Construction of Inverse Completion/Quotient Structure

Theorem
Let $\left({S, \circ}\right)$ be a commutative semigroup which has cancellable elements.

Let $C \subseteq S$ be the set of cancellable elements of $S$.

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:
 * $\circ \restriction_C$ is the restriction of $\circ$ to $C \times C$, and
 * $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ \restriction_C$ on $C$.

Let $\mathcal R$ be the relation $\mathcal R$ defined on $S \times C$ by:
 * $\left({x_1, y_1}\right) \mathcal R \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

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

Let the quotient structure defined by $\mathcal R$ be $\displaystyle \left({\frac {S \times C} {\mathcal R}, \oplus_\mathcal R}\right)$

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

Let us use $T'$ to denote the quotient set $\displaystyle \frac {S \times C} {\mathcal R}$.

Let us use $\oplus'$ to denote the operation $\oplus_\mathcal R$.

Thus $\displaystyle \left({\frac {S \times C} {\mathcal R}, \oplus_\mathcal R}\right)$ is now denoted $\left({T', \oplus'}\right)$.

Quotient Mapping 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:


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

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.

Quotient Mapping to Image is Isomorphism
Let $S'$ be the image $\psi \left({S}\right)$ of $S$.

Then:
 * $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$;
 * $\psi$ 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 Equivalence Relation on Semigroup Product with Cancellable Elements we have that:
 * $\left({x_1, y_1}\right) \mathcal R \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

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

We also have, from the same source, that


 * $(1) \quad \forall x, y \in S, a, b \in C: \left({x \circ a, a}\right) \mathcal R \left({y \circ b, b}\right) \iff x = y$


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


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

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

Proof that Quotient Mapping is Commutative Semigroup
The canonical epimorphism from $\left({S \times C, \oplus}\right)$ onto $\left({T', \oplus'}\right)$ is given by:

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

where, by definition:

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.

Proof that Quotient Mapping is Injective
We have:

Proof that Quotient Mapping is Monomorphism
From above, $\psi: S \to T'$ is an injection.

We now need to show that $\psi: S \to T'$ is a homomorphism.

Let $x, y \in S$. Then:

Thus we see that $\psi \left({x \circ y}\right) = \psi \left({x}\right) \oplus' \psi \left({y}\right)$, and the morphism property is proven.

Proof that Quotient Mapping to Image is Isomorphism

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

We have that $S'$ is the image $\psi \left({S}\right)$ of $S$.

For $\left({S', \oplus'}\right)$ to be a subsemigroup of $\left({T', \oplus'}\right)$, by Subsemigroup Closure Test we need to show that $\left({S', \oplus'}\right)$ is closed.

Let $x, y \in S'$.

Then $x = \phi \left({x'}\right), y = \phi \left({y'}\right)$ for some $x', y' \in S$.

But as $\phi$ is an isomorphism, it obeys the morphism property.

So $x \oplus' y = \phi \left({x'}\right) \oplus' \phi \left({y'}\right) = \phi \left({x' \circ y'}\right)$.

Hence $x \oplus' y$ is the image of $x' \circ y' \in S$ and hence $x \oplus' y \in S'$.

Thus by the Subsemigroup Closure Test, $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$


 * $\psi$ is an isomorphism from $S$ onto $S'$:

Because $S'$ is the image of $\psi \left({S}\right)$, by Surjection by Restriction of Codomain $\psi$ is a surjection.

From above, $\psi$ is an injection.

Therefore $\psi: S \to S'$ is a bijection.

From above, $\psi: \left({S, \circ}\right) \to \left({S', \oplus'}\right)$ is a monomorphism, therefore by definition a homomorphism.

A bijective homomorphism is an isomorphism.

Proof of Image of Cancellable Elements in Quotient Mapping
Homomorphism conserves cancellability.

Thus $c \in C \implies \psi \left({c}\right) \in C'$.

So by Subset of Image, $\psi \left({C}\right) \subseteq C'$.

From above, $\psi$ is an isomorphism.

Hence $c' \in C' \implies \psi^{-1} \left({c'}\right) \in C$, also because Homomorphism conserves cancellability.

So by Subset of Image, $\psi^{-1} \left({C'}\right) \subseteq C$.

Hence by definition of set equality, $\psi \left({C}\right) = C'$.