I just read a quite sketchy proof, where some details were skipped and it appears to me that the proof uses some of the following things. However a (probably very easy) question came up: 


Assume that we force with a Boolean algebra of the form $P(A) / I$, where $A$ is an arbitrary set and $I$ is an ideal over $A$; i.e. our forcing conditions are $I$-postive subsets of $A$ and $p \le q$ iff $p \supset q$.


Moreover assume that $P(A) /I$ forces that $\dot{f}$ is a function in $V$ from $A$ to $V$ (where $V$ denotes our ground model), then it follows that there exists a maximal antichain $(S\_{\alpha} : \alpha < \lambda)$ in $P(A)/I$ and corresponding functions (lying in $V$ ) $g\_{\alpha}$:$A$ $\to V$ such that $S\_{\alpha} \Vdash \dot{f} = g\_{\alpha}$.


Now if the antichain can be written in a way such that the representatives of the $S\_{\alpha}'s$ are even pairwise disjoint then the partial functions $g\_{\alpha}\upharpoonright S\_{\alpha}$ can be glued together to a new function $g$ and now
it should follow that $P(A) / I$ forces that $\dot{f} = g$. 


My question now is: How does one see that the $P(A) /I $ forces $\dot{f} =g$ ?



Gluing functions together in the generic extension
The generic subset $G\subseteq P(A)/I$ adjoined by your forcing can be viewed as an ultrafilter on $A$ (disjoint from the ideal $I$). Your glued-together $g$ will (be forced to) be equal to $f$ on a set in this ultrafilter; that is, $g$ and $f$ will represent the same element of the generic ultrapower of $V$. But, as Justin Palumbo noted, they won't generally be equal everywhere.


