Trivial Projection of Forcing Posets and Generics

Question

$\pi : Q \rightarrow P$ is a projection of two forcing poset if and only if

(1) $\pi$ is order preserving, onto, and maps $0_Q$ to $0_P$.

(2) For all $q \in Q$ and $p' \in P$, if $p' \geq \pi(q)$, then there exists a $q' \geq q$ such that $\pi(q') = p'$.

$\pi$ is a trivial projection if and only if $\pi(q_1) = \pi(q_2)$ implies $q_1 \not \perp q_2$.

In $\textit{Proper Forcing}$ by Uri Abraham, he claims if $G$ is $P$-generic, then $\pi^{-1}(G)$ is $Q$-generic, when $\pi$ is a trivial projection.

I can't seem to show that $\pi^{-1}(G)$ is a filter in $Q$. In particular, I can not show that two elements of $\pi^{-1}(G)$ has a compatible extension in $\pi^{-1}(G)$.

Abraham forcing notation is the opposite of Jech and Kunen; however, I will use his convention here.

One possible approach: let $q_1, q_2 \in \pi^{-1}(G)$. $\pi(q_1)$ and $\pi_1(q_2)$ are in $G$. Hence, there exists an extension $r \in G$ such that $r \geq \pi(q_1)$ and $r \geq \pi(q_2)$. By (2), there exists $q_1' \geq q_1$ and $q_2' \geq q_1$ such that $\pi(q_1') = r = \pi(q_2')$. By the definition of trivial, $q_1' \not \perp q_2'$. However, I do not see how to show that there exists a compatible extension in $\pi^{-1}(G)$.

Thanks for any help.

Answer 1

I hope you don't mind, but I will change your forcing order direction (i.e. I will extend downward).

Your approach works; it just needs a small step at the end. Let me just reiterate to fix my notation:

Let $q_1,q_2\in\pi^{-1}[G]$. Since $G$ is a filter, there is a $p\in G$ such that $p\leq q_1,q_2$. By (2) there are $q_1'\leq q_1$ and $q_2'\leq q_2$ such that $\pi(q_1')=\pi(q_2')=p$.

I now claim that the image of the set of common extensions of $q_1'$ and $q_2'$ is dense below $p$. Indeed, take $p^*\leq p$. By (2) there are $q_1^*\leq q_1'$ and $q_2^*\leq q_2'$ such that $\pi(q_1^*)=\pi(q_2^*)=p^*$ and these two are compatible, by triviality, with common extension $q^*$. But then, by monotonicity, we have $\pi(q^*)\leq p^*$.

By genericity there is a $q_0\leq q_1,q_2$ such that $\pi(q_0)\in G$, as required.

Answer 2

What's involved here is a special case of the following more general phenomenon. Suppose (as in the usual forcing set-up) $P$ is a poset in some transitive model $M$ and $G$ (usually not in $M$) is a subset of $P$ such that (1) $G$ is closed upward [I'm using the non-Jerusalem forcing convention: stronger = lower.] (2) every two members of $G$ are compatible and (3) $G$ meets every dense subset of $P$ that is in $M$. [These hypotheses almost say that $G$ is $M$-generic; the difference is that, in (2), any two elements of $G$ are required merely to have a lower bound in $P$, not necessarily in $G$.] Then $G$ is $M$-generic.

Proof: The only part of genericity that I haven't assumed is that, if $p_1,p_2\in G$, then they have a lower bound not merely in $P$ but in $G$. To prove this, consider the set $$ D=\{q\in P: \text{ Either }q\leq p_1\text{ and }q\leq p_2\text{ or }q\text{ is incompatible with at least one of }p_1,p_2\}. $$ $D$ is a subset of $P$ and an element of $M$. I'll show that it is dense in $P$. Once this is shown, I'll know that $G$ contains some $q\in D$. Being in $G$, this $q$ must be compatible with $p_1$ and with $p_2$, by my assumption (2). So the only way it can be in $D$ is to be an extension of both of them, which is what I needed.

So it remains only to prove that $D$ is dense, but this is easy. Let an arbitrary $r\in P$ be given; I want an extension $q$ in $D$. If $r$ is incompatible with $p_1$, then $r$ itself is in $D$, so I'm done, with $q=r$. Otherwise, let $r'$ be a common extension of $r$ and $p_1$. If $r'$ is incompatible with $p_2$, then $r'\in D$ so I'm done, with $q=r'$. Otherwise, let $q$ be a common extension of $r'$ and $p_2$, and again this $q$ does what I needed.