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) $$x \circ y$$ have a left inverse for $$\circ$$;
 * 2) $$y \circ x$$ 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, i.e. is invertible.


 * From the above, we have:

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, i.e. is invertible.