Construction of Inverse Completion/Invertible Elements in Quotient Structure

Theorem
Let $\left({S, \circ}\right)$ be a commutative semigroup which has cancellable elements.

Let $C \subseteq S$ be the set of cancellable elements of $S$.

Let $\left({S \times C, \oplus}\right)$ be the external direct product of $\left({S, \circ}\right)$ and $\left({C, \circ \restriction_C}\right)$, where:
 * $\circ \restriction_C$ is the restriction of $\circ$ to $C \times C$, and
 * $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ \restriction_C$ on $C$.

Let $\mathcal R$ be the congruence relation $\mathcal R$ defined on $\left({S \times C, \oplus}\right)$ by:
 * $\left({x_1, y_1}\right) \ \mathcal R \ \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

Let the quotient structure defined by $\mathcal R$ be:
 * $\displaystyle \left({T\,', \oplus'}\right) := \left({\frac {S \times C} {\mathcal R}, \oplus_\mathcal R}\right)$

where $\oplus_\mathcal R$ is the operation induced on $\displaystyle \frac {S \times C} {\mathcal R}$ by $\oplus$.

Every cancellable element of $S\,'$ is invertible in $T\,'$.

Proof
From Identity of Quotient Structure, $\left({T\,', \oplus'}\right)$ has an identity, and it is $\left[\!\left[{\left({c, c}\right)}\right]\!\right]_\mathcal R$ for any $c \in C$. Call this identity $e_{T\,'}$.

From Image of Cancellable Elements in Quotient Mapping, $C\,' = \psi \left({C}\right)$.

So:

The inverse of $x'$ is $\left[\!\left[{\left({a, a \circ x}\right)}\right]\!\right]_{\mathcal R}$, as follows:

... thus showing that the inverse of $\left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\mathcal R$ is $\left[\!\left[{\left({a, a \circ x}\right)}\right]\!\right]_\mathcal R$.