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

Theorem
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 $S$.

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 Image of Subset is Subset of Image/Corollary 2, $\psi \left({C}\right) \subseteq C\,'$.

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 Image of Subset is Subset of Image/Corollary 3, $\psi^{-1} \left({C\,'}\right) \subseteq C$.

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