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) There is one and only one homomorphism $$\psi$$ from $$\left({S', \circ'}\right)$$ into $$\left({T, *}\right)$$ extending $$\phi$$;
 * 2) $$\psi \left({x \circ' y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$$;
 * 3) 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:

$$ $$ $$ $$ $$

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

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, 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.