Let \\(r: y \to x\\) be a right inverse, and \\(s: x \to y\\) its left inverse, ie

\\[ (r \circ s)(x) = x \\]

then composition in the opposite direction produces an idempotent function,

\\[ (s\circ r)\circ (s\circ r)(y) = (s\circ r)(y) \\]

**Proof:**

\\[
\begin{align}
(s\circ r)\circ (s\circ r)(y) \\\\
= (s\circ (r\circ s)\circ r)(y) \\\\
= (s\circ (1\_x)\circ r)(y) \\\\
= (s\circ r)(y). \\\\
\end{align}
\\]