Given two forcing notions $\mathbb{P},\mathbb{Q}$ and $i:\mathbb{P}\rightarrow\mathbb{Q}$ a complete embedding in $V$ it is well-known that the following factorization holds
$$\mathbb{Q}\sim \mathbb{P}\ast \mathbb{Q}/G$$
where $\mathbb{Q}/G$ is the $V[G]$ forcing notion $\{q\in \mathbb{Q}: \forall p\in G\; q\;\text{is compatible with}\; i(p)\}$.
If we are given an elementary embedding $j: V\longrightarrow M$ and a forcing notion $\mathbb{P}\in V$ then $j$ is not necessarily a complete embedding between $\mathbb{P}$ and $j(\mathbb{P})$ so at first instance we do not have such factorization.
My question is the following. Suppose we were given $m\in j(\mathbb{P})$ a master condition. Recall that such a condition is a master condition if for every $D\subseteq \mathbb{P}$, $D\in V$ dense and every $q\in j(\mathbb{P})\downarrow m$ there exists $p\in D$ such that $j(p)$ and $q$ are compatible. Then, is it true that the following factorization holds?
$$j(\mathbb{P})\downarrow m\sim \mathbb{P}\ast j(\mathbb{P})/G\downarrow m$$
where $G\subseteq \mathbb{P}$ is $V-$generic.
Thanks in advance.
I think you need to replace $j(\mathbb{P})\downarrow m$ with $j(\mathbb{P})/m$ which is the subset of $j(\mathbb{P})$ consisting of elements compatible with $m$. Then since $m$ is a master condition, $m\Vdash^V_{j(\mathbb{P})} j^{-1}(\dot{H})$ is generic over $V$. Where $\dot{H}$ is the name for the generic filter. Hence there exists $q\in \mathbb{P}$ such that $j: \mathbb{P}/q \to j(\mathbb{P})/m$ is a complete embedding. Hence $j(\mathbb{P})/m\simeq \mathbb{P}/q *( j(\mathbb{P})/m)/\dot{G}$