# Construction of Inverse Completion/Quotient Mapping is Monomorphism

## 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:

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

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

## Proof

We have that this quotient mapping $\psi: S \to T'$ is an injection.

Let $x, y \in S$. Then:

 $\displaystyle \psi \left({x}\right) \oplus' \psi \left({y}\right)$ $=$ $\displaystyle \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes \oplus' \left[\!\left[{\left({y \circ a, a}\right)}\right]\!\right]_\boxtimes$ Definition of $\psi$ $\displaystyle$ $=$ $\displaystyle \left[\!\left[{\left({x \circ a, a}\right) \oplus' \left({y \circ a, a}\right)}\right]\!\right]_\boxtimes$ Definition of $\oplus'$ $\displaystyle$ $=$ $\displaystyle \left[\!\left[{\left({x \circ a \circ y \circ a, a \circ a}\right)}\right]\!\right]_\boxtimes$ Definition of $\boxtimes$ $\displaystyle$ $=$ $\displaystyle \left[\!\left[{\left({\left({x \circ y}\right) \circ \left({a \circ a}\right), a \circ a}\right)}\right]\!\right]_\boxtimes$ Commutativity of $\circ$ $\displaystyle$ $=$ $\displaystyle \psi \left({x \circ y}\right)$ as $a \circ a \in C$

So $\psi \left({x \circ y}\right) = \psi \left({x}\right) \oplus' \psi \left({y}\right)$, and the morphism property is proven.

Thus $\psi$ is an injective homomorphism, and so by definition a monomorphism.

$\blacksquare$