Can an almost injective quotient map remove interior?

Question

Let $X$ be a second countable compact Hausdorff space without isolated points and let $q: X\to Y$ be a quotient map onto another compact Hausdorff space. Suppose that $X$ has a dense $G_{\delta}$ subset $D$ such that the restriction of $q$ to $D$ is injective. Let $V$ be a closed subset of $Y$ without interior. Can I conclude that $W=q^{-1}(V)$ has no interior?

Answer 1

Consider the following counterexample. Let $X=[0,1]\times [0,1]\bigcup \{2\}\times [0,1]$ be a subspace of the space $\Bbb R^2$, $Y=[0,1]\times [0,1]$, and $q$ is a retraction of $X$ onto $y$ such that $f((2,y))=(1,y)$ for each $y\in [0,1]$. Since $q$ is a continuous map between compact Hausdorff spaces, it is closed and so quotient. Let $D=[0,1)\times [0,1]\bigcup \{2\}\times [0,1]$. Then $D$ is a dense $G_\delta$ subset of $X$ and $f|D$ is injective. Let $V=\{1\}\times [0,1]$. Then $V$ is a closed subset of $Y$ with empty interior, whereas $q^{-1}(V)= \{1,2\}\times [0,1]$ contains an open subset $\{2\}\times [0,1]$ of $X$.