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: {{begin-eqn}} {{eqn | l = \map \psi {x \circ' y^{-1} } * \map \psi y       | r = \map \psi {x \circ' y^{-1} \circ' y}       | c = {{Defof|Morphism Property}} }} {{eqn | r = \map \psi x       | c = {{Defof|Inverse Element}} }} {{eqn | r = \map \phi x       | c = $\psi$ extends $\phi$ }} {{eqn | r = \map \phi x * \paren {\map \phi y}^{-1} * \map \phi y       | c = by hypothesis: $\map \phi y$ is invertible in $\struct {T, *}$ }} {{eqn | r = \map \phi x * \paren {\map \phi y}^{-1} * \map \psi y       | c = $\psi$ extends $\phi$ }} {{end-eqn}} So as $\map \psi y$ is cancellable for $*$: :$(1): \quad \map \psi {x \circ' y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$ 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 $\phi$. {{Qed}} === 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 $\phi$. {{Qed}} === 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: {{begin-eqn}} {{eqn | l = \map \psi {x \circ' y^{-1} }       | r = \map \psi {z \circ' w^{-1} }       | c = {{Defof|Morphism Property}} }} {{eqn | ll= \leadsto       | l = \map \psi x * \map \psi {y^{-1} }       | r = \map \psi z * \map \psi {w^{-1} }       | c =  }} {{eqn | ll= \leadsto       | l = \map \psi x * \map \psi w       | r = \map \psi z * \map \psi y       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi x * \map \phi w       | r = \map \phi z * \map \phi y       | c = $\psi$ is an extension of $\phi$ }} {{eqn | ll= \leadsto       | l = \map \phi {x \circ w}       | r = \map \phi {z \circ y}       | c = {{Defof|Morphism Property}} }} {{eqn | ll= \leadsto       | l = x \circ w       | r = z \circ y       | c = $\phi$ is a monomorphism }} {{eqn | ll= \leadsto       | l = x \circ' y^{-1}       | r = z \circ' w^{-1}       | c =  }} {{end-eqn}} Thus $\psi$ is a monomorphism if $\phi$ is. {{Qed}}