Extension Theorem for Isomorphisms

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Theorem

Let the following conditions be fulfilled:

Let $\left({S, \circ}\right)$ be a commutative semigroup with cancellable elements
Let $\phi$ be an isomorphism from $\left({S, \circ}\right)$ into a semigroup $\left({T, *}\right)$
Let $\left({S', \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$
Let $\left({T', \circ'}\right)$ be an inverse completion of $\left({T, \circ}\right)$.


Then there is a unique isomorphism $\phi': S' \to T'$ extending $\phi$.


Proof

Let $C$ be the subsemigroup of cancellable elements of $S$.

It is proved that this is a semigroup by Cancellable Elements of Semigroup form Subsemigroup.


The set of cancellable elements of $T$ is $\phi \left({C}\right)$.

By the Extension Theorem for Homomorphisms, there is:

A unique homomorphism $\phi'$ from $S'$ into $T'$ extending $\phi$

and:

A unique homomorphism $\psi$ from $T'$ into $S'$ extending $\phi^{-1}$.


Therefore, $\psi \circ \phi'$ is an endomorphism of $S'$ whose restriction to $S$ is the inclusion mapping $\iota_S: S \to S'$:

$\forall x \in S: \iota \left({x}\right) = x$

From Inclusion Mapping is Monomorphism, $\iota_S$ is a monomorphism from $S$ into $S'$.


But by the Extension Theorem for Homomorphisms, the identity automorphism of $S'$ is the only endomorphism of $S'$ extending $\iota_S$ from $S$ into $S'$, so:

$\psi \circ \phi' = I_{S'}$


Similarly:

$\phi' \circ \psi = I_{T'}$


So by Bijection iff Left and Right Inverse, $\phi'$ is a bijection and therefore an isomorphism.

$\blacksquare$


Sources