Order Isomorphism between Wosets is Unique

Theorem
Let $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ be wosets.

Let $\left({S_1, \preceq_1}\right) \cong \left({S_2, \preceq_2}\right)$, i.e. let $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ be order isomorphic.

Then there is exactly one mapping $f: S_1 \to S_2$ such that $f$ is an order isomorphism.

Proof
Let $f: S_1 \to S_2$ and $g: S_1 \to S_2$ both be order isomorphisms.

By Inverse of Order Isomorphism, the inverse $f^{-1}$ is also an order isomorphism.

Let $h = f^{-1} \circ g$ be the composition of $f^{-1}$ and $g$, which, by Composite of Order Isomorphisms is Order Isomorphism, is itself an order isomorphism.

So, by Order Isomorphism from Woset onto Subset, $\forall x \in S_1: x \preceq_1 h \left({x}\right)$.

Now we apply $f$ to $S_1$ and see that:
 * $\forall x \in S_1: f \left({x}\right) \preceq_2 f \left({h \left({x}\right)}\right) = g \left({x}\right)$.

In a similar way we can show that $\forall x \in S_1: g \left({x}\right) \preceq_2 f \left({x}\right)$.

Hence $f = g$ and the proof is complete.