Let $L$ be the power set lattice ${\cal P}(\{0,1,2\})$. It is clear that there is an order-preserving injective map from $M_3$ into $L$, but no injective lattice homomorphism (because $L$ is distributive, and $M_3$ is not).

What is an example of distributive lattices $K, L$ such that there is an order-preserving injective map from $K$ into $L$, but no injective lattice homomorphism from $K$ into $L$?


Divisibility latticesLet $K=\{1,2,3,6,12,18,36\}$ ordered by divisibility.

Let $L=\{1,2,3,6,12,24,36,72\}$ ordered by divisibility.

Then $6=2\vee 3=12\wedge 18$ would have to be sent to both $6$ and $12$, but it can only be sent one place.

  • 2
    $\begingroup$ (Ignore the arrow tips in the drawing) $\endgroup$ – Bjørn Kjos-Hanssen Mar 15 '18 at 8:20

The example in the following picture is from "Burris and Sankappanavar, A Course in Universal Algebra".

enter image description here

It is actually one with a bijective map.
Clearly, the map is not an isomorphism.
Since the lattices are finite, any injective homomorphism would be an isomorphism, and so we can conclude there is none.

Note: This actually doesn't answer the title question, but rather the one in the body of the question. The difference is that in the title the OP asks for an order-embedding; in the text, for an order-preserving injective map, which is weaker.

  • $\begingroup$ Thanks for your example -- I don't think this map is an order embedding, as $L_1$ is not isomorphic to a sub-poset of $L_2$ $\endgroup$ – Dominic van der Zypen Mar 29 '18 at 8:00
  • $\begingroup$ @DominicvanderZypen You are absolutely right! It's not an order-embedding. It's just that, although you ask for an order-embedding in the title of the question, in the body you only ask for an order-preserving injective map. I overlooked the title. Sorry... $\endgroup$ – amrsa Mar 29 '18 at 8:14
  • $\begingroup$ Sorry -> this is my mistake! I will therefore upvote your answer. $\endgroup$ – Dominic van der Zypen Mar 29 '18 at 8:17
  • $\begingroup$ @DominicvanderZypen Ok, thanks. And I will add a note for future reference that this is not an answer for the title question. Otherwise it may confuse some future user reading this. $\endgroup$ – amrsa Mar 29 '18 at 8:23

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