Construction of Inverse Completion/Quotient Structure is Commutative Semigroup

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Theorem

Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.

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


Let $\struct {S \times C, \oplus}$ be the external direct product of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, 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:

$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$

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


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

$\struct {T', \oplus'} := \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$

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


Then:

$\struct {T', \oplus'}$ is a commutative semigroup.


Proof

The quotient epimorphism from $\struct {S \times C, \oplus}$ onto $\struct {T', \oplus'}$ is given by:

$q_\boxtimes: \struct {S \times C, \oplus} \to \struct {T', \oplus'}: \map {q_\boxtimes} {x, y} = \eqclass {\tuple {x, y} } \boxtimes$

where, by definition:

\(\ds \forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times C: \, \) \(\ds \) \(\) \(\ds \map {q_\boxtimes} {\tuple {x_1, y_1} \oplus \tuple {x_2, y_2} }\)
\(\ds \) \(=\) \(\ds \map {q_\boxtimes} {x_1, y_1} \oplus' \map {q_\boxtimes} {x_2, y_2}\)


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 $\struct {T', \oplus'}$ is closed, associative and commutative, and therefore a commutative semigroup.

$\blacksquare$


Sources