## If $\sigma$ is a reverse mapping such that $\sigma^2\geq \operatorname{id}$, then $\sigma^3=\sigma$.

Let $S$, $T$ be two partial order sets, and suppose $\sigma:S\to T$ and $\tau:T\to S$ are two maps such that \begin{eqnarray*} \tau\circ\sigma(s)\geq s, \forall s\in S\\ \sigma\circ\tau(t)\geq t, \forall t\in T. \end{eqnarray*} If $\sigma(s_1)\leq \sigma(s_2)\forall s_1\geq s_2$, then $\sigma\circ\tau\circ\sigma=\sigma$.