extending elementary embeddings

Question

Suppose $j : M \to N$ is an elementary embeddings between transitive models of ZFC. Everyone knows that if $G$ is $\mathbb{P}$-generic over $M$, $H$ is $j(\mathbb{P})$-generic over $N$, and $j[G] \subseteq H$, then the map can be extended to $\hat{j} : M[G] \to N[H]$.

Suppose now that $\mathbb{P}$ is a complete boolean algebra generated by some name $\tau$ for a subset of an ordinal $\kappa$. Suppose $A \subseteq \kappa$ is generic. In $M[A]$, we compute a generic for $\mathbb{P}$ corresponding to $A$. Now assume $B \subseteq j(\kappa)$ is $j(\mathbb{P})$-generic over $N$, and for all $\alpha < \kappa$, $\alpha \in A$ iff $j(\alpha) \in B$. Is it always true that we can extend $j$ to $\hat{j} : M[A] \to N[B]$? In other words, is it necessarily the case that the generic filter $G$ computed from $A$ has the property that $j[G] \subseteq H$, where $H$ is the generic filter computed from $B$?

Answer 1

The answer is no.

First, let's just consider the latter part of your question, whether the property $j[G]\subset H$ follows from your assumption on $A$ and $B$. It does not in general. Here is an example with forcing that is atomic. Let $\mathbb{P}$ be the Boolean algebra arising from the full support $\kappa$-product of the two-atom forcing. Thus, a condition is an element of $2^\kappa$, with these conditions forming an antichain of atoms. Let $\tau$ be the name of the subset of $\kappa$ where the generic condition has its $1$s. The Boolean values $[[\check\alpha\in\tau]]$ generate $\mathbb{P}$, since if you know all those values, you know which atom was selected. Let $G$ be the generic filter selecting the all-ones atom: $\vec 1=\langle 1,1,1,\cdots\rangle$. So $A=\kappa$. In $j(\mathbb{P})$, let $H$ be the filter selecting all ones up to $\kappa$, and then zeros. So $B=\kappa\subset j(\kappa)$. Since $\kappa$ is the critical point of $j$, we have $\alpha\in A\iff j(\alpha)=\alpha\in B$, since this only refers to $\alpha<\kappa$. But we don't have $j[G]\subset H$, since $j(\vec 1)$ is the all-ones condition in $j(\mathbb{P})$, which is not in $H$. So the pull-back property on the filters does not hold.

(Meanwhile, since both $G$ and $H$ are trivial forcing, the embedding does lift, so this shows that the two versions of your question are not equivalent.)

So let's now get a negative answer also to the lifting version of the question. We simply modify the forcing as follows, to get an example where the embedding does not lift. Let $\mathbb{P}$ be the forcing that either selects an eventually constant atom as above, or else adds a Cohen subset to $\kappa$. So the generic filter is determined by a subset $A\subset\kappa$, whose name $\tau$ will generate the Boolean algebra. Now, let $A$ be the eventually constant all-ones sequence, so $G$ is atomic, but this time, let $H$ be the filter adding a Cohen subset to $j(\kappa)$, but extending the condition having all ones up to $\kappa$, and after that being Cohen generic. Now, the argument as above shows that $\alpha\in A\iff j(\alpha)=\alpha\in B$, but this time the embedding does not lift to $j:M[G]\to N[H]$, since $G$ was trivial and $H$ was not.