Can you always reconstruct the underlying order of its induced topology? A total order in a set $X$ induces a topology by letting the open rays and open intervals be the basis of the topology. That is, there is a function $F$ that takes an ordered set to its topology. My question is, if $F(X)$ is homeomorphic to $F(Y)$ does it necessarily follow that $X$ is order-isomorphic to $Y$?
Equivalently, in more fancy terms, is the functor that takes a totally ordered set to its induced topology an embedding of the category of totally ordered sets to the category of topological spaces?
Edit: Sadly not a functor, see Alessandro's comment.
 A: No. For instance, let $\langle X,\le\rangle$ be any linear order, and let $\preceq$ be the lexicographic order on $Y=X\times\Bbb Z$; then the order topology on $Y$ is discrete irrespective of the order on $X$. However, if $\le_0$ and $\le_1$ are distinct linear orders on $X$, and $\preceq_0$ and $\preceq_1$ are the corresponding lexicographic orders on $Y$, $\langle Y,\preceq_0\rangle$ and $\langle Y,\preceq_1\rangle$ need not be isomorphic as linear orders. Here is a concrete example.

Let $\le_0$ be the usual order on $\Bbb R$, and let $\le_1$ be a well-order on $\Bbb R$ of type $2^\omega$. Then every $y\in Y$ has $2^\omega=\mathfrak{c}$ predecessors in $\preceq_0$ but fewer than $2^\omega$ predecessors in $\preceq_1$.

Here is another.

Let $X=\omega_1$, and let $\le_0$ be the usual well-order on $\omega_1$. Let $\preceq_1$ be the reverse ordering: $\alpha\le_1\beta$ iff $\beta\le_0\alpha$ for all $\alpha,\beta\in X$. Then every $y\in Y$ has countable many $\preceq_0$-predecessors and uncountably many (in fact $\omega_1$) $\preceq_1$-predecessors.

That second example is similar to a more general fact: if $\le_0$ is a linear order an a set $X$, and $\le_1$ is its reversal, then $\le_0$ and $\le_1$ generate the same topology on $X$. And if $\langle X,\le_0\rangle$ and $\langle X,\le_1\rangle$ are not order-isomorphic, we clearly cannot reconstruct the order generating that topology. One might hope that at least we could reconstruct it up to reversal, but the first example above clearly puts paid to that hope. (The second does, too, since $\preceq_0$ and $\preceq_1$ are not reversals of each other, though each is isomorphic to the reversal of the other.)
