Identity Mapping is Order Isomorphism

Theorem
Let $\left({S, \preceq}\right)$ be a poset.

The identity mapping $I_S$ is an order isomorphism from $\left({S, \preceq}\right)$ to itself.

Proof
By definition: $\forall x \in S: I_S \left({x}\right) = x$.

So $x \preceq y \implies I_S \left({x}\right) \preceq I_S \left({y}\right)$.

As $I_S$ is a bijection, we also have $I_S^{-1} \left({x}\right) = x$.

So $x \preceq y \implies I_S^{-1} \left({x}\right) \preceq I_S^{-1} \left({y}\right)$.