Extension Theorem for Homomorphisms

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Theorem

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

Let $\left({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of all cancellable elements of $S$

Let $\left({S', \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$

Let $\phi$ be a (semigroup) homomorphism from $\left({S, \circ}\right)$ into a semigroup $\left({T, *}\right)$ such that $\phi \left({y}\right)$ is invertible for all $y \in C$.


Then:

$(1): \quad$ There is one and only one homomorphism $\psi$ from $\left({S', \circ'}\right)$ into $\left({T, *}\right)$ extending $\phi$
$(2): \quad \forall x \in S, y \in C: \psi \left({x \circ' y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$
$(3): \quad$ If $\phi$ is a monomorphism, then so is $\psi$.


Proof

It is proved that $\left({C, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$ by Cancellable Elements of Semigroup form Subsemigroup.


Proof of at most one such homomorphism

To show there is at most one such homomorphism:

Let $\psi$ be a homomorphism from $\left({S', \circ'}\right)$ into $\left({T, *}\right)$ extending $\phi$.

It will be shown that for each element in the domain of $\psi$ , its image is uniquely determined by $\phi$.

It will therefore follow that $\psi$ is the only such extension of $\phi$.


Now $\phi \left({y}\right)$ is invertible and hence cancellable for $*$ by Invertible Element of Monoid is Cancellable.

As $\psi$ is an extension of $\phi$, it follows that $\psi \left({y}\right)$ is also cancellable for $*$.

Consider:

$\psi \left({x \circ' y^{-1} }\right) * \psi \left({y}\right)$

for arbitrary $x \in S, y \in C$.

We have:

\(\displaystyle \psi \left({x \circ' y^{-1} }\right) * \psi \left({y}\right)\) \(=\) \(\displaystyle \psi \left({x \circ' y^{-1} \circ' y}\right)\) $\quad$ Morphism Property $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \psi \left({x}\right)\) $\quad$ Inverse Element $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({x}\right)\) $\quad$ $\psi$ extends $\phi$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1} * \phi \left({y}\right)\) $\quad$ by hypothesis: $\phi \left({y}\right)$ is invertible in $\left({T, *}\right)$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1} * \psi \left({y}\right)\) $\quad$ $\psi$ extends $\phi$ $\quad$


So as $\psi \left({y}\right)$ is cancellable for $*$:

$(1): \quad \psi \left({x \circ' y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$


From Inverse Completion is Commutative Semigroup:

$S' = S \circ' C^{-1}$

thus $(1)$ holds for all $x \in S, y \in C$.

So, there is at most one homomorphism extending $\phi$.

$\blacksquare$


Proof of at least one such homomorphism

From Surjection iff Image equals Codomain, $\phi: S \to \phi \left({S}\right)$ is a surjection.

Therefore, by definition, $\phi$ is an epimorphism.

We have that $\left({S, \circ}\right)$ is a commutative semigroup.

By Epimorphism Preserves Semigroups and Epimorphism Preserves Commutativity, $\phi \left({S, \circ}\right)$ is also a commutative semigroup.

As $S \subseteq T$ it follows that $\phi \left({S, \circ}\right)$ is a subsemigroup of $\left({T, *}\right)$.

By Commutation with Inverse in Monoid, every element of $\phi \left({S}\right)$ commutes with every element of $\left({\phi\left({C}\right)}\right)^{-1}$.


Thus it follows that the mapping $\psi: S' \to T$ defined by:

$\forall x \in S, y \in C: \psi \left({x \circ' y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$

is a well-defined homomorphism extending $\phi$.

$\blacksquare$


Proof of Monomorphism

Let $\phi$ be a monomorphism.

Then:

$\forall x, y \in S: \phi \left({x}\right) = \phi \left({y}\right) \implies x = y$


Now let $x \circ y^{-1}, z \circ w^{-1} \in S'$ such that $\psi \left({x \circ y^{-1}}\right) = \psi \left({z \circ w^{-1}}\right)$.


Then:

\(\displaystyle \psi \left({x \circ' y^{-1} }\right)\) \(=\) \(\displaystyle \psi \left({z \circ' w^{-1} }\right)\) $\quad$ Morphism Property $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \psi \left({x}\right) * \psi \left({y^{-1} }\right)\) \(=\) \(\displaystyle \psi \left({z}\right) * \psi \left({w^{-1} }\right)\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \psi \left({x}\right) * \psi \left({w}\right)\) \(=\) \(\displaystyle \psi \left({z}\right) * \psi \left({y}\right)\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \phi \left({x}\right) * \phi \left({w}\right)\) \(=\) \(\displaystyle \phi \left({z}\right) * \phi \left({y}\right)\) $\quad$ $\psi$ is an extension of $\phi$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \phi \left({x \circ w}\right)\) \(=\) \(\displaystyle \phi \left({z \circ y}\right)\) $\quad$ Morphism Property $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle x \circ w\) \(=\) \(\displaystyle z \circ y\) $\quad$ $\phi$ is a monomorphism $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle x \circ' y^{-1}\) \(=\) \(\displaystyle z \circ' w^{-1}\) $\quad$ $\quad$

Thus $\psi$ is a monomorphism if $\phi$ is.

$\blacksquare$


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