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Good evening,

I have a question concerning the inverse of a measurable function.

Let $f\colon X\to Y$ be a Borel measurable mapping between two topological spaces. Suppose that $f$ is surjective.

My question : Does there exist always a measurable right inverse of $f$?

Any help is appreciated.Thanks in advance.

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up vote 3 down vote accepted

By a Borel measurable mapping $f$ you must mean that $f$ is a Borel map. (That is, I guess you're not assuming that there are any measures floating around.) In that case, the answer is no:

Let $f$ be the identity function $\mathbb{N}_d \to \mathbb{N}_i$, where $\mathbb{N}_d$ has the discrete topology and $\mathbb{N}_i$ has the indiscrete (or trivial) topology. Then $f$ is surjective, but its inverse is not Borel. For instance, $f^{-1}\{3\} = \{3\}$ is not a Borel set in $\mathbb{N}_i$, though $\{3\}$ is Borel in $\mathbb{N}_d$. (Of course, there's nothing special about $\mathbb{N}$ here: pick your favorite set with at least two elements.)

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The two definitions coincide so long as the two measure spaces are equipped with the Borel algebras don't they? –  Chris Janjigian Mar 3 '12 at 0:25
    
Sorry for my poor mathematics vocabulary in english. Borel measurable mapping means the mapping f is measurable with respect to the Borel $\sigma-$algebra of $X$ and $Y$. Thank you for your answer. Maybe I will have to pose a better question. So if we put additional conditions on X and Y, e.g., X and Y are compact Hausdorff spaces, does there exist always a measurable right inverse? –  Đức Anh Mar 3 '12 at 0:28
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Zach, since you can now post comments and actually have answered the question.. how about editing out that part which is a comment and make your answer cleaner? –  Asaf Karagila Mar 3 '12 at 10:39
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