Construction of Inverse Completion/Quotient Mapping/Image of Cancellable Elements
Theorem
Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.
Let $\struct {C, \circ {\restriction_C} } \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.
Let $\struct {S \times C, \oplus}$ be the external direct product of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, 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:
- $\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
This cross-relation is a congruence relation on $S \times C$.
Let the quotient structure defined by $\boxtimes$ be:
- $\struct {T', \oplus'} := \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
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:
- $\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
Let $S'$ be the image $\psi \sqbrk S$ of $S$.
The set $C'$ of cancellable elements of the semigroup $S'$ is $\psi \sqbrk C$.
Proof
We have Morphism Property Preserves Cancellability.
Thus:
- $c \in C \implies \map \psi c \in C'$
So by Image of Subset under Mapping is Subset of Image:
- $\psi \sqbrk C \subseteq C'$
From above, $\psi$ is an isomorphism.
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Hence, also from Morphism Property Preserves Cancellability:
- $c' \in C' \implies \map {\psi^{-1} } {c'} \in C$
So by Preimage of Subset is Subset of Preimage:
- $\psi^{-1} \sqbrk {C'} \subseteq C$
Hence by definition of set equality:
- $\psi \sqbrk C = C'$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers