Are filters in lattices exactly the homomorphic preimages $\varphi^{-1}(1)$ of top elements?

Question

Say I got a lattice $L$, a bounded lattice $K$ with top-element $1$ and a homomorphism $\varphi : L \to K$, then $\varphi^{-1}(1)$ is a filter in L.

I wondered whether I can represent every filter $F \subset L$ in that way, i.e. I find some other bounded-above lattice such that $F$ is some preimage of the top element. This should be some kind of contraction of $F$ into a new, single element.

My idea would be to define an equivalence relation on $L$

$$a \sim b :\Leftrightarrow a = b \text{ or } a, b \in F$$

and show that it's in fact a congruence. Then $L/\sim$ should be the lattice I look for. Is this proposition in fact true and is there an easier proof? Thanks

Answer 1

This lattice

(known as "the diamond"), has no proper congruence and three proper filters.

Answer 2

Not every ideal is a kernel of a congruence [Steven Roman, Lattices and Ordered Sets, Springer, 2008, p.78, Example 3.40]. Dually, not every filter is a preimage of the top element.