Here $Fn(I,J)$ denotes all partial functions from $I$ to $J$ with finite domain.
To check that if $\mathbb{P}=Fn(I,J)$ where $I,J\in M$ and are infinite and $G$ is a $\mathbb{P}$-generic filter over $M$ then $\cup G \in M[G]$ and $\cup G$ is a function from $I$ onto $J$ we define the following dense sets:
For every $i\in I$ let
$$D_i = \{q\in Fn(I,J) \mid i\in dom(q)\}$$
and for $j\in J$ let $$R_j = \{ q\in Fn(I,J) \mid j\in ran(q)\}$$
I understand how $G$ meeting all of these dense sets makes $\cup G$ a function from $I$ onto $J$, I also understand that if $G$ is $\mathbb{P}$-generic over $M$ and all of these sets are in $M$ then $G$ has to meet them all. What I can't prove is that all of these sets are in $M$ from just assuming $I,J\in M$. I'm sure this is an absoluteness argument but my course went quite quickly through that section and I can't figure it out on my own.
$\textbf{In brief}$: If $M$ is a ctm for ZFC and $I,J\in M$ are infinite sets, why are $Fn(I,J)$ and the $D_i$'s and $R_j$'s defined above necessarily elements of $M$.
Look at $Fn(I,J)^{M}$, i.e. look at the formula $\phi(x,p,q)$ such that $Fn(I,J)$ is (provably in ZFC) the unique set $x$ such that $\phi(x,I,J)$ holds and let $x \in M$ be such that $M \models \phi(x,I,J)$. This set $x$ is $Fn(I,J)^{M}$. So $Fn(I,J)^{M}$ exists (as an element) in $M$ (since $M$ is a model of ZFC and such an $x$ provably exists). Now, since 'being a function' and 'being finite' are absolute properties between any transitive models of the language of set theory, $M$ computes every element of $Fn(I,J)$ correctly, so $Fn(I,J)^{M} \subseteq Fn(I,J)$. On the other hand, all elements of $Fn(I,J)$ are elements of $M$ (essentially, because they are finite) and given any $p \in M$, $M$ will be able to tell whether $p \in Fn(I,J)^{M}$ (again, by absoluteness of 'being a function' and 'being finite'). Hence $Fn(I,J) \subseteq Fn(I,J)^{M}$.