Extension Theorem for Homomorphisms

Theorem

Let the following conditions be fulfilled:

• 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 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 $\phi$

$(2): \quad \psi \left({x \circ' y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$

$(3): \quad$ 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 a Semigroup.)

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 $\phi$.

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

So:

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

Hence:

	\(\displaystyle \)	\(\displaystyle \psi \left({x \circ' y^{-1} }\right) * \phi \left({y}\right)\)	\(=\)	\(\displaystyle \psi \left({x \circ' y^{-1} \circ' y}\right)\)	\(\displaystyle \)	Morphism Property
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \psi \left({x}\right)\)	\(\displaystyle \)	Inverse Element
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({x}\right)\)	\(\displaystyle \)	$\psi$ extends $\phi$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1} * \phi \left({y}\right)\)	\(\displaystyle \)	Inverse Element
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1} * \psi \left({y}\right)\)	\(\displaystyle \)	$\psi$ extends $\phi$

From the definition of inverse completion, we have $T = S \circ' C^{-1}$.

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

$\blacksquare$

Proof of Morphism Property

From the above, we saw in passing that $\psi \left({x \circ' y^{-1}}\right) * \phi \left({y}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1} * \phi \left({y}\right)$.

As (also from above) $\phi \left({y}\right)$ is cancellable for $*$, it follows that $\psi \left({x \circ' y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$.

$\blacksquare$

Proof of at least one such homomorphism

From Surjection iff Image equals Codomain, $\phi: S \to \phi \left({S}\right)$ is a surjection and therefore by definition is an epimorphism.

By Epimorphism Preserves Semigroups and Epimorphism Preserves Commutativity, as $\left({S, \circ}\right)$ is a commutative semigroup, then so is $\phi \left({S}\right)$, and is a subsemigroup of $\left({T, *}\right)$.

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

Thus it follows that the function $\psi: S' \to T$ 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:

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

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

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 20$: Theorem $20.4$