# Do $G_\delta$-measurable maps preserve dimension?

This question (in a bit different form) I leaned from Olena Karlova.

Question. Let $f:X\to Y$ be a bijective continuous map between metrizable separable spaces such that for every open set $U\subset X$ the image $f(U)$ of type $G_\delta$ in $Y$. Is $\dim(X)=\dim(Y)$? Does $\dim(Y)=0$ imply $\dim(X)=0$?

Comment. The Jane-Rogers Theorem implies that the answer to this question is affirmative if the space $X$ or $Y$ is analytic. In this case the inverse map $f^{-1}$ is piecewise continuous in the sense that $Y$ admits a countable closed cover $\{Y_n\}_{n\in\omega}$ such that $f^{-1}|Y_n$ is continuous and hence a homeomorphism of $Y_n$ and the closed subset $f^{-1}(Y_n)$ of $X$. So $$\dim(X)=\sup_{n\in\omega}\dim(f^{-1}(Y_n))=\sup_{n\in\omega}\dim(Y_n)=\dim(Y)$$ by the countable sum theorem for dimension. A similar argument works if the space $Y$ is hereditarily Baire.

All this shows that a counterexample to this question if exists should be somewhat exotic.

• Is it even clear that $\dim(Y)=0$ implies $\dim(X)\leq 1$? – D.S. Lipham Dec 7 '19 at 16:09