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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.)

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    $\begingroup$ Is $\mathbb{Q} $ a quotient of $\mathbb{R}\cup\left\{\pm\infty\right\} $ ? I don't know the answer, but it doesn't look like it works this way. (That said finite posets might be different.) $\endgroup$
    – darij grinberg
    Commented Oct 1, 2015 at 12:40
  • $\begingroup$ The DM completion of a finite lattice is the lattice itself. (The DM completion of a poset $P$ is the smallest complete lattice containing $P$ as a subposet. For finite posets, "lattice" and "complete lattice" are equivalent, so the DM completion of a lattice is itself.) $\endgroup$
    – David E Speyer
    Commented Oct 1, 2015 at 13:31
  • $\begingroup$ @darijgrinberg For the record, no, there is no order preserving surjection $\phi: \mathbb{R} \cup \{ \pm \infty \} \to \mathbb{Q}$. The image of such a map would land in $[\phi(- \infty), \phi(\infty) ]$, which is not all of $\mathbb{Q}$. There are also no order preserving surjections $\mathbb{R} \to \mathbb{Q}$, although it takes a bit more work. Might be a good problem for an analysis exam... $\endgroup$
    – David E Speyer
    Commented Oct 1, 2015 at 15:03
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    $\begingroup$ @Dominic Do you want the composition $P \to DM(P) \to P$ to be the identity? $\endgroup$
    – David E Speyer
    Commented Oct 1, 2015 at 15:03
  • $\begingroup$ For that matter, there is no surjective order-preserving map from the completion of the naturals N to N as a lattice. It should be clear that unbounded and countably (maybe uncountably?) cofinal lattices cannot have such maps from their completions. Gerhard "Where Does The Top Go?" Paseman, 2015.10.01 $\endgroup$
    – Gerhard Paseman
    Commented Oct 1, 2015 at 15:56

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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.

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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.

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