Construction of Inverse Completion/Quotient Mapping to Image is Isomorphism

From ProofWiki
Jump to navigation Jump to search

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:

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


Let $S'$ be the image $\psi \left({S}\right)$ of $S$.

Then $\psi$ is an isomorphism from $S$ onto $S'$.


Proof

From Quotient Mapping is Monomorphism, $\psi: \left({S, \circ}\right) \to \left({S', \oplus'}\right)$ is a monomorphism.

Therefore by definition:

$\psi$ is a homomorphism
$\psi$ is an injection.

Because $S'$ is the image of $\psi \left({S}\right)$, by Surjection by Restriction of Codomain $\psi$ is a surjection.

Therefore by definition $\psi: S \to S'$ is a bijection.

A bijective homomorphism is an isomorphism.

$\blacksquare$


Sources