# Construction of Inverse Completion/Invertible Elements in 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 $\oplus$.

Every cancellable element of $S'$ is invertible in $T'$.

## Proof

From Identity of Quotient Structure, $\struct {T', \oplus'}$ has an identity, and it is $\eqclass {\tuple {c, c} } \boxtimes$ for any $c \in C$.

Call this identity $e_{T'}$.

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

$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
$C' = \psi \sqbrk C$

So:

 $\ds x'$ $\in$ $\ds C'$ $\ds \leadsto \ \$ $\ds \exists x \in C: \,$ $\ds x'$ $=$ $\ds \map \psi x$ as $\psi$ is a surjection $\ds \leadsto \ \$ $\ds \forall a \in C: \,$ $\ds x'$ $=$ $\ds \eqclass {\tuple {x \circ a, a} } \boxtimes$ Definition of $\psi$

The inverse of $x'$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$, as follows:

 $\ds a \circ x$ $\in$ $\ds C$ Cancellable Elements of Semigroup form Subsemigroup $\ds a \circ a \circ x$ $\in$ $\ds C$ Cancellable Elements of Semigroup form Subsemigroup $\ds \leadsto \ \$ $\ds \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {a, a \circ x} } \boxtimes$ $=$ $\ds \eqclass {\tuple {x \circ a \circ a, a \circ a \circ x} } \boxtimes$ Definition of $\oplus'$ $\ds$ $=$ $\ds \eqclass {\tuple {a \circ a \circ x, a \circ a \circ x} } \boxtimes$ Commutativity of $\circ$ $\ds$ $=$ $\ds e_{T'}$ Identity of Quotient Structure

thus showing that the inverse of $\eqclass {\tuple {x \circ a, a} } \boxtimes$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$.

$\blacksquare$