Left and Right Inverses of Product

Theorem
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e_S$.

Let $x, y \in S$.

Let:
 * $(1): \quad x \circ y$ have a left inverse for $\circ$
 * $(2): \quad y \circ x$ have a right inverse for $\circ$.

Then both $x$ and $y$ are invertible for $\circ$.

Proof
Let $z_L$ be the left inverse of $x \circ y$ and $z_R$ be the right inverse of $y \circ x$. Then:

Thus $y$ has both a left inverse $z_L \circ x$ and a right inverse $x \circ z_R$.

From Left Inverse and Right Inverse is Inverse:
 * $z_L \circ x = x \circ z_R$

and $y$ has an inverse, that is, is invertible.

From the above, we have:

and:

Thus $x$ has both a left inverse $y \circ z_L$ and a right inverse $z_R \circ y$.

From Left Inverse and Right Inverse is Inverse:
 * $y \circ z_L = z_R \circ y$

and $x$ has an inverse, that is, is invertible.