Retracts and dense $G_{\delta}$ sets

Question

Suppose $X$ is a complete separable metric space and $Y\subset X$ is a retract of $X$ with retraction $r\colon X\longrightarrow Y$. I am interested in the following question: Given a dense $G_{\delta}$-set $G\subset X$. Is $r(G)$ a dense $G_{\delta}$ subset of $Y$ (in the relative topology)?

Answer 1

Of course not necessarily, since in general $G_\delta$-sets are not preserved by continuous (and even open) maps (and even retractions). Let $X=\mathbb R^2\times\mathbb N$ be the product of spaces $\mathbb R^2$ and $\mathbb N$ endowed with standard topologices. Put $Y=\mathbb R^2\times\{1\}\subset X$ and $r:X\to Y$ be the projection onto the first coordinate. Let $K\subset [0,1]$ be a Cantor set and $\{q_n\}$ be an enumeration of rationals $\mathbb Q$. For each $n$ put $A_n=q_n+K$, $B_n=\mathbb R^2\setminus(\mathbb R\setminus\{0\})\cup A_n\times\{0\}$, and $G=\bigcup B_n\times\{n\}$. Since each set $B_n$ is a dense $G_\delta$-subset of the space $\mathbb R^2$, $G$ is a dense $G_\delta$-subset of the space $X$. From the other hand, the set $r(G)=\bigcup \mathbb R^2\setminus(\mathbb R\setminus\{0\})\cup A_n\times\{0\}$ is not a $G_\delta$ set of the space $Y$, because the set $A=r(G)\cap\mathbb R\times\{0\}$ is not a $G_\delta$ set, see my previous answer.