Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms/Lemma 2

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Theorem

Let $\struct{R, + , \circ}$ be a ring with unity $1_R$.


Let $I_R : R \to R$ be the identity mapping on $R$.

For all $y \in R$, let $y * I_R : R \to R$ be the mapping defined by:

$\forall z \in R: \map {\paren {y * I_R} } z = y * \map {I_R} z$

For all $y \in R$, let $I_R * y : R \to R$ be the mapping defined by:

$\forall z \in R: \map {\paren {I_R * y} } z = \map {I_R} z * y$


Let $x \in R$ be a unit of $R$ with product inverse $x^{-1}$.


Then:

$x * I_R$ is a bijection and $x^{-1} * I_R$ is the inverse of $x * I_R$
$I_R * x$ is a bijection and $I_R * x^{-1}$ is the inverse of $I_R * x$


Proof

Consider the composite of $x * I_R$ with $x^{-1} * I_R$.

\(\ds \forall y \in R: \, \) \(\ds \map {\paren {\paren {x * I_R} \circ \paren {x^{-1} * I_R} } } y\) \(=\) \(\ds \map {\paren {x * I_R} } {\map {\paren {x^{-1} * I_R} } y}\)
\(\ds \) \(=\) \(\ds \map {\paren {x * I_R} } {x^{-1} * y}\)
\(\ds \) \(=\) \(\ds x * \paren {x^{-1} * y}\)
\(\ds \) \(=\) \(\ds \paren {x * x^{-1} } * y\)
\(\ds \) \(=\) \(\ds y\)
\(\ds \) \(=\) \(\ds \map {I_R} y\)

From Equality of Mappings, $\paren {x * I_R} \circ \paren {x^{-1} * I_R} = I_R$.


Consider the composite of $x^{-1} * I_R$ with $x * I_R$.

\(\ds \forall y \in R: \, \) \(\ds \map {\paren {\paren {x^{-1} * I_R} \circ \paren {x * I_R} } } y\) \(=\) \(\ds \map {\paren {x^{-1} * I_R} } {\map {\paren {x * I_R} } y}\)
\(\ds \) \(=\) \(\ds \map {\paren {x^{-1} * I_R} } {x * y}\)
\(\ds \) \(=\) \(\ds x^{-1} * \paren {x * y}\)
\(\ds \) \(=\) \(\ds \paren {x^{-1} * x} * y\)
\(\ds \) \(=\) \(\ds y\)
\(\ds \) \(=\) \(\ds \map {I_R} y\)

From Equality of Mappings, $\paren {x^{-1} * I_R} \circ \paren {x * I_R} = I_R$.


Hence $x * I_R$ has a left inverse and a right inverse.

From definition 2 of bijection, $x * I_R$ is a bijection and the inverse is $x^{-1} * I_R$.


Similarly, $I_R * x$ is a bijection and the inverse is $I_R * x^{-1}$.

$\blacksquare$

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