Unit of Ring of Mappings iff Image is Subset of Ring Units/Unit of Ring of Mappings implies 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 $R^S$.

Let $f \in R^S$ be a unit of $R^S$.


Then:

$\Img f \subseteq U_R$

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:

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


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.

$\blacksquare$