Quotient Mapping of Inverse Completion

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

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


 * $\forall x \in S, y \in C: \map f {x, y} = x \circ' y^{-1}$

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

Then:


 * $\tuple {x_1, y_1} \mathrel {\mathcal R_f} \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$

Proof
By the definition of $\mathcal R_f$:


 * $\tuple {x_1, y_1} \mathrel {\mathcal R_f} \tuple {x_2, y_2} \iff x_1 \circ' y_1^{-1} = x_2 \circ' y_2^{-1}$

Now:

which leads us to:


 * $\tuple {x_1, y_1} \mathrel {\mathcal R_f} \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$