Extension Theorem for Distributive Operations/Existence and Uniqueness

Theorem
Then:
 * There exists a unique operation $\circ'$ on $T$ which distributes over $*$ in $T$ and induces on $R$ the operation $\circ$.

Proof
We have that all the elements of $\struct {R, *}$ are cancellable.

Thus Inverse Completion of Commutative Semigroup is Abelian Group can be applied.

So $\struct {T, *}$ is an abelian group.

Existence
For each $m \in R$, we define $\lambda_m: R \to T$ as:


 * $\forall x \in R: \map {\lambda_m} x = m \circ x$

Then:

So $\lambda_m$ is a homomorphism from $\struct {R, *}$ into $\struct {T, *}$.

Now, by the Extension Theorem for Homomorphisms, every homomorphism from $\struct {R, *}$ into $\struct {T, *}$ is the restriction to $R$ of a unique endomorphism of $\struct {T, *}$.

This can be applied because $\struct {T, *}$ is abelian.

We have just shown that $\lambda_m$ is such a homomorphism.

Therefore there exists a unique endomorphism $\lambda'_m: T \to T$ which extends $\lambda_m$.

Now:

By Homomorphism on Induced Structure, $\lambda'_m * \lambda'_n$ is an endomorphism of $\struct {T, *}$ that, as we have just seen, coincides on $R$ with $\lambda'_{m * n}$.

Hence $\lambda'_{m * n} = \lambda'_m * \lambda'_n$.

Similarly, for each $z \in T$, we define $\rho_z: R \to T$ as:


 * $\forall m \in R: \map {\rho_z} m = \map {\lambda'_m} z$

Then:

Therefore $\rho_z$ is a homomorphism from $\struct {R, *}$ into $\struct {T, *}$.

Consequently there exists a unique endomorphism $\rho'_z: T \to T$ extending $\rho_z$.

By Homomorphism on Induced Structure, $\rho'_y * \rho'_z$ is an endomorphism on $\struct {T, *}$ that coincides (as we have just seen) with $\rho'_{y * z}$ on $R$.

Hence $\rho'_{y * z} = \rho'_y * \rho'_z$.

Now we define an operation $\circ'$ on $T$ by:


 * $\forall x, y \in T: x \circ' y = \map {\rho'_y} x$

Now suppose $x, y \in R$. Then:

so $\circ'$ is an extension of $\circ$.

Next, let $x, y, z \in T$. Then:

So $\circ'$ is distributive over $*$.

Uniqueness
To show that $\circ'$ is unique, let $\circ_1$ be any operation on $T$ distributive over $*$ that induces $\circ$ on $R$.

Since $\circ'$ and $\circ_1$ both distribute over $*$, for every $m \in R$, the mappings:

are endomorphisms of $\struct {T, *}$ that coincide on $R$, so must be the same mapping.

Therefore:
 * $\forall m \in R, y \in T: m \circ_1 y = m \circ' y$

Similarly, for every $y \in T$, the mappings:

are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have just proved.

So these two mappings must be the same mapping.

Hence:
 * $\forall x, y \in T: x \circ_1 y = x \circ' y$

Thus $\circ'$ is the only operation on $T$ which extends $\circ$ and distributes over $*$.