# Construction of Inverse Completion/Image of Quotient Mapping is Subsemigroup

## Theorem

Let $\left({S, \circ}\right)$ be a commutative semigroup which has cancellable elements.

Let $\left({C, \circ {\restriction_C}}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.

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 $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ {\restriction_C}$ on $C$.

Let $\boxtimes$ be the cross-relation on $S \times C$, defined as:

$\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

This cross-relation is a congruence relation on $S \times C$.

Let the quotient structure defined by $\boxtimes$ be:

$\left({T', \oplus'}\right) := \left({\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}\right)$

where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.

Let the mapping $\psi: S \to T'$ be defined as:

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

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

Then $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$.

## Proof

We have that $S'$ is the image $\psi \left({S}\right)$ of $S$.

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

Let $x, y \in S'$.

Then $x = \phi \left({x'}\right), y = \phi \left({y'}\right)$ for some $x', y' \in S$.

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

So $x \oplus' y = \phi \left({x'}\right) \oplus' \phi \left({y'}\right) = \phi \left({x' \circ y'}\right)$.

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, $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$

$\blacksquare$