Order Isomorphism is Surjective Order Monomorphism

Theorem

Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ 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.

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:

$\forall x, y \in S: x \preceq_1 y \implies f \left({x}\right) \preceq_2 f \left({y}\right)$

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:

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

$\Box$

Sufficient Condition

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

• $f$ is a surjection;
• $\forall x, y \in S: x \preceq_1 y \iff f \left({x}\right) \preceq_2 f \left({y}\right)$

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.

$\blacksquare$

Sources

• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 7$: Theorem $7.1$