Construction of Inverse Completion/Quotient Structure

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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:

$\displaystyle \left({T', \oplus'}\right) := \left({\frac {S \times C} \boxtimes, \oplus_\boxtimes}\right)$

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


Quotient Structure is Commutative Semigroup

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


Quotient Mapping is Injective

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

$\forall x \in S: \psi \left({x}\right) = \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes$


Then $\psi: S \to T'$ is an injection, and does not depend on the particular element $a$ chosen.


Quotient Mapping is Monomorphism

The mapping $\psi: S \to T'$ is a monomorphism.


Image of Quotient Mapping is Subsemigroup

Let $S'$ be the image $\psi \left({S}\right)$ of $S$.

Then $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$.


Quotient Mapping to Image is Isomorphism

Let $S'$ be the image $\psi \left({S}\right)$ of $S$.

Then $\psi$ is an isomorphism from $S$ onto $S'$.


Image of Cancellable Elements in Quotient Mapping

The set $C'$ of cancellable elements of the semigroup $S'$ is $\psi \left[{C}\right]$.


Proof

From the defined equivalence relation, we have that:

$\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

is a congruence relation on $\left({S \times C, \oplus}\right)$.


From the definition of the members of the equivalence classes:

$(1) \quad \forall x, y \in S, a, b \in C: \left({x \circ a, a}\right) \boxtimes \left({y \circ b, b}\right) \iff x = y$
$(2) \quad \forall x, y \in S, a, b \in C: \left[\!\left[{\left({x \circ a, y \circ a}\right)}\right]\!\right]_\boxtimes = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$


From the definition of the equivalence class of equal elements:

$(3) \quad \forall c, d \in C: \left({c, c}\right) \boxtimes \left({d, d}\right)$

where $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$ is the equivalence class of $\left({x, y}\right)$ under $\boxtimes$.

Hence we are justified in asserting the existence of the quotient structure:

$\displaystyle \left({T', \oplus'}\right) = \left({\frac {S \times C} \boxtimes, \oplus_\boxtimes}\right)$

$\blacksquare$


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