Order-preserving image of a complete lattice If $L$ is a complete lattice and $P$ is a poset and $f: L\to P$ is an order preserving surjective map, does this imply that $P$ is a (complete) lattice?
 A: No, clearly not, because you could put junk on top. 
But even if you avoid this by insisting that the map is surjective, there are counterexamples. Consider the map of Eric Wofsey's recent answer, where he considered the partial order $L(X)$, where $X$ is an antichain of pairwise incomparable elements, and we add $0$ and $1$ to bound it. This is a complete lattice, and it admits an order preserving surjection to any bounded partial order on $X$. Such an order may not be a complete lattice, and so this provides numerous counterexamples.
A: Not necessarily - there is even a finite counterexample. 
Let $X = \{1,2,3\}$ and set $L := \mathcal{P}(X)$. 
Let $P := \{b, t\} \cup \{(i,j): i,j \in \{0,1\}\}$, where $b$ will be the botton (least) element, $t$ is the top (greatest) element, and $\{(i,j): i,j \in \{0,1\}\}$ is ordered by $$(i, j) < (k, l) \text{  if and only if } i = 0 \text{ and } k = 1.$$ So $P$ is not a lattice, as $(0,0)$ and $(0,1)$ do not have a least upper bound.
As for the surjection $f: \mathcal{P}(X) \to P$, set $\emptyset \mapsto b$, $X\mapsto t$, and moreover $\{1\} \mapsto (0,0)$ and $\{2\},\{3\}\mapsto (0,1)$ and $X\setminus \{1\} \mapsto (1,0)$ and the rest maps to $(1,1)$.
It's a routine verification that $f$ is onto and order-preserving.
