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 $$\left({\frac {S \times C} {\mathcal R}, \oplus_{\mathcal R}}\right)$$

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

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

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

Thus $$\left({\frac {S \times C} {\mathcal R}, \oplus_{\mathcal R}}\right)$$ 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, $$\psi \left({C}\right) = C'$$.