In his paper Ensembles génériques d'entiers, Bukovský says the following (recall that if $M$ is a transitive model of ZF containing all ordinals and $x$ is a subset of $M$, then $M[X]$ denotes the least model such that that $M\subset M[x]$ and $x\in M[x]$):
Let $\mathcal{B}^{\#}$ be the set of all Borel subsets of the Cantor space $C^{\#}={}^{\omega}2$. Let $\mathcal{B},\mathcal{C}$ be the same sets as seen in $L$. It is well known that the natural injection $\#:\mathcal{B}\longrightarrow\mathcal{B}^{\#}$ preserve countable unions. Let $\chi_{x}\in\mathcal{C}^{\#}$ be the characteristic function of $x\subset\omega$. We define the ultrafilter $j(x)$ over $\mathcal{B}$ by: $u\in j(x)$ if and only if $\chi_{x}\in\#(u)$. Evidently, $j(x)$ is closed under countable intersections and $L[x]=L[j(x)]$.
In the slides of a talk that he gave in 2014 he says:
Let $M\subseteq V$ be transitive models of ZFC. Assume that $a\subseteq\omega$, $a\notin M$ and $a\in V$. Let $\mathcal{B}$ be the family of all Borel subsets of the Cantor space ${}^{\omega}2$. According to Solovay, there exists a mapping $\phi:\mathcal{B}^{M}\longrightarrow\mathcal{B}^{V}$ preserving complements and unions of countable families belonging to $M$. If $a\in{}^{\omega}2\cap V$, we set $j(a)=\{A\in\mathcal{B}^{M}:a\in\phi(A)\}$. Evidently, M[a]=M[j(a)].
We know after Solovay that if $G$ is $\mathcal{B}^{\ast}$-generic, with $\mathcal{B}^{\ast}$ the collection of Borel sets which are not null, then there is a unique $x\in {}^{\omega}{\omega}$ such that for any closed code $c\in V$, then $x\in A_{c}^{V[G]}$ if and only if $A_{c}^{V}\in G$, and $V[x]=V[G]$. Here, $A_{c}$ denotes the set coded by $c\in {}^{\omega}\omega$ (see Kanamori's The Higher Infinite, pag. 137). This result resembles the claim of Bukovský that I mention. However, its hypothesis have been relaxed (for instance, he doesn't consider only the Borel subsets which are not null, but the whole family of Borel subsets) and I cannot find a way to adapt Solovay's proof. So my question(s):
Why is it true (and why is it evident) that $M[a]=M[j(a)]$ with $a$ and $j(a)$ as above?
Thank you.
P.S. It is my first try with descriptive set theory arguments so it might be the case that I cannot find a solution of this due to the lack of background. If that's what you think, just a hint of what I should read before attacking this problem will be greatly appreciated.
For models $M\subseteq W$ let us write $\phi_{M, W}$ for the natural map $\mathcal B^M\rightarrow \mathcal B^W$. Here are a few nice properties of these maps:
a) $\phi_{M, W}(A)\cap M= A$ for any $A\in \mathcal B^M$
b) $\phi_{M, W}(N_s^M)=N_s^W$ where $N_s=\{x\in 2^\omega\mid s\subseteq x\}$ is the basic open set with base $s\in 2^{{<}\omega}$.
c) $\phi$ factors nicely through interemediate models: If $M\subseteq N\subseteq W$ are inner models then $\phi_{M, W}=\phi_{N, W}\circ\phi_{M, N}$.
$M[a]=M[j(a)]$ can be seen from these three properties alone. Give it a try yourself! Otherwise read the rest of the answer.
To prove $M[a]=M[j(a)]$ we only need to show
To see 1, note that for any $n<\omega$, $a\upharpoonright n$ is the unique $s\in 2^n$ with $a\in N_s^{V}$ which by b) is the unique $s\in 2^n$ with $N_s^M\in j(a)$. Thus $$a=\bigcup\{s\in 2^{<\omega}\mid N_s^M\in j(a)\}\in M[j(a)]$$
Next let's show 2.: From a) and c) it follows that $j(a)=\{A\in\mathcal B^M\mid a\in\phi_{M, M[a]}(A)\}$ and the letter set is in $M[a]$ as it can be defined from $\mathcal B^M, a, \phi_{M, M[a]}\in M[a]$.