Let $X$, $Y$ be topological spaces, and $f:X\rightarrow Y$ surjective. Why is $f$ closed if and only if the image of any Borel set is a Borel set? For one direction, it would clearly be sufficient to prove that, if $f$ is closed, the image of any open set is Borel, but I don't know how to proceed from there. For the other direction, I don't have any idea how to proceed.
2026-04-03 00:03:56.1775174636
a surjective function is closed iff it maps Borel sets to Borel sets
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Th statement is blatantly false. Let $f(x)=1/x$ if $x \neq 0$ and 1 if $x=0$. Then $f$ maps Borel sets to Borel sets but $f$ is not closed.