Let $L$ be a lattice and let $\textbf{DM}(\cdot)$ denote the [Dedekind-MacNeille completion](http://en.wikipedia.org/wiki/Dedekind%E2%80%93MacNeille_completion).


Is there a lattice $L$ that is not a quotient of $\textbf{DM}(L)$? And what if we generalise this question to posets: is every poset $P$ a quotient of $\textbf{DM}(L)$? (I just realise whether the second question makes sens depends on the answer to [this post](https://mathoverflow.net/questions/219741/quotients-of-posets).)



Are lattices quotients of their Dedekind-MacNeille completion?
In order to finally kill every possible version of this question: No, even if $P$ is finite. Let $P$ be two incomparable elements. Then $DM(P)$ is the diamond poset $0 < p,q < 1$. The only maps of posets from $DM(P)$ to $P$ are to send everything to $0$ or everything to $1$. In particular, there is no surjection $DM(P) \to P$ and, thus, $P$ is not a quotient of $DM(P)$. 


To summarize comments above: If $L$ is a a finite lattice, then yes, since $L = DM(L)$. If $L$ is an infinite lattice, even a totally ordered one, then no (simplest example is $\mathbb{Z}$). And the example above shows the answer is also "no" for finite posets.


