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Let $X$ and $Y$ be standard Borel spaces, and let $f:X\to Y$ be a surjective Borel map. Does there exist a Borel inverse of $f$, that is a Borel map $g:Y\to X$ such that $f\circ g = \mathrm{id}_Y$.

In case $f$ is injective, such map does exist, so I am interested in the case when $f$ is not injective. The fact that the graph $\mathrm{Gr}(f)$ is Borel and $g$ chooses over $\mathrm{Gr}(f)^{-1}$ means that there exists at least a universally measurable $g$. If we could show that its graph is Borel as well, that would mean $g$ is Borel. I'm not sure how to do this, and whether such result is true. I've tried to find a counterexample by means of Borel isomorphism, but without much success.

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If $f$ happens to be a countable to one Borel function, then the answer is yes. You can see Classical Descriptive Set Theory, Kechris 18.10. (Indeed, countable to one case is an exercise [18.14] in the book!) –  Burak Jul 19 at 20:01
    
I think you mean $\text{id}_Y$ rather than $\text{id}_Z$. –  Joel David Hamkins Jul 20 at 1:37
    
@JoelDavidHamkins: thanks, fixed that. –  Ilya Jul 20 at 16:59
    
@Burak: nice, I haven't heard of that selection theorem before. –  Ilya Jul 20 at 17:00

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up vote 6 down vote accepted

Your question is really about the uniformization problem, a major focus of descriptive set theory. A set $B\subset X\times Y$ is uniformized by a set $C\subset B$ if $C$ is the graph of a function with the same projection as $B$. In other words, the function selects one member of each nonempty section of $B$, and this function is a one-sided inverse of the projection function $\pi_X$ from from $B$ to $\pi_X(B)$.

There are a variety of uniformization theorems in descriptive set theory, which you can read about in the usual descriptive set-theoretic texts. For example, the Lusin-Novikoff result is that if $B$ is Borel and all sections are countable, then it has a Borel uniformation, as mentioned by Burak in the comments (see Exercise 4F.6 in Moschovakis's book. Alternatively, when the sections are large in various senses, then again there is a Borel uniformization (see Borel classes of uniformizations of sets with large sections by Petr Holicky and these further exercises in Moschovakis's book).

Meanwhile, the full Borel uniformazation property is not true (and indeed perhaps this is why the descriptive set theorists make so much effort to discover the circumstances when sets can be uniformized and to find out how simple the uniformizing sets can be). Namely, there is a Borel set $B\subset\mathbb{R}\times\mathbb{R}$ projecting to the whole of $\mathbb{R}$, but having no Borel uniformization. (You can find a discussion of the proof that such a $B$ exists in Example 5.1.7 of this text by Srivastava.) In particular, the projection map $\pi:B\to \mathbb{R}$ is a Borel surjection having no Borel inverse in your sense.

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I was exactly trying to use a Borel set which does not have a Borel uniformization, and relate it to a $\mathrm{Gr}(f)^{−1}$ via an isomorphism, however that did not work. Your idea of using a projection is really nice, thanks. Also, there is a survey of measurable selection theorems (Wagner 1977-79). Do you know if there was any progress after that, maybe another survey paper? Please tell me in case I should make it a separate question. –  Ilya Jul 20 at 20:06
    
I'm not quite sure, so if you have a specific question, I would encourage you to ask it. –  Joel David Hamkins Jul 21 at 0:23

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