# Preorder of forcing notions

(Throughout, fix a universe of set theory $V$ to serve as a base.) We can define a fairly natural preordering on forcing notions as follows. Let $\mathbb{P}_1$, $\mathbb{P}_2$ be two forcing notions. Then we have $\mathbb{P}_1\le\mathbb{P}_2$ iff for all $G$ $\mathbb{P}_2$-generic, there is some $\mathbb{P}_1$-generic $H$ such that $H\in V[G]$.

One question which occurred to me is the following: if $\mathbb{P}_1\le\mathbb{P}_2$, then is it the case that for every $\mathbb{P}_1$-generic $H$, there is some $\mathbb{P}_2$-generic $G$ such that $H\in V[G]$?

This is the wrong question to ask, since it has a trivial negative answer. Let $\mathbb{P}_2$ be, say, Sacks forcing, and let $\mathbb{P}_1$ consist of a copy of Sacks forcing and a copy of some countably closed forcing $\mathbb{Q}$ placed side by side, so that a generic for $\mathbb{P}_1$ is either Sacks generic or $\mathbb{Q}$ generic, but not both. Then clearly $\mathbb{P}_1\le\mathbb{P}_2$, but all the $\mathbb{P}_1$-generics which come from a $\mathbb{Q}$-generic need not live in any extension by Sacks forcing.

So my question has two parts. First, the following rephrasing of the wrong question:

Question 1: Suppose $\mathbb{P}_1\le\mathbb{P}_2$, and for all $p\in\mathbb{P}_1$, there exists a $\mathbb{P}_2$-generic $G$ and $\mathbb{P}_1$-generic $H$ with $p\in H\in V[G]$. Then is it the case that, for all $H$ $\mathbb{P}_1$-generic, there is some $\mathbb{P}_2$-generic $G$ such that $H\in V[G]$?

My second, more informal question is whether this is the "right" way to ask the question I'm trying to ask. I suppose an attempt to clarify this informal question would be:

Question 2: What sorts of assumptions on $\mathbb{P}_1$ and $\mathbb{P}_2$, besides $\mathbb{P}_1\le\mathbb{P}_2$, do we need to make to ensure that every $\mathbb{P}_1$-generic lives in the generic extension by some $\mathbb{P}_2$-generic?

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The basic situation for complete Boolean algebras is that $\mathbb{B}_1\leq \mathbb{B}_2$ just in case there is a condition $p\in\mathbb{B}_1$ such that the restriction of the forcing $\mathbb{B}_1\upharpoonright p$ below $p$ has a complete embedding into $\mathbb{B}_2$. This means that the forcing $\mathbb{B}_2$ is isomorphic to first forcing with $\mathbb{B}_1\upharpoonright p$, and then forcing with the quotient forcing of $\mathbb{B}_2$ over that subalgebra.
From this, it follows that two forcing notions $\mathbb{P}_1$ and $\mathbb{P}_2$ can have a forcing extension $V[G]=V[H]$ in common if and only if there are conditions $p\in G\subset\mathbb{P}_1$ and $q\in H\subset\mathbb{P}_2$, below which the forcing notions are isomorphic: $\text{RO}(\mathbb{P}_1\upharpoonright p)\cong\text{RO}(\mathbb{P}_2\upharpoonright q)$.