Let $X$ and $Y$ be compact Hausdorff spaces and let $\varphi:X\to Y$ be continuous with a property that if $A$ is a nowhere dense zero-set in $Y$, then $\varphi^{-1}(A)$ is nowhere dense in $X$. Let $Z=\varphi(X)$.
Does $\varphi$ still have the analogous property as a map into $Z$?
Note that the condition implies that the set $J_{Z}$ of $f\in C(Y)$ which vanish on $Z$ is a $\sigma$-ideal in $C(Y)$, which means that it contains all existing supremums of countable sets in $J_{Z}$. Such $Z$'s could be considered countable analogues of regular closed sets, because $Z$ is regular if and only if $J_{Z}$ contains existing supremums of all sets.
Best Answer
The answer here is negative: for $Y$ take the remainder $\beta\omega\setminus\omega$ of the Stone-Cech remander of the discrete space $\omega$ of finite ordinals. In the space $Y$ take any countable discrete subspace $D$ and let $Z$ be the closure of $D$. Since $Y$ has no isolated points, the space $D$ is nowhere dense in $Y$ and so is its closure $Z$. Since $D$ is countable and discrete in the compact space $Z$, the remainder $R=Z\setminus D$ is a nonempty functionally closed nowhere dense set in $Z$. Consider the space $X=(Z\times\{0\})\cup (R\times\{1\})$ and the natural projection $\varphi:X\to Z\subseteq Y$.
Observe that the set $R$ is functionally closed and nowhere dense in $Z$ and its preimage $\varphi^{-1}[R]$ contains the nonempty clopen subset $R\times\{1\}$ of $X$.
On the other hand, each nonempty $G_\delta$-subset of the space $Y=\beta\omega\setminus\omega$ has nonempty interior in $Y$. So, $Y$ contains no functionally closed nowhere dense subsets and hence the function $\varphi:X\to Y$ has the desired property: for every nowhere dense functionally closed set $A$ in $Y$ the preimage $\varphi^{-1}(A)$ has any desired property, in particular is nowhere dense in $X$.