# Obtaining elements of a generic extension from a Boolean-valued model of ZFC

Let $$\mathcal{M}$$ be a countable transitive standard-model of ZFC.
Let $$B \in \mathcal{M}$$ be a boolean algebra that is complete in $$\mathcal{M}$$.
Further, let $$\mathcal{M}^{(B)}$$ be the corresponding boolean model of ZFC.

Now we consider an $$\mathcal{M}$$-generic ultrafilter $$U$$ on $$B$$.
According to Jech (Set Theory, p.216), the elements of the generic extension $$\mathcal{M}[U]$$ can be obtained via the map $$\cdot^U: x \mapsto x^U := \lbrace{y^U \mid x(y) \in U \rbrace}$$ where $$x \in \mathcal{M}^{(B)}$$. But when reading Bell (Set Theory Boolean-Valued models and Independence Proofs, p.91) the following map is given: $$i(x) = \lbrace i(y) \mid \Vert y \in x\Vert \in U \rbrace$$ where $$x \in \mathcal{M}^{(B)}$$. I assume that the following holds: $$\forall x \in \mathcal{M}^{(B)}: x^U = i(x). \ \ \ [*]$$

But at the same time I am not yet fully convinced as $$u(x) \leq \Vert x \in u \Vert$$ for $$x \in \text{dom}(u)$$.

Can anyone verify that $$[*]$$ (or a variant of that statement) holds (and give a short reason why)?

Simply use induction on $$\mathbb{B}$$-names (or the rank of $$\mathbb{B}$$-names, if you are not familiar with applying induction directly to $$\mathbb{B}$$-names.) Suppose that $$i(z)=z^U$$ holds for all $$z\in\operatorname{dom} x$$.
Since $$x(z)\le \|z\in x\|$$, we have $$x^U\subseteq i(x)$$. Conversely, assume that $$\|y\in x\|\in U$$. Then $$\sum_{z\in\operatorname{dom}x}x(z)\cdot \|z=y\| \in U.$$ By genericity of $$U$$, we can find $$z\in\operatorname{dom} x$$ such that $$x(z)\cdot\|z=y\|\in U$$. This proves
1. $$i(z)=z^U\in x^U$$, and
2. $$i(z)=i(y)$$.
Hence $$i(y)\in x^U$$. This proves $$i(x)\subseteq x^U$$ (since $$i(y)\in i(x)$$ means $$\|y\in x\|\in U$$.)