$f: X \to Y$ order preserving implies $Ord(X) \leq Ord(Y)$ Let $X,Y$ be well-ordered sets and let $f:X \to Y$ be a map that preserves the strict order. I would like to prove that $Ord(X) \leq Ord(Y)$.
You can assume I know the 'basic' results about maps on well-ordered sets and segments etc.
I feel like there should be an easy, short solution.
Attempt:
Composing with isomorphisms $X \cong Ord(X), Y \cong Ord(Y)$ we get a map $f: Ord(X) \to Ord(Y)$ that preserved the strict order. So it suffices to prove:
If $f: \alpha \to \beta$ is a map between ordinals that preserves the strict order, then $\alpha \leq \beta$.
Assume to the contrary that $\beta < \alpha$. Then $\beta$ is a segment of $\alpha$. Thus there is $a \in \alpha$ with $\beta =\alpha_a$. Then I'm stuck.
Thanks in advance.

If $\beta<\alpha$, you can view $f$ as a map from $\alpha$ into $\alpha$. $\{\xi\in\alpha:f(\xi)<\xi\}\ne\varnothing$, since clearly $f(\beta)<\beta$. Let $\eta=\inf\{\xi\in\alpha:f(\xi)<\xi\}$, and derive a contradiction with the assumption that $f$ is strictly order-preserving.
