Let $A$ and $B$ be complete Boolean algebras and $f:A\to B$ a homomorphism. Recall that $f$ is complete if $f$ it preserves arbitrary (including possibly infinite) joins.
Does there exist a non-complete endomorphism of $2^{\mathbb{N}}$ that preserves aribitrary disjoint unions?
If a homomorphism $f:2^{\mathbb{N}}\to 2^{\mathbb{N}}$ preserves disjoint unions, then it is determined by what it does on singletons by the formula $f(A)=\bigcup_{n\in A}f(\{n\})$, since each set is the disjoint union of the singletons it contains. But if $f$ is given by this formula, then it clearly preserves arbitrary unions as well, since $f(\bigcup_i A_i)=\bigcup_{n\in \bigcup_i A_i} f(\{n\})=\bigcup_i \bigcup_{n\in A_i}f(\{n\})=\bigcup_i f(A_i)$.