Order Isomorphism is Surjective Order Embedding

Theorem
Let $$\left({S, \preceq_1}\right)$$ and $$\left({T, \preceq_1}\right)$$ be posets.

Let $$f: S \to T$$ be a mapping.

Then $$f$$ is an order isomorphism iff:
 * $$f$$ is a surjection;
 * $$\forall x, y \in S: x \preceq_1 y \iff f \left({x}\right) \preceq_2 f \left({y}\right)$$

That is, iff $$f$$ is an order monomorphism which is also a surjection.

Necessary Condition
Suppose $$f$$ is an order isomorphism.

Then by definition $$f$$ is a bijection and so a surjection.

Also by definition, $$f$$ is increasing, and so:
 * $$\forall x, y \in S: x \preceq_1 y \implies f \left({x}\right) \preceq_2 f \left({y}\right)$$

Also by definition $$f^{-1}$$ is also a bijection which is increasing, and so:
 * $$\forall x, y \in S: f \left({x}\right) \preceq_2 f \left({y}\right) \implies x = f^{-1} \left({f \left({x}\right)}\right) \preceq_1 f^{-1} \left({f \left({y}\right)}\right) = y$$

and so:
 * $$\forall x, y \in S: x \preceq_1 y \iff f \left({x}\right) \preceq_2 f \left({y}\right)$$

Sufficient Condition
Suppose $$f: S \to T$$ is a mapping such that:
 * $$f$$ is a surjection;
 * $$\forall x, y \in S: x \preceq_1 y \iff f \left({x}\right) \preceq_2 f \left({y}\right)$$

From Order Monomorphism is Injection we have that $$f$$ is an injection.

As, by hypothesis, it is also surjective, it follows that it is a bijection.

Now, suppose $$a, b \in T$$.

As $$f$$ is surjective:
 * $$\exists x, y \in S: f \left({x}\right) = a, f \left({y}\right) = b$$

As $$f$$ is bijective, then:
 * $$x = f^{-1} \left({a}\right), y = f^{-1} \left({b}\right)$$

So by hypothesis:
 * $$a \preceq_2 b \implies f \left({x}\right) \preceq_2 f \left({y}\right) \implies f^{-1} \left({a}\right) = x \preceq_1 y = f^{-1} \left({b}\right)$$

Hence, by definition, $$f$$ is an order isomorphism.