Extension Theorem for Isomorphisms

Theorem
Let the following conditions be fulfilled:


 * Let $$\left({S, \circ}\right)$$ be a commutative semigroup with cancellable elements;
 * Let $$\phi$$ be an isomorphism from $$\left({S, \circ}\right)$$ into a semigroup $$\left({T, *}\right)$$;
 * Let $$\left({S', \circ'}\right)$$ be an inverse completion of $$\left({S, \circ}\right)$$;
 * Let $$\left({T', \circ'}\right)$$ be an inverse completion of $$\left({T, \circ}\right)$$.

Then there is a unique isomorphism $$\phi': S' \to T'$$ extending $$\phi$$.

Proof
Let $$C$$ be the subsemigroup of cancellable elements of $$S$$.

It is proved that this is a semigroup by Cancellable Elements of a Semigroup.

The set of cancellable elements of $$T$$ is $$\phi \left({C}\right)$$.

By the Extension Theorem for Homomorphisms, there is:


 * A unique homomorphism $$\phi'$$ from $$S'$$ into $$T'$$ extending $$\phi$$, and
 * A unique homomorphism $$\psi$$ from $$T'$$ into $$S'$$ extending $$\phi^{-1}$$.

Therefore, $$\psi \circ \phi'$$ is an endomorphism of $$S'$$ whose restriction to $$S$$ is the identity monomorphism from $$S$$ into $$S'$$.

But by the Extension Theorem for Homomorphisms, the identity automorphism of $$S'$$ is the only endomorphism of $$S'$$ extending the identity monomorphism from $$S$$ into $$S'$$, so $$\psi \circ \phi' = I_{S'}$$.

Similarly, $$\phi' \circ \psi = I_{T'}$$.

Therefore, by Two-Sided Inverse, $$\phi'$$ is a bijection and therefore an isomorphism.