Are lattices quotients of their Dedekind-MacNeille completion? Let $L$ be a lattice and let $\textbf{DM}(\cdot)$ denote the Dedekind-MacNeille 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.)
 A: 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.
A: I claim that if $X$ is a poset and $f:DM(X)\rightarrow X$ is a surjective mapping where $f|_{X}$ is the identity mapping, then $X$ is already a complete lattice. 
Recall that the Dedekind-Macneille of a poset $X$ is the unique extension of $X$ to a complete lattice $C$ such that for all $c\in C$ there are $R,S\subseteq X$ with $c=\bigvee^{C}R=\bigwedge^{C}S$.
Suppose that $c\in C$. Then there are subsets $R,S\subseteq X$ such that $c=\bigvee^{C}R=\bigwedge^{C}S$. If $r\in R$, then $f(c)\geq f(r)=r$, so
$f(c)\geq\bigvee^{C}R=c$ and $f(c)\leq\bigwedge^{C}S=c$. Therefore $f(c)=c$. Thus $c\in X$ after all. Therefore since $X=C$ and $X$ is complete, then $X$ is also complete.
