Construction of Inverse Completion/Quotient Mapping is Monomorphism

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

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

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

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

The mapping $\psi: S \to T\,'$ is a monomorphism.

Proof
We have that this quotient mapping $\psi: S \to T\,'$ is an injection.

Let $x, y \in S$. Then:

So $\psi \left({x \circ y}\right) = \psi \left({x}\right) \oplus' \psi \left({y}\right)$, and the morphism property is proven.

Thus $\psi$ is an injective homomorphism, and so by definition a monomorphism.