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 itself an order isomorphism.

So, by Order Isomorphism 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.