# Order Isomorphism iff Strictly Increasing Surjection

## Theorem

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be totally ordered sets.

A mapping $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ is an order isomorphism if and only if:

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

## Proof

### Necessary Condition

Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ 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:

- $x \mathop {\prec_1} y \implies \map \phi x \mathop {\preceq_2} \map \phi y$

as $\phi$ is an order isomorphism.

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

Thus:

- $\map \phi x = \map \phi y \implies x = y$

and so it follows that:

- $x \mathop {\prec_1} y \implies \map \phi x \mathop {\prec_2} \map \phi y$

$\Box$

### Sufficient Condition

Suppose $\phi$ is a mapping which satisfies the conditions:

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

From $(2)$ and Strictly Increasing Mapping is Increasing we have:

- $x \mathop {\preceq_1} y \implies \map \phi x \mathop {\preceq_2} \map \phi y$

Now suppose $\map \phi x \mathop {\preceq_2} \map \phi y$.

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

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

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:

- $x \mathop {\preceq_1} y \iff \map \phi x \mathop {\preceq_2} \map \phi y$

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

$\blacksquare$

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 7$: Theorem $7.2$