Construction of Inverse Completion/Quotient Mapping/Image of Cancellable Elements

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 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:
 * $\forall x \in S: \psi \left({x}\right) = \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\mathcal R$

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

The set $C\,'$ of cancellable elements of the semigroup $S\,'$ is $\psi \left({C}\right)$.

Proof
Homomorphism preserves cancellability.

Thus $c \in C \implies \psi \left({c}\right) \in C\,'$.

So by Subset of Image, $\psi \left({C}\right) \subseteq C\,'$.

From above, $\psi$ is an isomorphism.

Hence $c' \in C\,' \implies \psi^{-1} \left({c'}\right) \in C$, also because homomorphism preserves cancellability.

So by Subset of Image, $\psi^{-1} \left({C\,'}\right) \subseteq C$.

Hence by definition of set equality, $\psi \left({C}\right) = C\,'$.