Unit of Ring of Mappings iff Image is Subset of Ring Units/Image is Subset of Ring Units implies Unit of Ring of Mappings

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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 $x \in S$:

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

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 $f^{-1}$.

$\blacksquare$