Quotient Structure 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:


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

Then the mapping $g: \left({S \times C}\right) / \mathcal R_f \to T$ defined by $g \left({\left[\!\left[{x, y}\right]\!\right]_{\mathcal R_f}}\right) = x \circ' y^{-1}$,

where $\left({S \times C}\right) / \mathcal R_f$ is a quotient structure, is an isomorphism.

Proof
$T$ is commutative, from Inverse Completion Commutative Semigroup.

The mapping $f \left({x, y}\right) = x \circ' y^{-1}$ is an epimorphism from the cartesian product of $\left({S, \circ}\right)$ and $\left({C, \circ \restriction_C}\right)$ onto $\left({T, \circ'}\right)$.

By the Quotient Theorem for Epimorphisms, the proof follows.