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 $$S$$.

= Theorem 1 =


 * Let $$\left({S \times C, \oplus}\right)$$ be the external direct product of $$\left({S, \circ}\right)$$ and $$\left({C, \circ|_C}\right)$$, where $$\oplus$$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ|_C$ on $C$.

That is:

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

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

Proof 1
By Cancellable Elements of a Semigroup, $$\left({C, \circ|_C}\right)$$ is a subsemigroup of $$\left({S, \circ}\right)$$, where $$\circ|_C$$ is the restriction of $$\circ$$ to $$C$$.

By Restriction of Operation Commutativity, as $$\left({C, \circ|_C}\right)$$ 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.

= Theorem 2 =

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

Transitive
= Theorem 3 =

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

Proof 3
From Theorem 2 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:

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

Thus this congruence defines an equivalence between pairs of elements which have the same Unique Minus.

= Theorem 4 =

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

Proof 4
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 $$C \times C$$.

= Theorem 5 =

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

Proof 5
= Theorem 6 =

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 $\left({\frac {S \times C} \mathcal{R}, \oplus_{\mathcal{R}}}\right)$ 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 6
The canonical epimorphism from $$\left({S \times C, \oplus}\right)$$ onto $$\left({T', \oplus'}\right)$$ is given by:

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

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.

= Theorem 7 =

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 7
We have:

= Theorem 8 =

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

Proof 8
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.

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.

= Theorem 9 =

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 9

 * $$\left({S', \oplus'}\right)$$ 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 Range $$\psi$$ is a surjection.

From Theorem 7 above, $$\psi$$ is an injection.

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

From Theorem 8 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.

= Theorem 10 =

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

Proof 10
Homomorphism conserves cancellability. Thus:

= Theorem 11 =

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

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

Proof 11
= Theorem 12 =

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

Proof 12
From Theorem 11 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 Theorem 10 above, $$C' = \psi \left({C}\right)$$. So:


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

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

= Theorem 13 =

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

Proof 13
= Theorem 14 =

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

Proof 14

 * Every cancellable element of $$S'$$ is invertible in $$T'$$, from Theorem 12 above.


 * $$T' = S' \cup \left({C'}\right)^{-1}$$ is a generator for the semigroup $$T'$$ is a generator for the semigroup $$T'$$, from Theorem 13 above.