Construction of Inverse Completion/Quotient Mapping to Image is Isomorphism

Theorem
Let $\left({S, \circ}\right)$ be a commutative semigroup which has cancellable elements.

Let $C \subseteq S$ be the set of cancellable elements of $S$.

Let $\left({S \times C, \oplus}\right)$ be the external direct product of $\left({S, \circ}\right)$ and $\left({C, \circ \restriction_C}\right)$, where:
 * $\circ \restriction_C$ is the restriction of $\circ$ to $C \times C$, and
 * $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ \restriction_C$ on $C$.

Let $\boxtimes$ be the congruence relation $\boxtimes$ defined on $\left({S \times C, \oplus}\right)$ by:
 * $\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

Let the quotient structure defined by $\boxtimes$ be:
 * $\displaystyle \left({T\,', \oplus'}\right) := \left({\frac {S \times C} \boxtimes, \oplus_\boxtimes}\right)$

where $\oplus_\boxtimes$ is the operation induced on $\displaystyle \frac {S \times C} \boxtimes$ by $\oplus$.

Let the mapping $\psi: S \to T\,'$ be defined as:
 * $\forall x \in S: \psi \left({x}\right) = \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes$

Let $S\,'$ be the image $\psi \left({S}\right)$ of $S$.

Then $\psi$ is an isomorphism from $S$ onto $S\,'$.

Proof
From Quotient Mapping is Monomorphism, $\psi: \left({S, \circ}\right) \to \left({S\,', \oplus'}\right)$ is a monomorphism.

Therefore by definition:
 * $\psi$ is a homomorphism
 * $\psi$ is an injection.

Because $S\,'$ is the image of $\psi \left({S}\right)$, by Surjection by Restriction of Codomain $\psi$ is a surjection.

Therefore by definition $\psi: S \to S\,'$ is a bijection.

A bijective homomorphism is an isomorphism.