Inverse Element in Inverse Completion of Commutative Monoid

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Theorem

Let $\left({S, \circ}\right)$ be a commutative monoid.

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


Proof

Let the identity of $\left({S, \circ}\right)$ be $e$.

Let $z$ be the inverse of $y$ for $\circ$:

$z \circ y = e$
$y \circ z = e$

From Identity of Inverse Completion of Commutative Monoid:

$z \circ' y = e$
$y \circ' z = e$

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

$\blacksquare$


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