Sweet amalgamation extends partial Boolean isomorphisms

First record the one-sided extension from the source result [F5]. Given a sweetness model on $P$, complete subalgebras $C_0,C_1\le BA(P)$ and a complete isomorphism $g:C_0\to C_1$, take a disjoint copy $P^{\mathrm{\prime }}$ with its copy isomorphism $c:BA(P)\to BA(P^{\mathrm{\prime }})$. Amalgamate the old copy and the new copy over $C_1$, using the two complete embeddings $C_1\hookrightarrow BA(P),b⟼c(g^{-1}(b))\in BA(P^{\mathrm{\prime }})\mathrm{.}$ If $j_0,j_1$ are the canonical embeddings into the amalgam, then $j_0(g(a))=j_1(c(a))(a\in C_0)\mathrm{.}$ Consequently the map $g^{\mathrm{\prime }}(j_0(x))=j_1(c(x))$ is a complete isomorphism from the entire old $BA(P)$ onto the second canonical copy and satisfies $g^{\mathrm{\prime }}(j_0(a))=j_0(g(a))$ for $a\in C_0$. The one-sided source conclusion recorded in [F5] equips this amalgam with a sweetness model extending the first, old factor. It places no extension requirement on the disjoint second factor. This is the twisted two-embedding construction; an untwisted identity amalgam would not extend $g$.

Starting from $(U_0,h_0)=(P,h)$, alternate this one-sided construction with the same construction applied to the inverse. Thus obtain an increasing $\omega$-chain of sweetness models $U_l$, complete subalgebras $B_l^1,B_l^2\le BA(U_l)$ and coherent complete isomorphisms $h_l:B_l^1\to B_l^2$ such that $B_{2l+1}^1=BA(U_{2l}),B_{2l+2}^2=BA(U_{2l+1}),$ and every $h_{l+1}$ extends $h_l$. Each successor sweetness model extends the previous fixed model, not merely its forcing-equivalence class.

Coherence is the displayed identity in step 1.1 at each even step and its inverse analogue at each odd step. The canonical old copy is complete at every successor and its sweetness presentation is fixed by the one-sided extension conclusion in [F5].

Each $U_l$ is sweet and $U_{l+1}$ extends the model of $U_l$. Therefore the direct union forcing $P^{\ast }={\bigcup }_l{U}_l$, with the union dense set and stabilised relations, is sweet, and every $U_l$ is complete in $P^{\ast }$ by [F2].

The same construction is compatible with a previously fixed sweetness-model embedding: start with that presentation and use [F5]'s first-factor extension conclusion at every successor.

Put $C={\bigcup }_{l}BA(U_l)$ through the coherent complete embeddings and $f_0={\bigcup }_l{h}_l$. The algebra $C$ is a Boolean algebra, but is not asserted to be complete: every finite Boolean calculation occurs in one stage, whereas an arbitrary subset of $C$ need not. Coherence makes $f_0$ well defined. The domain is all of $C$ because every element of $BA(U_{2l})$ lies in $B_{2l+1}^1$, and the range is all of $C$ because every element of $BA(U_{2l+1})$ lies in $B_{2l+2}^2$. Thus $f_0$ is a Boolean automorphism of $C$ extending $h$.

Let $B^{\ast }=BA(P^{\ast })$. The canonical image of $P^{\ast }$ is order-dense in $B^{\ast }$ and lies in $C$, so $C$ is order-dense in $B^{\ast }$. The source completion clause recorded in [F5] gives the unique extension of $f_0$ to a complete Boolean automorphism $f$ of $B^{\ast }$. Explicitly it is determined by $f(b)=\bigvee \{f_0(c):c\in C,c\le b\},$ and the corresponding formula for $f_0^{-1}$ supplies the inverse. This is a completion step after the alternating direct union; it does not identify $C$ with a complete direct limit.

The complete algebra $B^{\ast }$ has the sweet dense forcing presentation $P^{\ast }$; no unsupported transport of the equivalence relations through the possibly noninjective completion map is used. Completeness of the stage embeddings puts the original $B$ completely inside $B^{\ast }$. Hence $B^{\ast }$ is a sweet complete Boolean algebra in this retained-presentation sense, and $f$ is the required automorphism, compatible with the previously fixed model.