Let $L$ be an complete lattice. A lattice homomorphism $f: L\to L$ is said to be *join-incomplete* if there is an infinite set $S \subseteq L$ such that $f(\bigvee\_L S) \neq \bigvee\_L f(S).$


Suppose $L$ contains an ideal $J$ such that $\bigvee\_L J \notin J$. Does this imply that there is a join-incomplete lattice homomorphism $f: L\to L$?


(Joseph Van Name gave a positive answer to [a similar question](https://mathoverflow.net/questions/211131/incomplete-lattice-homomorphisms-between-complete-lattices-2). The difference to this question here is that we considered homomorphisms $f:L\to K$ where $K$ was any complete lattice, whereas here we focus on endomorphisms $f:L\to L$.)



Join-incomplete lattice endomorphisms
Here is a way to build examples of complete lattices with no join-incomplete endomorphisms. Suppose $L$ is 


1. complete,
2. simple, and
3. each proper principal ideal of $L$ has ACC, but $L$ does not have ACC.


Then $L$ has a nonprincipal ideal but no join-incomplete endomorphism.


Reason: If $L$ does not have ACC, then the ideal generated by any increasing $\omega$-chain is nonprincipal. Suppose that $f\colon L\to L$ is an endomorphism. By item 2, $f$ is constant or an embedding. If $f$ is constant it preserves complete joins, so we are done. If $f$ is an embedding, then it must preserve $1$, since $1$ is the only element that dominates an increasing $\omega$-chain. Now, assuming still that $f$ is an embedding, if $J\subseteq L$ is a nonprincipal ideal it contains an increasing $\omega$-chain, so by item 3 the join of $J$ must $1$, but by the same argument the set $f(J)$ has join $1$. Thus $f$ preserves infinite joins.


What remains to show is that there really is a lattice satisfying 1, 2 and 3.


Start with the lattice $L\_0$ of finite dimensional subspaces of an infinite dimensional vector space. $L\_0$ satisfies items 2 and 3. Add three new elements to $L\_0$: $1$ (a top element), and $a$ and $b$ which are incomparable with all other elements except $0$ (bottom of $L\_0$) and $1$. I want $a$ and $b$ to be incomparable with each other, as well.
$L$ has the property that, for every element $\ell<1$, there is a finite bound on the lengths of chains below $\ell$. This implies that $L$ has DCC. Since $L$ has DCC and a top element, it is complete. $L$ inherits item 3 from $L\_0$. I leave it to you to check item 2, that $L$ is a simple lattice.


