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

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

Symmetric
$$ $$ $$ $$

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 Members of Equivalence Classes
$$ $$ $$

$$ $$ $$

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 $$C \times C$$.

$$ $$