Identity of Inverse Completion of Commutative Monoid

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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 $\circ'$.


Proof

Let $e$ be the identity for $\circ$.

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


Then:

\(\displaystyle y\) \(=\) \(\displaystyle e \circ y\) Definition of Identity
\(\displaystyle \) \(=\) \(\displaystyle \left({x \circ' y^{-1} }\right) \circ' y\) Definition of Inverse
\(\displaystyle \) \(=\) \(\displaystyle x \circ' \left({y^{-1} \circ' y}\right)\) $\circ'$ is associative
\(\displaystyle \) \(=\) \(\displaystyle x \circ' e\) Definition of Inverse


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

$\blacksquare$


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