Construction of Inverse Completion/Generator for Quotient Structure

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Theorem

Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.

Let $C \subseteq S$ be the set of cancellable elements of $S$.


Let $\struct {S \times C, \oplus}$ be the external direct product of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, where:

$\circ {\restriction_C}$ is the restriction of $\circ$ to $C \times C$

and:

$\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ \restriction_C$ on $C$.


Let $\boxtimes$ be the congruence relation $\boxtimes$ defined on $\struct {S \times C, \oplus}$ by:

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

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 $\oplus$.


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


Proof

Let $\tuple {x, y} \in S \times C$. Then:

\(\ds \map \psi x \oplus' \paren {\map \psi y}^{-1}\) \(=\) \(\ds \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {a, a \circ y} } \boxtimes\) Invertible Elements in Quotient Structure above
\(\ds \) \(=\) \(\ds \eqclass {\tuple {x \circ a \circ a, a \circ a \circ y} } \boxtimes\) Definition of $\oplus'$
\(\ds \) \(=\) \(\ds \eqclass {\tuple {x \circ a \circ a, y \circ a \circ a} } \boxtimes\) Commutativity of $\circ$
\(\ds \) \(=\) \(\ds \eqclass {\tuple {x, y} } \boxtimes\) Cancellability of $a \in C$

$\blacksquare$


Sources