I've seen the following result:
For elementary $j$, if $G$ is $P$-generic over $V$ and $G^*$ is $j(P)$-generic over $M$ and $j''G\subseteq G^*$ then $j$ can be extended to elementary $j:V[G]\rightarrow M[G^*]$.
This can be proven as:
For $x\in V[G]$ let $\tau$ by a name such that $\tau_G=x$ and define $j(x)=j(\tau)_{G^*}$. Let $x_1,..,x_n\in V$ and $\tau_1,..\tau_n$ be names for the $x_i$. $$V[G]\models\phi(x_1,..x_n)$$ $$\implies p\Vdash \phi(\tau_1,..,\tau_n)\text{ for some } p\in G$$ $$\implies j(p)\Vdash \phi(j(\tau_1),..,j(\tau_n))$$ $$\implies M[G^*]\models \phi(j(x_1),..,j(x_n))$$
The last implication using the fact that $j(p)\in G^*$. Then substituting $\neg\phi$ for $\phi$ we get the iff.
I can't find where the fact that $G^*$ is $j(P)$ generic is used in the proof. As it seems to me that finding a $G^*$ that is $j''P$ generic would be enough for this proof to work and so we could always extend $j$ to $V[G]$ as $j''G$ will always be $j''P$ generic over $M$.
My question is: where is this fact used in the proof above?
I can see however that the claim:
If $G$ is $P$-generic over $V$ and $G^*$ is $j''P$-generic over $M$ then $j$ can be extended to $V[G]\rightarrow M[G^*]$
can't be true as $V[G]$ will satisfy that some filter intersects every dense set in $P$ and $M[G^*]$ will probably not satisfy that some filter intersects every dense set in $j(P)$.
(Like for example if $crit(j)=\kappa$ and $P$ is the Cohen forcing $Fn(\kappa,2).$ Where $j''P$ is just $P$ but $j(P)$ is $Fn(j(\kappa),2)$ and $j''G$ is definitely not $j(P)$-generic over $M$.)
Well, how can you justify that $j''P$ is even an element of $M$ in order for $G^*$ to be $M$-generic for it?
Moreover, how do you justify that $M[G^*]\models\phi(j(x))$ without appealing to the truth lemma? If $G^*$ is not $M$-generic, then why should $M[G^*]$ be a transitive [class] model of set theory?
Finally, we know that $V$ is definable in $V[G]$, and therefore $M$ should be definable in $M[G^*]$ by elementarity. So $V[G]$ satisfies that it is a generic extension of $V$ by the forcing $P$ with the filter $G$. Therefore $M[G^*]$ should also satisfy something similar, but applying $j$ to the whole thing gives us $j(P)$ and not $j''P$.