Suppose $P$ and $Q$ are forcing notions and that $f:P\rightarrow Q$ is a function such that:

• $p_1\leq_P p_2$ implies $f(p_1)\leq_Q f(p_2)$
• If $p_1,p_2\in P$ are incomparable then $f(p_1),f(p_2)$ are incomparable
• $f[P]$ is dense 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

If $q_1\in H$ and $q_1\leq q_2$ 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 $f(p_1)\leq q_1$ and $f(p_2)\leq q_2$ . Then there is some $p_3\leq p_1,p_2$ such that $p_3\in G$ , and since $f$ is order preseving $f(p_3)\leq f(p_1)\leq q_1$ and $f(p_3)\leq f(p_2)\leq q_2$ .

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 $f(p)\leq d$ . For each $d\in D$ , assign (using the axiom of choice) some $d_p\in P$ such that $f(d_p)\leq d$ , 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 $d\leq f(p)$ , and so some $d_p\in D_P$ such that $f(d_p)\leq d$ . If $d_p\leq p$ then $D_P$ is dense, so suppose $d_p\nleq p$ . If $d_p\leq p$ 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 $p\leq d_p$ , 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 $f(p)\leq d$ . But since $p\in G$ , $d\in H$ , so $H$ intersects $D$ .

$G$ can be recovered from $F(G)$

$F^{-1}(H)=G$ where $G$ is generic

Consider $D$ , the set of elements of $Q$ which are $f(p)$ for some $p\in P$ and either $f(p)\leq q$ or there is no element greater than both $f(p)$ and $q$ . This is dense, since given any $q_1\in Q$ , if $q_1\leq q$ then (since $f[P]$ is dense) there is some $p$ such that $f(p)\leq q_1\leq q$ . If $q\leq q_1$ then there is some $p$ such that $f(p)\leq q\leq q_1$ . If neither of these and $q$ there is some $r\leq q_1,q$ then any $p$ such that $f(p)\leq r$ suffices, and if there is no such $r$ then any $p$ such that $f(p)\leq q$ suffices.

There is some $f(p)\in D\cap H$ , and so $p\in G$ . Since $H$ is directed, there is some $r\leq f(p),q$ , so $f(p)\leq q\leq f(p_1),f(p_2)$ . If it is not the case that $f(p)\leq f(p_1)$ 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 $p\leq q$ , and some $d\in D$ such that $d\leq p\leq q$ . So there is some $f(p)\in f[D]\cap H$ , and so $p\in D\cap G$ .

$H$ can be recovered from $F^{-1}(H)$