forcings are equivalent if one is dense in the other
Suppose $P$ and $Q$ are forcing^{} notions and that $f:P\to Q$ is a function such that:

•
${p}_{1}{\le}_{P}{p}_{2}$ implies $f({p}_{1}){\le}_{Q}f({p}_{2})$

•
If ${p}_{1},{p}_{2}\in P$ are incomparable then $f({p}_{1}),f({p}_{2})$ are incomparable

•
$f[P]$ is dense (http://planetmath.org/DenseInAPoset) in $Q$
then $P$ and $Q$ are equivalent^{}.
Proof
We seek to provide two operations^{} (computable in the appropriate universes^{}) which convert between generic^{} subsets of $P$ and $Q$, and to prove that they are inverses^{}.
$F(G)=H$ where $H$ is generic
Given a generic $G\subseteq P$, consider $H=\{q\mid f(p)\le q\}$ for some $p\in G$.
If ${q}_{1}\in H$ and ${q}_{1}\le {q}_{2}$ then ${q}_{2}\in H$ by the definition of $H$. If ${q}_{1},{q}_{2}\in H$ then let ${p}_{1},{p}_{2}\in P$ be such that $f({p}_{1})\le {q}_{1}$ and $f({p}_{2})\le {q}_{2}$. Then there is some ${p}_{3}\le {p}_{1},{p}_{2}$ such that ${p}_{3}\in G$, and since $f$ is order preseving $f({p}_{3})\le f({p}_{1})\le {q}_{1}$ and $f({p}_{3})\le f({p}_{2})\le {q}_{2}$.
Suppose $D$ is a dense subset of $Q$. Since $f[P]$ is dense in $Q$, for any $d\in D$ there is some $p\in P$ such that $f(p)\le d$. For each $d\in D$, assign (using the axiom of choice^{}) some ${d}_{p}\in P$ such that $f({d}_{p})\le d$, and call the set of these ${D}_{P}$. This is dense in $P$, since for any $p\in P$ there is some $d\in D$ such that $d\le f(p)$, and so some ${d}_{p}\in {D}_{P}$ such that $f({d}_{p})\le d$. If ${d}_{p}\le p$ then ${D}_{P}$ is dense, so suppose ${d}_{p}\nleqq p$. If ${d}_{p}\le p$ then this provides a member of ${D}_{P}$ less than $p$; alternatively, since $f({d}_{p})$ and $f(p)$ are compatible, ${d}_{p}$ and $p$ are compatible, so $p\le {d}_{p}$, and therefore $f(p)=f({d}_{p})=d$, so $p\in {D}_{P}$. Since ${D}_{P}$ is dense in $P$, there is some element $p\in {D}_{P}\cap G$. Since $p\in {D}_{P}$, there is some $d\in D$ such that $f(p)\le d$. But since $p\in G$, $d\in H$, so $H$ intersects $D$.
$G$ can be recovered from $F(G)$
Given $H$ constructed as above, we can recover $G$ as the set of $p\in P$ such that $f(p)\in H$. Obviously every element from $G$ is included in the new set, so consider some $p$ such that $f(p)\in H$. By definition, there is some ${p}_{1}\in G$ such that $f({p}_{1})\le f(p)$. Take some dense $D\in Q$ such that there is no $d\in D$ such that $f(p)\le d$ (this can be done easily be taking any dense subset and removing all such elements; the resulting set is still dense since there is some ${d}_{1}$ such that ${d}_{1}\le f(p)\le d$). This set intersects $f[G]$ in some $q$, so there is some ${p}_{2}\in G$ such that $f({p}_{2})\le q$, and since $G$ is directed, some ${p}_{3}\in G$ such that ${p}_{3}\le {p}_{2},{p}_{1}$. So $f({p}_{3})\le f({p}_{1})\le f(p)$. If ${p}_{3}\nleqq p$ then we would have $p\le {p}_{3}$ and then $f(p)\le f({p}_{3})\le q$, contradicting the definition of $D$, so ${p}_{3}\le p$ and $p\in G$ since $G$ is directed.
${F}^{1}(H)=G$ where $G$ is generic
Given any generic $H$ in $Q$, we define a corresponding $G$ as above: $G=\{p\in P\mid f(p)\in H\}$. If ${p}_{1}\in G$ and ${p}_{1}\le {p}_{2}$ then $f({p}_{1})\in H$ and $f({p}_{1})\le f({p}_{2})$, so ${p}_{2}\in G$ since $H$ is directed. If ${p}_{1},{p}_{2}\in G$ then $f({p}_{1}),f({p}_{2})\in H$ and there is some $q\in H$ such that $q\le f({p}_{1}),f({p}_{2})$.
Consider $D$, the set of elements of $Q$ which are $f(p)$ for some $p\in P$ and either $f(p)\le q$ or there is no element greater than both $f(p)$ and $q$. This is dense, since given any ${q}_{1}\in Q$, if ${q}_{1}\le q$ then (since $f[P]$ is dense) there is some $p$ such that $f(p)\le {q}_{1}\le q$. If $q\le {q}_{1}$ then there is some $p$ such that $f(p)\le q\le {q}_{1}$. If neither of these and $q$ there is some $r\le {q}_{1},q$ then any $p$ such that $f(p)\le r$ suffices, and if there is no such $r$ then any $p$ such that $f(p)\le q$ suffices.
There is some $f(p)\in D\cap H$, and so $p\in G$. Since $H$ is directed, there is some $r\le f(p),q$, so $f(p)\le q\le f({p}_{1}),f({p}_{2})$. If it is not the case that $f(p)\le f({p}_{1})$ then $f(p)=f({p}_{1})=f({p}_{2})$. In either case, we confirm that $H$ is directed.
Finally, let $D$ be a dense subset of $P$. $f[D]$ is dense in $Q$, since given any $q\in Q$, there is some $p\in P$ such that $p\le q$, and some $d\in D$ such that $d\le p\le q$. So there is some $f(p)\in f[D]\cap H$, and so $p\in D\cap G$.
$H$ can be recovered from ${F}^{1}(H)$
Finally, given $G$ constructed by this method, $H=\{q\mid f(p)\le q\}$ for some $p\in G$. To see this, if there is some $f(p)$ for $p\in G$ such that $f(p)\le q$ then $f(p)\in H$ so $q\in H$. On the other hand, if $q\in H$ then the set of $f(p)$ such that either $f(p)\le q$ or there is no $r\in Q$ such that $r\le q,f(p)$ is dense (as shown above), and so intersects $H$. But since $H$ is directed, it must be that there is some $f(p)\in H$ such that $f(p)\le q$, and therefore $p\in G$.
Title  forcings are equivalent if one is dense in the other 

Canonical name  ForcingsAreEquivalentIfOneIsDenseInTheOther 
Date of creation  20130322 12:54:43 
Last modified on  20130322 12:54:43 
Owner  Henry (455) 
Last modified by  Henry (455) 
Numerical id  6 
Author  Henry (455) 
Entry type  Result 
Classification  msc 03E35 
Classification  msc 03E40 