Boolean-valued model and the use of generic ultrafilter in ZFC

Question

So I asked the question about generic filter; but I was also reading http://math.mit.edu/~tchow/forcing.pdf which is a forcing (in ZFC) guide for dummies. Then I was struck with the part where it basically declared that use of generic ultrafilter $U$ makes $M^{\mathbb{B}}/U$ isomorphic to standard transitive model of ZFC. I don't get why this would be the case - as of my knowledge in basic set theory and some model theory knowledge I know of, this must be something related to element relation being standard, but I just don't see how the use of generic ultrafilter would preserve original element relation.

By the way, I know what generic filter is, so no need for defining it :)

Answer 1

I don't think it's supposed to be obvious that this is the case. Chow is just asserting that it happens to be, and you'll need to refer to a more thorough treatment of forcing to learn why it is so.

Fair warning of this is given in the introduction:

This paper does not solve this open exposition problem, but I believe it is a step in the right direction. My goal is to give a rapid overview of the subject, emphasizing the broad outlines and the intuitive motivation while omitting most of the proofs.

(My emphasis).

Answer 2

Standard models, as you may recall, are transitive models whose $\in$ membership is the true $\in$ membership of the universe.

To verify that $M^\Bbb B/U$ is a standard model, one has to verify that it is transitive and that its $\in$ relation is the same as the universe. Recall that every element in $M^\Bbb B/U$ is an interpretation of a name by the filter $U$. Since names have a naturally assigned rank, it is usually the wise thing to use it and prove things by induction.

Both proofs are essentially that. For transitivity, suppose that $\dot x$ is a name such that for all $\dot y$ whose rank is below that of $\dot x$, we have that $\dot y^U\in M^\Bbb B/U$. Therefore $\dot x^U=\{\dot y^U\mid [\![\dot y\in\dot x]\!]\in U\}\in M^\Bbb B/U$, and all $\dot y^U\in\dot x^U$ are already assumed to be in $M^\Bbb B/U$. And therefore $\dot x^U\subseteq M^\Bbb B/U$.

In a similar fashion one proves that the $\in$ relation agrees with the original one, by unwinding the definition of $y\in x$ and $[\![\dot y\in\dot x]\!]\in U$.