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Here's a counterexample. Let $L=\{0,1,x_0,x_1,x_2,\dots,y\}$, where $x_0<x_1<x_2<\dots$ and $y$ is incomparable with every $x_n$. Then the only non-principal ideal in $L$ is $I=\{0,x_0,x_1,\dots\}$; …

No; in fact, we can canonically recover $L$ from $\mathcal{Id}(L)$ as the sublattice of compact elements (that is, elements $x$ such that whenever $x=\bigvee S$, there is a finite subset $F\subseteq S …

For a simple example with complete total orders, take $L=\{0\}\cup[1,2]$ and $K=\{-1,0\}\cup[1,2]$.

A trivial necessary condition for such a surjection to exist is that $P$ is bounded, and this is sufficient. For any set $X$, let $F(X)$ be the free bounded poset on $X$ (i.e., $F(X)=X\sqcup \{0,1\}$ …

Yes, in fact, there is a surjective lattice-homomorphism. Your two posets are better known as the Boolean algebras $P(\omega)/fin$ and $P(\omega)$ ($fin$ being the ideal of finite sets). Let $\omega …

Yes, this is apparently a fairly hard theorem of Pudlak and Tuma (or at least I assume it is hard, because it seems to have been an open problem for decades before they finally proved it in 1980).

The following characterization follows easily from the general theory of spectral spaces, though it isn't exactly the most explicit criterion to apply in practice. Theorem (Hochster, Proposition 1 …

Here's a counterexample: on $\mathbb{Z}^2$, $f(x,y)=(x,y+x)$. More generally, the order-preserving automorphisms of $\mathbb{Z}^n$ are exactly the upper triangular matrices with 1s on the diagonal ( …