse of $f$.

Let $x \in R$.

Then:

\(\displaystyle 1_R\) | \(=\) | \(\displaystyle \map {f_{1_R} } x\) | Structure Induced by Ring with Unity Operations is Ring with Unity | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {\paren {f \circ' f^{-1} } } x\) | Definition of product inverse | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map f x \circ \map {f^{-1} } x\) | Definition of Pointwise Operation $\circ'$ |

Similarly, $1_R = \map {f^{-1} } x \circ \map f x$

Hence $\map {f^{-1}} x$ is the product inverse of $\map f x$.

So $\map f x$ is a unit in $R$ and $\map f x^{-1} = \map {f^{-1} } x$.

The result follows.

$\Box$

### Image is Subset of Ring Units implies Unit of Ring of Mappings

By assumption:

- $\forall x \in S: \exists \map f x^{-1} : \map f x \circ \map f x^{-1} = \map f x^{-1} \circ \map f x = 1_R$

Let $f^{-1} : S \to U_R$ be defined by:

- $\forall x \in S : \map {f^{-1}} {x} = \map f x^{-1}$

Consider the mapping $f \circ’ f^{-1}$.

For all $x \in S$:

\(\displaystyle \map {\paren {f \circ’ f^{-1} } } x\) | \(=\) | \(\displaystyle \map f x \circ’ \map {f^{-1} } x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map f x \circ \map f x^{-1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1_R\) |

Hence $f \circ’ f^{-1} = f_{1_R}$.

Similarly, $f^{-1} \circ’ f = f_{1_R}$.

Hence $f$ is a unit of $\struct {R^S, +', \circ'}$ and the inverse of $f$ is the mapping $f^{-1}$.

$\blacksquare$