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.



Images of Borel subsets of non-metric compact spaces
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.


