Justification of the assumption that a generic ultrafilter exists. I do not understand why the assumption of existence of a genric ultrafilter $G$ is justified.
For the transitive model $V=\{x:x=x\}$ of $ZFC$ and a complete Boolean algebra $B$ in $V$, can we prove that if $V[G]\vDash\lnot\varphi$, then in the $V^{B}$ the Boolean value of $\varphi$ is not $1$? I think if we can do, then this fact may applied to justisfy the assumption that a generic ultrafilter on $B$ exists.
 A: When forcing over the universe, we can't literally have a generic ultrafilter $G\subseteq B$ and a transitive extension $V[G]\supsetneq V$ (hopefully it's at least clear that the latter is absurd).
So the question is not whether or not we can rightfully assume one exists, it's whether it's safe operationally to pretend. The answer to that is that it is, at least if we're just doing run-of-the-mill consistency proofs. That is, given forcing notion, if we assume a generic object and forcing extension exist, then the properties we can show hold in the "forcing extension" on this basis are relatively consistent with whatever we assumed of the universe.
The usual procedure is "just don't worry about it". Imagining a real transitive generic extension is generally the most intuitive way to lay the argument out, and for any informal argument we can make in this manner, there are several well-known ways of converting into a rigorous proof of relative consistency. It would be a waste to go through the rigamarole every time when it's always the same argument... think of it as a piece of the proof that is 'factored out'. It may be necessary to go into the subtleties in certain specialized research on forcing if there's something new there, but not for just doing usual consistency proofs via generic extension.
As for why this works, the easiest way to see it is probably the countable transitive model approach. The specific structure of the ground model doesn't usually matter (as opposed to what axioms it satisfies), so there's not really any disadvantage to just assuming the ground model is a countable transitive model. Then we can proceed rigorously with the generic extension, since these things provably exist. This method tends to be favored as an approach for beginners since it is tangible in this way. However it's not quite enough to just plop 'countable' in front of 'transitive model' at every point in the argument. There is a different metamathematical issue that gets factored out here: the existence of a transitive set-model of the theory is not relatively consistent with the theory. Again, it is well-known how to patch this up, so it is often patched up silently.
Within the Boolean-valued model (BVM) approach (or a forcing-relation-only approach) we don't really need the extension since we prove the relevant preservation theorems directly, but it's often more intuitive to imagine we have one. One approach is to simply think of it as a convenient short-hand for what's really going on under the hood since everything is readily translatable into a pure BVM argument. But if we really want to make contact, we can see that in fact in the BVM, there is a generic ultrafilter over the embedded $V$ and $B.$ Thus we can imagine the construction of the generic extension takes place in the BVM. Of course the Boolean-valued model thinks that its embedded $V$ is not the whole universe, so the obvious inconsistency of the "extension" is no longer a problem. Then we can get the consistency proof from the extension in the usual way (an inconsistency in the target theory would bubble up from the forcing extension, into the Boolean-valued model, and finally into the ground model where it yields an inconsistency in the base theory).
Perhaps more compelling, we can define a quotient $V^B/U$, which will be a (not-necessarily-well-founded) two-valued model that embeds $V.$ This model has $V^B/U\models \varphi\iff \Vert \varphi\Vert_B\in U$ (note $U$ is not assumed generic) and as such thinks there is a generic ultrafilter for its embedded $V$ and $B,$ so we can construct the generic extension there too.
