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

Theorem
Let the mapping $\psi: S \to T'$ be defined as:
 * $\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$

Let $S'$ be the image $\psi \sqbrk S$ of $S$.

The set $C'$ of cancellable elements of the semigroup $S'$ is $\psi \sqbrk C$.

Proof
We have Morphism Property Preserves Cancellability.

Thus:
 * $c \in C \implies \map \psi c \in C'$

So by Image of Subset under Mapping is Subset of Image:
 * $\psi \sqbrk C \subseteq C'$

From above, $\psi$ is an isomorphism.

Hence, also from Morphism Property Preserves Cancellability:
 * $c' \in C' \implies \map {\psi^{-1} } {c'} \in C$

So by Preimage of Subset is Subset of Preimage:
 * $\psi^{-1} \sqbrk {C'} \subseteq C$

Hence by definition of set equality:
 * $\psi \sqbrk C = C'$