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 $$\circ'$$.