Unit of Ring of Mappings iff Image is Subset of Ring Units
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:
| \(\ds 1_R\) | \(=\) | \(\ds \map {f_{1_R} } x\) | Structure Induced by Ring with Unity Operations is Ring with Unity | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {\paren {f \circ' f^{-1} } } x\) | Definition of product inverse | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map f x \circ \map {f^{-1} } x\) | 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 $x \in S$:
| \(\ds \map {\paren {f \circ’ f^{-1} } } x\) | \(=\) | \(\ds \map f x \circ’ \map {f^{-1} } x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map f x \circ \map f x^{-1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1_R\) |
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 $f^{-1}$.
$\blacksquare$