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 $I_R * x^{-1}$.