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Let $X,Y$ be compact Hausdorff spaces, fix a continuous surjection $f\colon X\to Y$.

Suppose $B\subseteq Y$ has the property that $f^{-1}[B]$ is Borel. Is $B$ necessarily Borel?

Note that if $X$ is metrisable, then this is true by Suslin's theorem, as then $f^{-1}[B]$ is analytic, and its complement is analytic as well, which easily implies that the same is true about $B$.

Since continuous maps between compact Hausdorff spaces are closed, it is true if the preimage is $F_\sigma$, and this implies the same for $G_\delta$ by complementation. So in a counterexample, we would need $f^{-1}[B]$ to be at best $F_{\sigma\delta}$ (or $G_{\delta\sigma}$).

I think it should be false in general, but I see no obvious counterexample.

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  • $\begingroup$ I thought that any analytic set is the continuous image of a Borel set... $\endgroup$ – Asaf Karagila May 12 '18 at 7:45
  • $\begingroup$ Can you please give me a reference for reading Suslin's theorem? Thanks. $\endgroup$ – caffeinemachine May 12 '18 at 8:35
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    $\begingroup$ @caffeinemachine: Theorem 14.11 in Kechris' Classical descriptive set theory. $\endgroup$ – tomasz May 12 '18 at 14:44
  • $\begingroup$ @AsafKaragila: Outside of Polish spaces, I guess it really depends how you define an analytic set (and probably where your Borel set is). How would it help? $\endgroup$ – tomasz May 12 '18 at 14:46
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    $\begingroup$ The answer is yes. See Steprans' post here: mathoverflow.net/questions/67858/… $\endgroup$ – hot_queen May 18 '18 at 0:31

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