Jech's Set Theory, 3rd edition, has the following result:
Lemma 14.13: (i) In the ground model M, let Q be the separative quotient of P and let h map P onto Q such that (14.5) holds. If $G \subset P$ is generic over M then $h(G) \subset Q$ is generic over M
The (14.5) condition referenced is i: $x \leq y$ implies $h(x) \leq h(y)$ and ii: $x$ and $y$ are compatible in $P$ if and only if $h(x)$ and $h(y)$ are compatible in $Q$.
In particular, since every generic set over a model $M$ is a filter, this implies that $h(G)$ must be a filter as well, and hence that it's closed above in the partial ordering on $Q$. But I can't for the life of me figure out how to prove this.
In other words: given $p \in h(G), p \leq q$, how can we verify that $q \in h(G)$? Thanks in advance for any help!
To solve this we will prove something stronger:
If $[p] \in h(G) \wedge [p] \le [q] \rightarrow q \in G$ this is stronger than the main statement because this actually implies that every representative of $[q]$ lies in $G$. Now it follows from the definition that $[p] \le [q]$ implies that for all $r$, $r$ is compatible with $p \rightarrow r$ is compatible with $q$. Now consider the set $D = \{r \in P: r$ is incompatible with $p$ or $r \le q\}$. You can easily check that $D$ is dense in $P$ and so $G \cap D \neq \emptyset$. Then there exists $r$ such that $r\in G$ and notice that $r$ can't be incompatible with $p$ so $r \le q$ and lastly we have $q \in G$ as needed.