# Construction of Inverse Completion/Quotient Structure is Commutative Semigroup

## 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 $\oplus$.

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:

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

where, by definition:

 $\displaystyle \forall \left({x_1, y_1}\right), \left({x_2, y_2}\right) \in S \times C:$  $\displaystyle q_\boxtimes \left({\left({x_1, y_1}\right) \oplus \left({x_2, y_2}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle q_\boxtimes \left({\left({x_1, y_1}\right)}\right) \oplus' q_\boxtimes \left({\left({x_2, y_2}\right)}\right)$

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.

$\blacksquare$