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

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 $x^{-1} * I_{_R}$.

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 $x * I_{_R}$.

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 $I_{_R} * x^{-1}$.