Right Inverse for All is Left Inverse

Theorem

Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$ such that:

$\forall x \in S: \exists x_R: x \circ x_R = e_R$

That is, every element of $S$ has a right inverse with respect to the right identity.

Then $x_R \circ x = e_R$, that is, $x_R$ is also a left inverse with respect to the right identity.

Let $y = x_R \circ x$. Then:

$\ds y \circ e_R$

$=$

$\ds y \circ \paren {y \circ y_R}$

$\ds $

$=$

$\ds \paren {y \circ y} \circ y_R$

$\ds $

$=$

$\ds y \circ y_R$

$\ds $

$=$

$\ds e_R$

So $x_R \circ x = e_R$, and $x_R$ behaves as a left inverse as well as a right inverse with respect to the right identity.

$\blacksquare$