Identity of Inverse Completion of Commutative Monoid

Theorem
Let $\left({S, \circ}\right)$ be a commutative monoid whose identity is $e$.

Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$.

Let $\left({T, \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$.

Then:
 * $e \in T$ is the identity for $\circ'$;


 * The inverse of an element of $S$ which is invertible for $\circ$ is also its inverse for $\circ'$.

Proof
Let $e$ be the identity of $\circ$.

Let $e = x \circ' y^{-1}$, where $x \in S, y \in C$.

Then:

Thus $e = y^{-1} \circ' y$, and $y^{-1} \circ' y$ is the identity for $\circ'$.

Let $z \circ y = e$ and $y \circ z = e$ be the inverse of $y$ for $\circ$.

It follows directly from the construction that $z \circ' y = e$ and $y \circ' z = e$.

Hence $z$ is the inverse of $y$ for $\circ'$.