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?



Order-preserving image of a complete lattice
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.


