Let $L$ be a lattice.
Let $\mathcal L$ denote a partition of $L$, and let $[a]$ denote the equivalence class presented by $a\in L$.
Further let $\wedge$ and $\vee$ be binary operations on $\mathcal L$ such that:
- $[a]\wedge[b]=[a\wedge b]$
- $[a]\vee[b]=[a\vee b]$
Then it is obvious that equipped with these operations $\mathcal L$ is a lattice itself and the function $\nu:L\to\mathcal L$ is a surjective homomorphism that has $$ as kernel.
Further $$ will serve as least element of $\mathcal L$ and $$ as largest element of $\mathcal L$.
It is easy to prove that:$$\exists i\in [a\vee i=b\vee i]\implies[a]=[b]\tag1$$
My question is: is the opposite of $(1)$ also true?
I hope the answer is "yes", and that I am only bothered by a blind spot for that.
If it would be "no" then this would lead to a partition that is properly finer by stating that: $$\exists i\in [a\vee i=b\vee i]\iff a\sim b\tag1$$
which would also be a lattice. Then again there is a surjective homomorphism having $$ as kernel.
So another way of asking would be:
are surjective homomorphisms completely determined by their kernel?