Construction of Inverse Completion/Quotient Structure is Inverse Completion

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Theorem

Let $\left({S, \circ}\right)$ be a commutative semigroup which has cancellable elements.

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


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:

$\circ {\restriction_C}$ 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 $\left({S \times C, \oplus}\right)$ by:

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

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 $\oplus$.


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


Proof

Every cancellable element of $S'$ is invertible in $T'$, from Invertible Elements in Quotient Structure.

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

Hence the result, by definition of inverse completion

$\blacksquare$


Sources