Isomorphisms of well ordered sets

Question

Let $A$, $B$ be well ordered sets. If there exist order-preserving functions $f : A \to B$ and $g : B \to A$ (do not need to preserve initial segments), are $A$ and $B$ isomorphic? I know this is not true in general, but well-order is quite rigid.

Answer 1

Assuming "order-preserving" means "strictly order-preserving" (i.e., it preserves the relation $<$ rather than the relation $\leq$), then the existence of an order-preserving map $f:A\to B$ implies the existence of an order-preserving map $A\to B$ whose image is an initial segment. Indeed, if no such map exists, then the standard theory of well-ordered sets implies there exists an $a\in A$ and an isomorphism $h:I\to B$, where $I=\{b\in A:b<a\}$. You can then prove by induction on $b$ that $h(b)\leq f(b)$ for each $b\in I$. But then $h(b)<f(a)$ for all $b\in I$, so $f(a)$ cannot be in the image of $h$, which is a contradiction.

Thus if there exist order-preserving maps $f:A\to B$ and $g:B\to A$, there exist such maps whose images are initial segments. It follows that $A$ and $B$ are isomorphic.

Answer 2

The answer is Yes if by "order-preserving" you mean injective and order-preserving ("strictly order-preserving"). Let $\alpha$ be the unique ordinal isomorphic to $A$ and $\beta$ be the unique ordinal isomorphic to $B$. So there is an order-preserving map $f:\alpha \to \beta$, which implies $\alpha\leq \beta$. Analogously we get $\beta\leq\alpha$ implying $\alpha=\beta$ and therefore $A\cong B$.

If you allow for non-injective order-preserving maps, then the answer is No. Let $A$ equal the ordinal $1$ and $B= 2$, let $f:A\to B$ be the inclusion map and $g:B\to A$ be the function mapping everything to $0\in A$.