Order Isomorphism iff Strictly Increasing Surjection

Theorem
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be totally ordered sets.

A mapping $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is an order isomorphism iff:


 * $(1): \quad \phi$ is a surjection
 * $(2): \quad \forall x, y \in S: x \mathop{\prec_1} y \implies \phi \left({x}\right) \mathop{\prec_2} \phi \left({y}\right)$

Necessary Condition
Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be an order isomorphism.

Then by definition $\phi$ is a bijection and so a surjection.

Suppose $x \mathop{\prec_1} y$.

That is:
 * $x \mathop{\preceq_1} y$
 * $x \ne y$

Then:
 * $x \mathop{\prec_1} y \implies \phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$

as $\phi$ is an order isomorphism.

But as $\phi$ is a bijection it is also an injection.

Thus:
 * $\phi \left({x}\right) = \phi \left({y}\right) \implies x = y$

and so it follows that:
 * $x \mathop{\prec_1} y \implies \phi \left({x}\right) \mathop{\prec_2} \phi \left({y}\right)$

Sufficient Condition
Suppose $\phi$ is a mapping which satisfies the conditions:
 * $(1): \quad \phi$ is a surjection
 * $(2): \quad \forall x, y \in S: x \mathop{\prec_1} y \implies \phi \left({x}\right) \mathop{\prec_2} \phi \left({y}\right)$

From $(2)$ and Strictly Increasing Mapping is Increasing we have:
 * $x \mathop{\preceq_1} y \implies \phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$

Now suppose $\phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$.

Suppose $y \mathop{\prec_1} x$.

Then from $(2)$ it would follow that $\phi \left({y}\right) \mathop{\prec_2} \phi \left({x}\right)$.

So it is not the case that $y \mathop{\prec_1} x$.

So from the Trichotomy Law $x \mathop{\preceq_1} y$.

Thus it follows that:
 * $x \mathop{\preceq_1} y \iff \phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$

It follows from Order Isomorphism is Surjective Order Monomorphism that $\phi$ is an order isomorphism.