Extension Theorem for Homomorphisms

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 $\phi$
 * $(2): \quad \psi \left({x \circ' y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$
 * $(3): \quad$ 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 $\phi$.

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

From Inverse Completion Commutative Semigroup, we have $S' = S \circ' C^{-1}$.

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

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

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 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 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 $\phi$.

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:

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