# Order Isomorphism is Surjective Order Embedding

## Theorem

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

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

Then $f$ is an order isomorphism if and only if:

- $(1): \quad f$ is a surjection
- $(2): \quad \forall x, y \in S: x \preceq_1 y \iff \map f x \preceq_2 \map f y$

That is, if and only if $f$ is an order embedding which is also a surjection.

## Proof

### 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:

- $\forall x, y \in S: x \preceq_1 y \implies \map f x \preceq_2 \map f y$

Also by definition $f^{-1}$ is also a bijection which is increasing, and so:

- $\forall x, y \in S: \map f x \preceq_2 \map f y \implies x = \map {f^{-1} } {\map f x} \preceq_1 \map {f^{-1} } {\map f y} = y$

and so:

- $\forall x, y \in S: x \preceq_1 y \iff \map f x \preceq_2 \map f y$

$\Box$

### Sufficient Condition

Suppose $f: S \to T$ is a mapping such that:

- $f$ is a surjection;
- $\forall x, y \in S: x \preceq_1 y \iff \map f x \preceq_2 \map f y$

From Order Embedding 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:

- $\exists x, y \in S: \map f x = a, \map f y = b$

As $f$ is bijective, then:

- $x = \map {f^{-1} } a, y = \map {f^{-1} } b$

So by hypothesis:

- $a \preceq_2 b \implies \map f x \preceq_2 \map f y \implies \map {f^{-1} } a = x \preceq_1 y = \map {f^{-1} } b$

$\Box$

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

$\blacksquare$

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 7$: Theorem $7.1$ - 1996: Winfried Just and Martin Weese:
*Discovering Modern Set Theory. I: The Basics*... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $19$