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Question 1: Suppose that $V[G]$ is a set generic extension of $V$ by some forcing notion $P\in V,$ and suppose that $W$ is a model of $ZFC, V \subset W \subset V[G].$ Can we find a forcing notion $Q\in V$ and a map $\pi: P\to Q$ such that:

1) $|Q|\leq |P|,$

2) $\pi$ and $G$ generate a filter $H$ which is $Q-$generic over $V$ (for example if $\pi$ is a projection, then we can take $H$ to be the filter generated by $\pi''[G]$),

3) $W=V[H].$

Remark. Let $B=r.o(P).$ Then for some $\bar{G}, B-$generic over $V$, we have $V[G]=V[\bar{G}].$ Also for some complete subalgebra $C$ of $B, W=V[\bar{G}\cap C].$ If $P$ satisfies the $\kappa-c.c.$, then $|C|\leq |B| \leq |P|^{<\kappa},$ so clearly the answer is yes if $|P|=|P|^{<\kappa}.$

Question 2: Suppose that $P, Q\in V$ are two forcing notions such that for any $G$ which is $P-$generic over $V$, there is $H\in V[G]$ which is $Q-$generic over $V$. Is there a map $\pi: P\to Q$ such that for any $G$ as above, we can choose $H$ to be the filter generated by $\pi[G]$?

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  • $\begingroup$ I can answer question 2 affirmatively in the case that either $P$ or $Q$ is a complete boolean algebra. But perhaps these cases are clear already. $\endgroup$
    – Monroe Eskew
    May 30, 2014 at 18:47
  • $\begingroup$ It would be nice if you give your argument for these cases. $\endgroup$
    – Mohammad Golshani
    Jun 2, 2014 at 15:55

1 Answer 1

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The answer to your first question is yes. Let $C \subseteq B$ be as in your remark. Let $\pi : B \to C$ be the standard projection map, $\pi(b) = \inf \{ c \in C : c \geq b \}$. The restriction of $\pi$ to $P$ maps a dense subset of $B$ onto a dense subset $Q \subseteq C$, it is a projection, and $|Q| \leq |P|$.

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  • $\begingroup$ Monroe, I don't think you intend that $\pi=\text{id}$, since $C$ is contained in $B$, so the identity map from $B$ to $C$ doesn't make sense. Rather, don't you want to use the projection $\pi(b)=\bigvee \{ c\in C\mid c\leq b\}$ or something like that? $\endgroup$
    – Joel David Hamkins
    May 27, 2014 at 17:22
  • $\begingroup$ Does this idea also answer a recent question that we had had here about intermediate models and projections of partial orders? Probably Mohammad can post the link to the question. $\endgroup$
    – Joel David Hamkins
    May 27, 2014 at 17:41
  • $\begingroup$ @Joel: I think that Mohammad deleted that question, and posting this one in its stead. $\endgroup$
    – Asaf Karagila ♦
    May 27, 2014 at 18:33
  • $\begingroup$ By weak projection do you mean for all $p \in P$ and all $q \leq \pi(p)$ there is $r \leq p$ such that $\pi(r) \leq q$? (Rather than $\pi(r) = q$.) I'm pretty sure this does the same job so I call it a projection. It should get you the three properties you desire. $\endgroup$
    – Monroe Eskew
    May 28, 2014 at 3:12
  • $\begingroup$ Well you get the stronger property in this situation. Take $q \in Q$, $q \leq \pi(p)$. There is $b \in B$, $b \leq p$, such that $\pi(b) = q$. Then take $r \leq b$, $r \in P$; we have $\pi(r) \leq q$. $\endgroup$
    – Monroe Eskew
    May 28, 2014 at 3:30

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