Let $X$ and $Y$ be standard Borel spaces, $Y$ uncountable, and $f : X \to Y$ a surjective Borel map. Is it possible that there is a countable ordinal $\alpha$ such that for each Borel set $B \subseteq Y$, the Borel rank of the set $f^{-1}(B) \subseteq X$ is at most $\alpha$?



Can there be an upper bound on the Borel rank of the preimages of Borel sets under a surjective Borel map?
How about this: Construct a Cantor set $E \subseteq X$ on which $f$ is bijective. Then $f$ is a homeomorpism of $E$ onto $f(E)$, and $f(E)$ has Borel subsets of arbitrarily high rank.


