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:

which leads us to:


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