Unit of Ring of Mappings iff Image is Subset of Ring Units/Image is Subset of Ring Units implies Unit of Ring of Mappings
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 $R^S$.
Let $\Img f \subseteq U_R$ where $\Img f$ is the image of $f$.
Then:
- $f \in R^S$ is a unit of $R^S$
and 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:
- $\forall x \in S: \map {f_{1_R}} x = 1_R$
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 $f^{-1}$.
$\blacksquare$