# Construction of Inverse Completion/Quotient Mapping is Injective

## Theorem

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 $C$.

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 $C$.

Let $\boxtimes$ be the cross-relation on $S \times C$, defined as:

$\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

This cross-relation is a congruence relation on $S \times C$.

Let the quotient structure defined by $\boxtimes$ be:

$\left({T', \oplus'}\right) := \left({\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}\right)$

where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.

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.

## Proof

 $\displaystyle \psi \left({x}\right)$ $=$ $\displaystyle \psi \left({y}\right)$ $\displaystyle \implies \ \$ $\displaystyle \forall a \in C: \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes$ $=$ $\displaystyle \left[\!\left[{\left({y \circ a, a}\right)}\right]\!\right]_\boxtimes$ Definition of $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$ $\displaystyle \implies \ \$ $\displaystyle x$ $=$ $\displaystyle y$ from $(1)$ above

The result follows by the definition of injection.

$\blacksquare$