# Construction of Inverse Completion/Identity of Quotient Structure

## 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 $\oplus$.

Let $c \in C$ be arbitrary.

Then:

$\eqclass {\tuple {c, c} } \boxtimes$

is the identity of $T'$.

## Proof

 $\ds \paren {x \circ c} \circ y$ $=$ $\ds x \circ \paren {c \circ y}$ Semigroup Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds x \circ \paren {y \circ c}$ Definition of Commutative Semigroup $\ds \leadsto \ \$ $\ds \eqclass {\tuple {x, y} } \boxtimes \oplus' \eqclass {\tuple {c, c} } \boxtimes$ $=$ $\ds \eqclass {\tuple {x \circ c, y \circ c} } \boxtimes$ Definition of $\oplus'$ $\ds$ $=$ $\ds \eqclass {\tuple {x, y} } \boxtimes$ Cancellability of elements of $C$

Hence the result, by definition of identity element.

$\blacksquare$