Keith - yes, that's a nice observation! You can prove this in any category, not just the category of sets. If \\(r: y \to x\\) and \\(s: x \to y\\) are morphisms in any category, and \\(r \circ s = 1_x\\), then \\(s \circ r\\) is idempotent:

\\[

\begin{array}{cl}

(s\circ r)\circ (s\circ r) &= s\circ (r\circ s) \circ r\\\\

&= s\circ 1\_x\circ r \\\\

&= s\circ r. \\\\

\end{array}

\\]

It's fun to think about what this means in various categories.

\\[

\begin{array}{cl}

(s\circ r)\circ (s\circ r) &= s\circ (r\circ s) \circ r\\\\

&= s\circ 1\_x\circ r \\\\

&= s\circ r. \\\\

\end{array}

\\]

It's fun to think about what this means in various categories.