Left-invertible elements of monoid form submonoid

Statement

Let $M$ be a monoid with neutral element $e$ . Then, the set of left-invertible elements of $M$ form a submonoid of $M$ .

Inverse map is involutive: Specifically the reversal law for inverses.

Given: A monoid $M$ with neutral element $e$

To prove: $e$ is left-invertible, and further, if $a_{1},a_{2}$ are left-invertible elements of $M$ , so is $a_{1}*a_{2}$

Proof: Clearly, $e$ is left-invertible, since $e*e=e$ .

Suppose $a_{1},a_{2}$ are left-invertible and $b_{1},b_{2}$ are left inverses for $a_{1},a_{2}$ . Then consider:

$(b_{2}*b_{1})*(a_{1}*a_{2})=((b_{2}*b_{1})*a_{1})*a_{2}=(b_{2}*(b_{1}*a_{1}))*a_{2}=(b_{2}*e)*a_{2}=b_{2}*a_{2}=e$

Thus, $b_{2}*b_{1}$ is a left inverse for $a_{1}*a_{2}$ , so $a_{1}*a_{2}$ is left-invertible, completing the proof.