Let $P$ and $Q$ be complete boolean algebras. Suppose that $\dot H$ is a $P$-name such that $1_P\Vdash\dot H$ is $Q$-generic. For each $p\in P$, let $A_p$ be the set of $q\in Q$ such that $p\Vdash q\in\dot H$. Then the map $\sigma:P\rightarrow Q$, given by $\sigma(p)=\prod A_p$, is a projection, in the sense that $\sigma$ is order-preserving and for all $p,q$ with $q\leq\sigma(p)$, there is $p'\leq p$ such that $\sigma(p')\leq q$. I wonder if this projection has any connection with any other arbitrary projection (between $P$ and $Q$). More specifically, suppose $\pi:P\rightarrow Q$ is a projection and let $p_1,p_2\in P$ (possibly $p_1\nleq p_2$). If $\sigma(p_1)\leq\sigma(p_2)$, is it true that $\pi(p_1)\leq\pi(p_2)$?

  • $\begingroup$ Perhaps you should refine the question by asking what if $\sigma$ is defined using the $P$-name $\pi(\dot{G})$. $\endgroup$ Apr 24 at 18:25

1 Answer 1


The answer is no. Let $P$ arise from product forcing $Q\times Q$. So forcing with $P$ adds two mutually generic filters for $Q$, one on each factor. Let $\sigma$ be the projection onto the first coordinate, and $\pi$ be the projection onto the second coordinate. Let $p_1=(q,1)$ and $p_2=(1,q)$ for some nontrivial $q\in Q$. Note that $\sigma(p_1)=q\leq 1=\sigma(p_2)$, but $\pi(p_1)=1\not\leq q=\pi(p_2)$.


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