Construction of Inverse Completion/Equivalence Relation

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 $C$.

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

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

Members of Equivalence Classes
We have:

$\forall x, y \in S, a, b \in C:$


 * $\left({x \circ a, a}\right) \mathcal R \left({y \circ b, b}\right) \iff x = y$

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

Equivalence Class of Equal Elements
We also have:
 * $\forall c, d \in C: \left({c, c}\right) \mathcal R \left({d, d}\right)$

Proof of Equivalence
First we show that $\mathcal R$ is an equivalence relation:

Reflexive

 * $x_1 \circ y_1 = x_1 \circ y_1 \implies \left({x_1, y_1}\right) \mathcal R \left({x_1, y_1}\right)$

Transitive
Thus we define the equivalence class $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\mathcal R$.

Proof of Congruence Relation
We now need to show that:

So:

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

Proof of Equivalence Class of Equal Elements
Note that as $C \subseteq S$, it is clear that $C \times C \subseteq S \times C$ from Cartesian Product of Subsets.

Thus we need only consider elements $\left({x, y}\right)$ of $C \times C$.