# complete embeddings of boolean algebras and preservation of stationarity

Define a complete embedding of Boolean algebra as an homomorphism of Boolean algebras which preserves also the sup and inf operations. Notice that if $\mathbb{B}$ and $\mathbb{D}$ are complete boolean algebras, $i:\mathbb{B}\to\mathbb{D}$ is a complete embedding and $G$ is $V$-generic for $\mathbb{D}$, then $H=i^{-1}[G]$ is $V$-generic for $\mathbb{B}$.

I'm curious to know if the following can be the case:

Assume $\mathbb{B}$ and $\mathbb{D}$ are complete stationary set preserving boolean algebras. Can there be two distinct complete embeddings $i_0:\mathbb{B}\to\mathbb{D}$, $i_1:\mathbb{B}\to\mathbb{D}$ such that if $G$ is $V$-generic for $\mathbb{D}$ and $H_j=i_j^{-1}[G]$ are the corresponding $V$-generic filters for $\mathbb{B}$ induced by the respective $i_j$, we can have that there is a name $\tau$ in the forcing language for $\mathbb{B}$ such that:

$\|\tau$ is a stationary subset of $\omega_1\|=1_{\mathbb{B}}$

$V[G]\models\sigma_{H_0}(\tau)$ is a stationary subset of $\omega_1$

$V[G]\models\sigma_{H_1}(\tau)$ is non-stationary

-
I should say that my interest in this question lies in the fact that a positive answer would provide an example of two substatntially different ways to embed $\mathbb{B}$ into $\mathbb{D}$ as a complete boolean sub-algebra. So it relates to the problem of under which conditions it is possible to have inside $\mathbb{D}$ two distinct complete subalgebras which are isomorphic. – matteo viale Jul 22 '11 at 4:06
Hi Matteo! Nice to see you in MO. – Andrés E. Caicedo Jul 22 '11 at 7:02
What is $\sigma$? Or $\sigma_{H_i}(\tau)$? – Stefan Geschke Jul 22 '11 at 7:33
Matteo, when you say "stationary-set preserving", are you talking just about subsets of $\omega_1$, or do you mean stationary subsets of any cardinal, or generalized stationarity? Stefan, I think that is Matteo's notation for the value of a name by a filter. i.e. what is elsewhere denoted $\text{val}(\tau,H_i)$ or $(\tau)_{H_i}$ or $(\tau)^{H_i}$. – Joel David Hamkins Jul 22 '11 at 9:41
I apologize for my unclear notation, it is exactly meaning what Joel was suggesting..... – matteo viale Jul 22 '11 at 16:17

Let $\mathbb{B}=\text{Add}(\omega_1,1)$ be the forcing to add a Cohen subset $S\subset \omega_1$, and let $\mathbb{D}$ be the forcing that first adds such a set $S$, and then shoots a club through it $C\subset S$. Note that $\mathbb{B}$ is countably closed in $V$ and therefore stationary-set preserving, and the generic Cohen set $S$ that is added is both stationary and co-stationary. Further, the forcing $\mathbb{D}$ is stationary-set preserving over $V$, because by a bootstrap argument we may find a dense set of conditions $(s,c)$, where $s\subset \omega_1$ is bounded and $c\subset s\cup\text{sup}(s)$ is closed, and the set of such conditions in $\mathbb{D}$ is countably closed. Thus, $V[S][C]$ is stationary-set-preserving over $V$, even though it is not stationary-set-preserving over $V[S]$.
Notice that we may completely embed $\mathbb{B}$ into $\mathbb{D}$ in the natural way, since $\mathbb{D}$ was described as first adding $S$, and then shooting a club through it.
But we may also embed $\mathbb{B}$ into $\mathbb{D}$ in a different way: by first applying the automorphism of $\mathbb{B}$ that flips all bits. This automorphism in effect replaces $S$ with its complement, so that under this embedding, the club gets added to the complement of $S$.
Thus, if $\tau$ is the name of the generic set $S$ added by $\mathbb{B}$, then $1_{\mathbb{B}}$ forces that $\tau$ is stationary, and with the first embedding we have that $\text{val}(\tau,H_0)=S$, which remains stationary and in fact containing a club in $V[S][C]$, but with the second embedding we have $\text{val}(\tau,H_1)=\omega_1\setminus S$, which is non-stationary in $V[S][C]$.