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

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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 $R^S$.


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 which 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 induced 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 $f^{-1}$.


$\blacksquare$