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 $[0]$ as kernel.
Further $[0]$ will serve as least element of $\mathcal L$ and $[1]$ as largest element of $\mathcal L$.
It is easy to prove that:$$\exists i\in[0] [a\vee i=b\vee i]\implies[a]=[b]\tag1$$
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: $$\exists i\in[0] [a\vee i=b\vee i]\iff a\sim b\tag1$$
which would also be a lattice. Then again there is a surjective homomorphism having $[0]$ as kernel.
So another way of asking would be:
are surjective homomorphisms completely determined by their kernel?
Apparently, the property that you're asking about is congruence-regularity.
A congruence is regular if one class determines the others;
an algebra is congruence-regular if each of its congruences is regular;
a variety is congruence-regular if each of its members is congruence-regular.
The archetypical example is the one of groups.
Lattices are not congruence-regular.
For example, take $C$ to be the three-element chain, $C = \{0,a,1\}$, with $0<a<1$.
Then $$\theta_1 = \Delta_C = \{(x,x):x \in C\}$$ is a congruence with $[0]_{\theta_1}=\{0\}$; but $$\theta_2 = \{(0,0),(a,a),(a,1),(1,a),(1,1)\}$$ is another congruence with $[0]_{\theta_2} = \{0\}$.
So the class of $0$ doesn't determine the congruence.
Notice also that $0$ doesn't play a special role here, so if we had a longer chain, and take a "middle" element $x$, in the sense that there would be at least two elements below $x$ or at least two elements above, again we could construct two different congruences such that the class of $x$ would be a singleton.
For lattices which are not chains, counter-examples could also be given, but of course, there are cases in which these congruences are indeed regular (for example, the four-element lattice that is not a chain is congruence-regular).
This question is related, and in particular, the answer given by Keith Kearnes points to characterizations of congruence-regular varieties, and related results.