Construction of Inverse Completion/Quotient Structure is Commutative Semigroup

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 $\mathcal R$ be the congruence relation $\mathcal R$ defined on $\left({S \times C, \oplus}\right)$ by:
 * $\left({x_1, y_1}\right) \ \mathcal R \ \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

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 $\oplus$.

Then:
 * $\left({T\,', \oplus'}\right)$ is a commutative semigroup.

Proof
The canonical epimorphism from $\left({S \times C, \oplus}\right)$ onto $\left({T\,', \oplus'}\right)$ is given by:


 * $q_\mathcal R: \left({S \times C, \oplus}\right) \to \left({T\,', \oplus'}\right): q_\mathcal R \left({x, y}\right) = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\mathcal R$

where, by definition:

By Morphism Property Preserves Closure, as $\oplus$ is closed, then so is $\oplus'$.

By Epimorphism Preserves Associativity, as $\oplus$ is associative, then so is $\oplus'$.

By Epimorphism Preserves Commutativity, as $\oplus$ is commutative, then so is $\oplus'$.

Thus $\left({T\,', \oplus'}\right)$ is closed, associative and commutative, and therefore a commutative semigroup.