# Are surjective homomorphisms in lattices completely determined by their kernel?

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?

• Are you claiming that the quotient of a lattice by any equivalence class is again a lattice? That's not true. Only if the equivalence is a congruence. Any way, the class of 0 doesn't determine the other classes (as it would in groups, for example). As a counter-example take a chain with more than two elements, say $0<a<1$; you can have a two different congruences in which $=\{0\}$: the trivial congruence, in which each class is a singleton, an the the congruence that relates $a$ and $1$. But perhaps I'm misunderstanding you. – amrsa Jul 29 '18 at 10:48
• @amrsa No, I am not claiming that. The fact that $\wedge$ and $\vee$ on $\mathcal L$ are well-defined assures that we are dealing with a congruence. I am not giving definitions there but properties. I will have a look at the counterexample that you gave. Thank you for that. – drhab Jul 29 '18 at 11:13
• If my counter-example is useful, then this question is related; Keith Kearnes gives a very nice answer from which it follows that lattices are not congruence-regular (that is, congruences are not determined by each class, or any class, in general). – amrsa Jul 29 '18 at 11:18
• @amrsa Your counterexample is effective as answer to my question. My request is that you write an answer to my question. That enables me to accept it. – drhab Jul 29 '18 at 11:58

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 $_{\theta_1}=\{0\}$; but $$\theta_2 = \{(0,0),(a,a),(a,1),(1,a),(1,1)\}$$ is another congruence with $_{\theta_2} = \{0\}$.
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.