# Construction of Inverse Completion

## Theorem

To access the proofs for the individual theorems, click on the links which form the titles of each major section.

## Initial Definitions

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 $C$.

## Cartesian Product with Cancellable Elements

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 $C$.

That is:

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

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

## Congruence Relation

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

### Members of Equivalence Classes

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

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

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

### Equivalence Class of Equal Elements

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

### Natural Number Difference

In the context of the natural numbers, the difference is defined as:

$n - m = p \iff m + p = n$

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

$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1 \iff x_1 - y_1 = x_2 - y_2$

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

## Quotient Structure

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 $\psi \left[{C}\right]$.

## Properties of Quotient Structure

### Identity of Quotient Structure

Let $c \in C$ be arbitrary.

Then:

$\left[\!\left[{\left({c, c}\right)}\right]\!\right]_\boxtimes$

is the identity of $T'$.

### Invertible Elements in Quotient Structure

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

### Generator for Quotient Structure

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

### Quotient Structure is Inverse Completion

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

## 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 $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$ 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 division, hence:

$\dfrac x y \equiv x \times y^{-1}$

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:

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

In particular:

$\psi \left({e_S}\right) = \left[\!\left[{\left({e_S, e_S}\right)}\right]\!\right]_\boxtimes$