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The following question was noted by Jan Pachl in connection with the study of Arens products and he has not received a satisfactory answer from the various experts he has asked. Let $X$ and $Y$ be compact Hausdorff spaces and let $F$ be a continuous mapping from $X$ onto $Y$. Let $A\subseteq Y$ and suppose that $F^{-1}A$ is Borel. Does it follow that $A$ is also Borel.

Certainly if $X$ and $Y$ are metric then the answer is positive; in this case both $Y\setminus A = F(X\setminus F^{-1}A)$ and $A = F(F^{-1}A)$ are analytic and hence Borel. But even for $X=Y=2^{\omega_1}$ the argument that disjoint analytic sets can be separated by Borel sets does not seem to be available.

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up vote 4 down vote accepted

It is true that if $f:K \to L$ is a continuous mapping from a compact space $K$ onto $L$ and $A \subseteq L$ has Borel preimage in $K$ then $A$ is Borel in $L$. Jan Pachl points out that this is a very special case of a general theorem (Theorem 10) by P. Holický and J. Spurný in "Perfect images of absolute Souslin and absolute Borel Tychonoff spaces", Topology Appl. 131 (2003), 281--294.

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Juris, do you now know of compact Hausdorff counterexample that shows that disjoint analytic sets need not be separated by Borel sets? –  Ali Enayat Jul 14 '11 at 13:21
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