Construction of Inverse Completion/Image of Quotient Mapping is Subsemigroup

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 $S$.

Then $\struct {S', \oplus'}$ is a subsemigroup of $\struct {T', \oplus'}$.

Proof
We have that $S'$ is the image $\psi \sqbrk S$ of $S$.

For $\struct {S', \oplus'}$ to be a subsemigroup of $\struct {T', \oplus'}$, by Subsemigroup Closure Test we need to show that $\struct {S', \oplus'}$ is closed.

Let $x, y \in S'$.

Then $x = \map \phi {x'}, y = \map \phi {y'}$ for some $x', y' \in S$.

But as $\phi$ is an isomorphism, it obeys the morphism property.

So $x \oplus' y = \map \phi {x'} \oplus' \map \phi {y'} = \map \phi {x' \circ y'}$.

Hence $x \oplus' y$ is the image of $x' \circ y' \in S$ and hence $x \oplus' y \in S'$.

Thus by the Subsemigroup Closure Test, $\struct {S', \oplus'}$ is a subsemigroup of $\struct {T', \oplus'}$