Is every surjective lattice homomorphism a split epimorphism?

Question

A morphism $f:A \twoheadrightarrow B$ is said to be split by a morphism $g:B \rightarrow A$, if $f \circ g = 1_{B}$. If such a $g$ exists for a given $f$, then $f$ is said to be a split epimorphism.

Is every surjective lattice homomorphism a split epimorphism?

Answer 1

No.

Consider the lattice $P(\aleph_0)$, the powerset of a countable set. Introduce a congruence $\sim$ on $P(\aleph_0)$ given by $x\sim y$ if and only if the symmetric difference of $x$ and $y$ is a finite set.

Then $\sim$ is a lattice congruence (so $x\mapsto[x]_\sim$ is a surjective lattice homomorphism) and it is easy to see that $P(\aleph_0)/\sim$ is a dense poset: if $[x]_\sim<[z]_\sim$, then there is $[y]_\sim$ in $P(\aleph_0)/\sim$ such that $[x]_\sim<[y]_\sim<[z]_\sim$. In particular, all maximal chains in $P(\aleph_0)/\sim$ are dense.

However, there is no dense maximal chain in $P(\aleph_0)$, hence there is no injective mapping $g\colon (P(\aleph_0)/\sim)\to P(\aleph_0)$.

EDIT

You asked for a finite counterexample. Here is one (I think that this is the smallest possible). Let $A$ be this

All of the equivalence classes of the congruence are singletons, except for $\{c,d\}$. It should be obvious that the quotient of $A$ is a lattice that cannot be embedded into $A$.