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 $f(\bigvee\_L S) \neq \bigvee\_L f(S).$


Is the following statement correct?



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> > Suppose $L$ is complete with smallest element $0$, and let $f:L\to L$ be a join-incomplete lattice homomorphism. Then there is an infinite set $S\subseteq L$ and a lattice homomorphism $f\_0:L\to L$ such that $\text{im}(f\_0|\_S)=\{0\}$ and $f\_0(\bigvee\_L S) > 0$.
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Simplyfing join-incomplete lattice endomorphisms
No.


Let $K$ be any infinite lattice with DCC, bottom element $b$, and no top element. Create $L$ by adding to four elements to $K$: elements $u$ and $v$ that are incomparable to every element of $K$ and to each other, a new bottom element $0$, and a new top element $1$. (So $0 < u < 1$, $0 < v < 1$, and $0<k<1$ for any element of $K$, but no comparabilities between $u, v, k$ for any $k\in K$.) The DCC is used only to guarantee that $L$ is complete.


$L$ has a join-incomplete endomorphism $f\colon L\to L$, which maps every element of $K$ to its bottom element $b$, and maps every other element of $L$ to itself. But if $S\subseteq L$ is infinite, then it must contain infinitely many members of $K$. If $f\_0\colon L\to L$ satisfies $f\_0(S)=\{0\}$, then $f\_0(b)=0$. But this forces $f\_0(u)=f\_0(u)\vee f\_0(b) = f\_0(u\vee b) = f\_0(1)$ and similarly $f\_0(v)=f\_0(1)$, hence $0 = f\_0(0) = f\_0(u\wedge v) = f\_0(1)\wedge f\_0(1) = f\_0(1)$. That is, any endomorphism of $L$ that maps an infinite set to $0$ is constant.


