Isomorphisms between complete lattices are complete.

Question

An exercise from "Universal Algebra: Fundamentals and Selected Topics" of Clifford Bergman.

(a) Define a complete homomorphism between complete lattices. Give an example of a homomorphism between complete lattices that is not a complete homomorphism.

(b) Prove or disprove: If $L$ and $M$ are complete lattices and $h: L \to M$ is a lattice isomorphism, then $h$ is a complete isomorphism.

For (a) I've taken $L=Sb(\omega)$ and $M=Sb_\omega(\omega) \cup\{\omega\}\cup \{E\}$, where $E=\{n: n \in \omega \;\&\; n \text{ is even}\}$, and:

$f:L \to M$

$$f(X)= \begin{cases} X &\text{ if } |X|<\omega \\ \omega & \text{ otherwise}\end{cases}$$

Now we get:

$f(\bigvee_{n \in E}\{n\})=f(E)=\omega$

$\bigvee_{n \in E}f(\{n\})=\bigvee_{n \in E}\{n\}=E$

Showing that $f$ is a lattice homomorphism but isn't a complete lattice homomorphism. I bet this construction was possible by taking advantage of the fact that $f$ was not surjective, and i think that we can prove that (b) is true.

Furthermore, I have the impression that this question may be topological in nature. Can we define a topology over a complete lattice where the property $f(\bigvee A_i)=\bigvee f(A_i)$ can be seen as a limit operator that commutes with a continuous function?

For example I know that an isotone map between posets can be seen as a continuous function in the downsets topology.

Answer 1

For (b), if $f:L\to K$ is a lattice isomorphism between complete lattices, then you want to show that $f(\bigvee A) = \bigvee\{f(a):a\in A\}$, for any $A \subseteq L$.

Now, for $a \in A$, we have that $a \leq \bigvee A$, whence $f(a) \leq f(\bigvee A)$. Therefore, $\bigvee\{f(a):a\in A\} \leq f(\bigvee A)$.

Conversely, since $K$ is complete and $f$ is surjective, there exists $a^* \in L$ such that $f(a^*)=\bigvee\{f(a):a \in A\}$.
Since $f(\bigvee A) \geq f(a^*)$ (which we have seen in the previous paragraph) and $f$ is an order-isomorphism, it follows that $a^* \leq \bigvee A$.
For each $b \in A$, we have $$f(b) \leq \bigvee \{f(a) : a \in A\} = f(a^*),$$ yielding $b \leq a^*$ (because $f$ is an order-isomorphism), and so $\bigvee A \leq a^*$.
We conclude that $a^*=\bigvee A$, and thus, $$f(\bigvee A) = f(a^*) = \bigvee\{ f(a) : a \in A \}.$$