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 an equivalence relation on $\left({S \times C, \oplus}\right)$.

Proof

Reflexivity

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

So $\mathcal R$ is a reflexive relation.

$\Box$

Symmetry

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x_1, y_1}\right)\)	\(\mathcal R\)	\(\displaystyle \left({x_2, y_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x_1 \circ y_2\)	\(=\)	\(\displaystyle x_2 \circ y_1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x_2 \circ y_1\)	\(=\)	\(\displaystyle x_1 \circ y_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$\circ$ is commutative
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({x_2, y_2}\right)\)	\(\mathcal R\)	\(\displaystyle \left({x_1, y_1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $\mathcal R$ is a symmetric relation.

$\Box$

Transitivity

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x_1, y_1}\right)\)	\(\mathcal R\)	\(\displaystyle \left({x_2, y_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \land\)	\(\displaystyle \)	\(\displaystyle \left({x_2, y_2}\right)\)	\(\mathcal R\)	\(\displaystyle \left({x_3, y_3}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x_1 \circ y_2\)	\(=\)	\(\displaystyle x_2 \circ y_1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \land\)	\(\displaystyle \)	\(\displaystyle x_2 \circ y_3\)	\(=\)	\(\displaystyle x_3 \circ y_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x_1 \circ y_3 \circ y_2\)	\(=\)	\(\displaystyle x_1 \circ y_2 \circ y_3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$\circ$ is commutative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x_2 \circ y_1 \circ y_3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	substituting $x_2 \circ y_1$ for $x_1 \circ y_2$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x_2 \circ y_3 \circ y_1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$\circ$ is commutative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x_3 \circ y_2 \circ y_1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	substituting $x_3 \circ y_2$ for $x_2 \circ y_3$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x_3 \circ y_1 \circ y_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$\circ$ is commutative
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x_1 \circ y_3\)	\(=\)	\(\displaystyle x_3 \circ y_1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $y_2 \in C$
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({x_1, y_1}\right)\)	\(\mathcal R\)	\(\displaystyle \left({x_3, y_3}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $\mathcal R$ is a transitive relation.

$\Box$

All the criteria are therefore seen to hold for $\mathcal R$ to be an equivalence relation.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 20$