# Borel bijection between Polish spaces [closed]

Let $X$ and $Y$ be two subsets of Polish spaces. Let $f:X\rightarrow Y$ be a bijection and borel measurable. How can one show that $f^{-1}$ is also borel measurable?

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Check that for Polish spaces the usual definition of Borel measurability of a function (inverse images of Borel sets are Borel) is equivalent to saying that the graph of the function is a Borel subset of the product space. –  Andreas Blass Aug 16 '11 at 22:36
See Kechris, Classical Descriptive Set Theory, Thm 14.12, p.88 books.google.com/books?id=pPv9KCEkklsC&pg=PA88 Note also that injective Borel maps map Borel sets to Borel sets, see Thm 15.1 p.89. –  Theo Buehler Aug 16 '11 at 22:52
I'd like to point out the site math.stackexchange.com to you, as one among some other sites mentioned in the FAQ which might be a bit more appropriate for this question (and your last one). –  Theo Buehler Aug 16 '11 at 22:56
Theo is right (A: the bijective Borel image of a Borel set is Borel). Andreas is also right (B: a map is Borel iff its graph is a Borel set), but I think you need (A) result to do it that way. The key thing in the proof for (A) is the separation theorem: two disjoint analytic sets can be separated by a Borel set. (An analytic set is a continuous image of a Borel set in a Polish space.) –  Gerald Edgar Aug 17 '11 at 0:14
Gerald, I think Andreas's suggestion is better than my second sentence, which was just intended as a complement that Arnold shouldn't miss. Kechris proves that f is Borel iff the graph is analytic from the separation theorem: if the graph is analytic and $A$ in $Y$ is Borel then $x \in f^{-1}(A)$ iff $\exists y \in A : f(x) = y$ (so $f^{-1}(A)$ is analytic) iff $\forall y : f(x) = y \implies y \in A$ (so $f^{-1}(A)$ is co-analytic). –  Theo Buehler Aug 17 '11 at 1:04