I'm having some trouble parsing the following passage from the Handbook of Set Theory:
I'm having trouble checking the statement that is "routine to check." My thought is that since $D,\mathbb{P}\in M$ and $k:M\to N$ is elementary, we should get that $k(D)\in N$ is dense in $k(\mathbb{P})\in N$. Since $H$ is $(\mathbb{P},N)-$generic, it must then meet $k(D)$ -- but I don't see how to guarantee that the points of $k(D)$ that $H$ meets are in the image of $k$. It seems like this would be the time to apply the fact that $q\in H$ is a master condition, but I don't see how $q$ being compatible with an element of $k(D)$ implies that $H$ reaches below this element. It seems like $q$ could be compatible with multiple elements of $k(D)$ and that $H$ may "pick a path" that only contains points outside the image of $k$. My best guess is that we would need to use the fact that $k$ is elementary to finish the argument, but I'm not sure how to proceed. Any advice would be appreciated!
Edit: Should the definition state that for every $r\leq q$ there is some $p\in D$ such that $r$ is compatible with $k(p)$? In this case the set $\{r\in k(\mathbb{P})| r\Vdash k^{-1}(H)\cap D\neq\emptyset\}$ is dense below $q$, so $q$ would force this desired property. Conversely, if there were some $r\leq q$ that was not compatible with any $k(p)$ for $p\in D$, it would be the case that $r\Vdash k^{-1}(H)\cap D=\emptyset$ and so $q$ would not force the desired property.
In particular, if $q$ did not force that $k^{-1}(H)\cap D\neq\emptyset$ or that $k^{-1}(H)\cap D=\emptyset$, then there would be some $r_1,r_2\leq q$ with $r_1\Vdash k^{-1}(H)\cap D\neq\emptyset$ and $r_2\Vdash k^{-1}(H)\cap D=\emptyset$. $r_1$ would show that $q$ is compatible with some $k(p)$ for $p\in D$, as the given definition states, but $r_2$ would show that $q$ does not force the fact that $k^{-1}(H)$ meets $D$ and makes it the case that any $G$ with $k(G)\subset H$ is not $(\mathbb{P},M)-$generic.