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 of Cancellable 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 $$\left({C, \circ|_C}\right)$$ onto $$\left({T, \circ'}\right)$$.

By the Quotient Theorem for Epimorphisms, the proof follows.