Proving isomorphism between $X$ and $Y$

Question

let $(X,\leq)$ and $(Y,\preceq)$ be well-order sets and they are isomorphic.

Prove that there is one isomorphism between $X$ and $Y$.

Notice that $X$ and $Y$ are sets, so by isomorphism I mean isomorphism between sets.

Answer 1

I would be surprised if you meant "isomorphism between sets" and not "isomorphism between ordered sets" (cause otherwise the claim of uniqueness wouldn't be true... which I assume is what you mean when you say 'prove there is one isomorphism').

If $f$ and $g$ are isomorphisms, are then they must respect least elements. So $f$ and $g$ both map the least element of $X$ to the least element of $Y.$ And so on, by induction. Will leave the details to you.

Answer 2

Every homomorphism between ordered sets $(X, \leq) \to (Y, \preceq)$ has an underlying homomorphism of sets $X \to Y$.

The usual definition of what it means for a homomorphism of ordered sets to be an isomorphism almost literally contains the condition that the underlying homomorphism of sets is an isomorphism.

(that the orders are well-orders is irrelevant)

(also note that the obvious proof proves that one exists; but once you know one exists it's easy to show that there are many more, at least in most cases)

So, I imagine one of the three things are true:

• You are overthinking it, and making the problem more complicated than it really is.
• You are using unusual definitions for which this is not immediate — if so you really need to include those definitions so that the rest of us know the actual problem you're facing
• You haven't written the question you mean to ask.

Answer 3

We do need an isomorphism between ordered sets here...

One method would be a proof by induction of the following statement: Let $f: X \rightarrow Y$ and $g \rightarrow Y$ be two isomorphisms. Then $\forall x \in X$ $f(x)=g(x)$

It's worth filling in the details yourself as a good exercise, but one key step would be the following:

If $\perp$ denotes the bottom element, then $f(\perp)=g(\perp)=\perp$. This is true because if either of them were a greater element then nothing less, including $\perp$, could be hit by them because the functions preserve order and all remaining inputs are greater than $\perp$.

Fill in the induction step of this proof and you'll have a proof of the entire proposition.