Construction of Inverse Completion/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 $\oplus$.

Proof
From the defined equivalence relation, we have that:
 * $\left({x_1, y_1}\right) \ \mathcal R \ \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

is a congruence relation on $\left({S \times C, \oplus}\right)$.

From the definition of the members of the equivalence classes:
 * $(1) \quad \forall x, y \in S, a, b \in C: \left({x \circ a, a}\right) \ \mathcal R \ \left({y \circ b, b}\right) \iff x = y$


 * $(2) \quad \forall x, y \in S, a, b \in C: \left[\!\left[{\left({x \circ a, y \circ a}\right)}\right]\!\right]_\mathcal R = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\mathcal R$

From the definition of the equivalence class of equal elements:
 * $(3) \quad \forall c, d \in C: \left({c, c}\right) \ \mathcal R \ \left({d, d}\right)$

where $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\mathcal R$ is the equivalence class of $\left({x, y}\right)$ under $\mathcal R$.

Hence we are justified in asserting the existence of the quotient structure:
 * $\displaystyle \left({T\,', \oplus'}\right) = \left({\frac {S \times C} {\mathcal R}, \oplus_\mathcal R}\right)$