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 $\alpha$?
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$\begingroup$ My previous comment misunderstood the question. "For each Borel set $B$..." $\endgroup$– Gerald EdgarCommented Mar 4, 2021 at 12:42
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$\begingroup$ @GeraldEdgar Looks like we need our combined efforts to both read the whole question! $\endgroup$– Alessandro CodenottiCommented Mar 4, 2021 at 12:48
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$\begingroup$ Hopefully I can make a sensible comment this time, but I believe that there are no examples if $f$ is injective. Because in that case $f(U)$ is Borel for all open $U\subseteq X$ $\endgroup$– Alessandro CodenottiCommented Mar 4, 2021 at 12:55
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3$\begingroup$ To clarify, is "Polish space" intended in the question instead of "standard Borel space"? The Borel rank of a set depends on which topology is used to define the Borel sets. If $X$ and $Y$ are Polish, then we can put a finer Polish topology on $X$ with the same Borel sets in which $f$ continuous (see Theorem 13.11 of Kechris's Classical Descriptive Set Theory). Since the new topology on $X$ is finer, the Borel rank of any $f^{-1}(B)$ can only go down, so if for every continuous function $f$, the sets $f^{-1}(B)$ have unbounded Borel rank, this also holds for all Borel maps $f$. $\endgroup$– Robert FurberCommented Mar 5, 2021 at 1:58
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1$\begingroup$ @RobertFurber: The reduction to continuous $f$ helps. Exercise 8.8 (ii) of Kechris claims that for continuous $f$ with uncountable image, there is a Cantor set on which $f$ is injective. It's right after the Baire category theorem so maybe the proof is not too hard (I didn't try it). $\endgroup$– Nate EldredgeCommented Mar 5, 2021 at 5:45
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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.
answered Mar 4, 2021 at 13:03
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$\begingroup$ @ArkadiPredtetchinski: "Cantor set" means it's compact, right? So the Borel rank of $f^{-1}(B) \cap E$ is at most 1 greater than that of $f^{-1}(B)$. $\endgroup$ Commented Mar 4, 2021 at 16:08
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$\begingroup$ I think I see how to build such a Cantor set by reducing to the case that $Y$ is perfect and building up a tree of open sets with disjoint closures, like in the usual proof that every perfect Polish space contains a Cantor set. Is this what is intended, or is there a simpler way that I'm missing? $\endgroup$ Commented Mar 5, 2021 at 4:12
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1$\begingroup$ @RobertFurber ...Yes, that is the method I had in mind. Choose complete metrics, and make sure the diameters in your tree go to zero. $\endgroup$ Commented Mar 5, 2021 at 17:03
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1$\begingroup$ @RobertFurber you can also reduce to continuous $f$ as in the comments under the question and show that $f$ is injective on a generic element of $K(X)$ (and a generic element of $K(X)$ is perfect). Of course this is more work than your approach, but you get a stronger result in the end $\endgroup$ Commented Mar 6, 2021 at 9:31
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$\begingroup$ @AlessandroCodenotti This seems (from context) to be the method of proof intended in Kechris's exercise mentioned by Nate Eldredge. This looks like a useful fact to know. $\endgroup$ Commented Mar 6, 2021 at 17:40