Quotient Mapping of Inverse Completion

Theorem

Let $\left({T, \circ'}\right)$ be an inverse completion of a commutative semigroup $\left({S, \circ}\right)$, where $C$ is the set of cancellable elements of $S$.

Let $f: S \times C: T$ be the mapping defined as:

$\forall x \in S, y \in C: f \left({x, y}\right) = x \circ' y^{-1}$

Let $\mathcal R_f$ be the equivalence relation induced by $f$.

Then:

$\left({x_1, y_1}\right) \mathcal R_f \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

Proof

By the definition of $\mathcal R_f$:

$\left({x_1, y_1}\right) \mathcal R_f \left({x_2, y_2}\right) \iff x_1 \circ' y_1^{-1} = x_2 \circ' y_2^{-1}$

Now:

	\(\displaystyle \)	\(\displaystyle x_1 \circ' y_1^{-1}\)	\(=\)	\(\displaystyle x_2 \circ' y_2^{-1}\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle x_1 \circ' y_1^{-1} \circ' y_1 \circ' y_2\)	\(=\)	\(\displaystyle x_2 \circ' y_2^{-1} \circ' y_1 \circ' y_2\)	\(\displaystyle \)	Elements $y_1$ and $y_2$ are cancellable
	\(\displaystyle \implies\)	\(\displaystyle x_1 \circ' y_2\)	\(=\)	\(\displaystyle x_2 \circ' y_1\)	\(\displaystyle \)	Definition of Inverse, and commutativity of $\circ'$
	\(\displaystyle \implies\)	\(\displaystyle x_1 \circ y_2\)	\(=\)	\(\displaystyle x_2 \circ y_1\)	\(\displaystyle \)	$\circ'$ extends $\circ$

... which leads us to:

$\left({x_1, y_1}\right) \mathcal R_f \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

$\blacksquare$