Construction of Inverse Completion

This page consists of a series of linked theorems, each of which builds towards one result.


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


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

= Commutative Semigroup by its Cancellable Elements is a Commutative Semigroup =

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 $C$.

That is:

$$\forall \left({x, y}\right), \left({u, v}\right) \in S \times C: \left({x, y}\right) \oplus \left({u, v}\right) = \left({x \circ u, y \circ \restriction_C v}\right)$$

Then $$\left({S \times C, \oplus}\right)$$ is a commutative semigroup.

Proof of Commutative Semigroup by its Cancellable Elements is a Commutative Semigroup
By Cancellable Elements of a Semigroup, $$\left({C, \circ \restriction_C}\right)$$ is a subsemigroup of $$\left({S, \circ}\right)$$, where $$\circ \restriction_C$$ is the restriction of $$\circ$$ to $$C$$.

By Restriction of Operation Commutativity, as $$\left({C, \circ \restriction_C}\right)$$ is a substructure of a commutative structure, it is also commutative.

From:


 * the external direct product preserves the nature of semigroups;
 * the external direct product preserves commutativity,

we see that $$\left({S \times C, \oplus}\right)$$ is a commutative semigroup.

= Equivalence Relation on Semigroup Product with Cancellable Elements =

The relation 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 an equivalence relation.

Reflexive
$$x_1 \circ y_1 = x_1 \circ y_1 \implies \left({x_1, y_1}\right) \mathcal{R} \left({x_1, y_1}\right)$$

Symmetric
$$ $$ $$ $$

Transitive
$$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$

Thus we define the equivalence class $$\left[\!\left[{\left({x, y}\right)}\right]\!\right]_{\mathcal{R}}$$.

= This Equivalence Relation is a Congruence =

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

Proof that this Equivalence Relation is a Congruence
From Definition of $\mathcal{R}$ above, we have that $$\mathcal{R}$$ is an equivalence on $$S \times C$$.

We now need to show that:

$$ $$

So:

$$ $$ $$ $$ $$ $$

Comment
In the context of the naturally ordered semigroup, the Unique Minus is defined:


 * $$n \ominus m = p \iff m \circ p = n$$

from which it can be seen that the above congruence can be understood as:


 * $$\left({x_1, y_1}\right) \mathcal{R} \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1 \iff x_1 \ominus y_1 = x_2 \ominus y_2$$

Thus this congruence defines an equivalence between pairs of elements which have the same Unique Minus (or, informally at this stage, "difference").

= Equivalence Class of Equal Elements =

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

Proof of Equivalence Class of Equal Elements
Note that as $$C \subseteq S$$, it is clear that $$C \times C \subseteq S \times C$$ from Cartesian Product of Subsets.

Thus we need only consider elements $$\left({x, y}\right)$$ of $$C \times C$$.

$$ $$

= Members of Equivalence Classes =

$$\forall x, y \in S, a, b \in C:$$


 * $$\left({x \circ a, a}\right) \mathcal{R} \left({y \circ b, b}\right) \iff x = y$$
 * $$\left[\!\left[{x \circ a, y \circ a}\right]\!\right]_{\mathcal{R}} = \left[\!\left[{x, y}\right]\!\right]_{\mathcal{R}}$$

Proof of Members of Equivalence Classes
$$ $$ $$

$$ $$ $$

= Quotient Structure =

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 $$\left({T', \oplus'}\right)$$.

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

Proof of Quotient Structure
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.

= 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.

Proof that Quotient Mapping is Injective
We have:

$$ $$ $$

= Quotient Mapping is Monomorphism =

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

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

So all we need to do now is 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.

= 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 $$S'$$.

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 Quotient Mapping is Injective above, $$\psi$$ is an injection.

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

From Quotient Mapping is Monomorphism 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.

= Image of Cancellable Elements in Quotient Mapping =

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

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, $$\psi \left({C}\right) = C'$$.

= Identity of Quotient Structure =

$$\forall c \in C$$, $$\left[\!\left[{\left({c, c}\right)}\right]\!\right]_{\mathcal{R}}$$ is the identity of $$T'$$.

We denote the identity of $$T'$$ as $$e_{T'}$$, as usual.

Proof of Identity of Quotient Structure
$$ $$ $$

= Invertible Elements in Quotient Structure =

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

Proof of Invertible Elements in Quotient Structure
From Identity of Quotient Structure above, $$\left({T', \oplus'}\right)$$ has an identity, and it is $$\left[\!\left[{\left({c, c}\right)}\right]\!\right]_{\mathcal{R}}$$ for any $$c \in C$$. Call this identity $$e_{T'}$$.


 * First we note that, from Image of Cancellable Elements in Quotient Mapping above, $$C' = \psi \left({C}\right)$$. So:

$$ $$ $$


 * The inverse of $$x'$$ is $$\left[\!\left[{\left({a, a \circ x}\right)}\right]\!\right]_{\mathcal{R}}$$, as follows:

$$ $$ $$ $$ $$

... thus showing that the inverse of $$\left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_{\mathcal{R}}$$ is $$\left[\!\left[{\left({a, a \circ x}\right)}\right]\!\right]_{\mathcal{R}}$$.

= Generator for Quotient Structure =

$$T' = S' \cup \left({C'}\right)^{-1}$$ is a generator for the semigroup $$T'$$.

Proof of Generator for Quotient Structure
Let $$\left({x, y}\right) \in S \times C$$. Then:

$$ $$ $$ $$

= Quotient Structure is Inverse Completion =

$$T'$$ is an inverse completion of its subsemigroup $$S'$$.

Proof that Quotient Structure is Inverse Completion

 * Every cancellable element of $$S'$$ is invertible in $$T'$$, from Invertible Elements in Quotient Structure above.


 * $$T' = S' \cup \left({C'}\right)^{-1}$$ is a generator for the semigroup $$T'$$, from Generator for Quotient Structure above.

= Notes =


 * Elements of $$T'$$ are equivalence classes of ordered pairs $$\left({x, y}\right)$$ where $$x \in S, y \in C$$.

Each of the elements of $$\left[\!\left[{\left({x, y}\right)}\right]\!\right]_{\mathcal{R}}$$ are such that $$x \circ y^{-1}$$ have the same value, where $$y^{-1} \in C^{-1}$$.

Hence is it a natural progression to define an operation $$\odot$$, say, such that:
 * $$x \odot y \equiv x \circ y^{-1}$$

In the context of the integers, this operation is "minus", hence:
 * $$x - y \equiv x + \left({-y}\right)$$

In the context of the rational numbers, this operation is "divided by", hence:
 * $$\frac x y \equiv x \times y^{-1}$$


 * Each element of $$S$$, and hence in $$C$$, is identified (via the isomorphism $$\psi$$) with one of these equivalence classes.

If $$S$$ is a monoid, then it has an identity $$e_S$$, say, which is in $$C$$.

Hence $$\forall x \in C: \psi \left({x}\right) = \left[\!\left[{\left({x, e_S}\right)}\right]\!\right]_{\mathcal{R}}$$.

In particular, $$\psi \left({e_S}\right) = \left[\!\left[{\left({e_S, e_S}\right)}\right]\!\right]_{\mathcal{R}}$$