Given a poset $(P, \leq)$, is there a complete Boolean lattice $B$ and an order-preserving map $i_P: P\to B$ such that for any complete Boolean lattice $B'$ and order-preserving map $f: P\to B'$ there is an order-preserving map $g: B\to B'$ such that $f = g\circ i_P$?
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asked Apr 1 '18 at 7:56
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$\begingroup$ You probably want to require that the poset is separative. $\endgroup$ – Asaf Karagila Apr 1 '18 at 11:24
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The answer is yes. Take $B := 2^P$ to be the power set of $P$, and $i_P(x)$ to be the principal down-set generated by $x$. Now for $A\subseteq P$, put $g(A) := \sup f(A)$. This makes $f = g\circ i_P$ hold by construction.
However, I don't think that this is a universal property, since the $g$ may not be unique. In fact, using maps which preserve all joins and suprema as the usual morphisms of complete Boolean algebras, the existence of a universal complete Boolean algebra generated by $P$ is not guaranteed, and fails even for $P$ being the countable antichain.
answered Apr 1 '18 at 9:16
