# Extension Theorem for Homomorphisms

## Theorem

Let $\struct {S, \circ}$ be a commutative semigroup with cancellable elements

Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of all cancellable elements of $S$

Let $\struct {S', \circ'}$ be an inverse completion of $\struct {S, \circ}$

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

Then:

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

## Proof

It is proved that $\struct {C, \circ}$ is a subsemigroup of $\struct {S, \circ}$ 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 $\struct {S', \circ'}$ into $\struct {T, *}$ 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 $\map \phi y$ is invertible and hence cancellable for $*$ by Invertible Element of Monoid is Cancellable.

As $\psi$ is an extension of $\phi$, it follows that $\map \psi y$ is also cancellable for $*$.

Consider:

$\map \psi {x \circ' y^{-1} } * \map \psi y$

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

We have:

 $\ds \map \psi {x \circ' y^{-1} } * \map \psi y$ $=$ $\ds \map \psi {x \circ' y^{-1} \circ' y}$ Definition of Morphism Property $\ds$ $=$ $\ds \map \psi x$ Definition of Inverse Element $\ds$ $=$ $\ds \map \phi x$ $\psi$ extends $\phi$ $\ds$ $=$ $\ds \map \phi x * \paren {\map \phi y}^{-1} * \map \phi y$ by hypothesis: $\map \phi y$ is invertible in $\struct {T, *}$ $\ds$ $=$ $\ds \map \phi x * \paren {\map \phi y}^{-1} * \map \psi y$ $\psi$ extends $\phi$

So as $\map \psi y$ is cancellable for $*$:

$(1): \quad \map \psi {x \circ' y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$
$S' = S \circ' C^{-1}$

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

By Restriction of Mapping to Image is Surjection, $\phi: S \to \phi \sqbrk S$ is a surjection.

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

We have that $\struct {S, \circ}$ is a commutative semigroup.

By Epimorphism Preserves Semigroups and Epimorphism Preserves Commutativity, $\phi \sqbrk {\struct {S, \circ} }$ is also a commutative semigroup.

As $S \subseteq T$ it follows that $\phi \sqbrk {\struct {S, \circ} }$ is a subsemigroup of $\struct {T, *}$.

By Commutation with Inverse in Monoid, every element of $\phi \sqbrk S$ commutes with every element of $\paren {\phi \sqbrk C}^{-1}$.

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

$\forall x \in S, y \in C: \map \psi {x \circ' y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$

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

$\blacksquare$

### Proof of Monomorphism

Let $\phi$ be a monomorphism.

Then:

$\forall x, y \in S: \map \phi x = \map \phi y \implies x = y$

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

Then:

 $\ds \map \psi {x \circ' y^{-1} }$ $=$ $\ds \map \psi {z \circ' w^{-1} }$ Definition of Morphism Property $\ds \leadsto \ \$ $\ds \map \psi x * \map \psi {y^{-1} }$ $=$ $\ds \map \psi z * \map \psi {w^{-1} }$ $\ds \leadsto \ \$ $\ds \map \psi x * \map \psi w$ $=$ $\ds \map \psi z * \map \psi y$ $\ds \leadsto \ \$ $\ds \map \phi x * \map \phi w$ $=$ $\ds \map \phi z * \map \phi y$ $\psi$ is an extension of $\phi$ $\ds \leadsto \ \$ $\ds \map \phi {x \circ w}$ $=$ $\ds \map \phi {z \circ y}$ Definition of Morphism Property $\ds \leadsto \ \$ $\ds x \circ w$ $=$ $\ds z \circ y$ $\phi$ is a monomorphism $\ds \leadsto \ \$ $\ds x \circ' y^{-1}$ $=$ $\ds z \circ' w^{-1}$

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

$\blacksquare$