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 $\dot{f} =g$ ?