Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 19 / 4 / iii / Solution

For a maximal forcing antichain $A\in M$, the set $D_A=\{q:\exists a\in A(q\ge a)\}$ is dense. Maximality ensures each condition is compatible with some $a\in A$, and a common strengthening lies in $D_A$. A generic filter meets $D_A$ and hence, by closure toward weaker conditions, meets $A$. Directedness makes any two of its conditions compatible, so it contains at most one member of an forcing antichain. Thus $|G\cap A|=1$.

Conversely, suppose this equality holds for every ground-model maximal forcing antichain. For any dense $D\in M$, use Choice in $M$ to select an forcing antichain $A\subseteq D$ maximal among forcing antichains contained in $D$. It is maximal in $P$ as well: a condition incompatible with every member of $A$ would have an extension in $D$ still incompatible with every member, enlarging $A$. The hypothesis gives a member of $G\cap A\subseteq G\cap D$. Thus $G$ meets every ground dense set and is generic.

This proves the maximal-antichain criterion for genericity in both directions.