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Suppose $E,F$ are Borel equivalence relations on Polish spaces $X,Y$, respectively. Under strong enough determinacy axioms, is it true that $E$ Borel reduces to $F$ iff there is an injective map from the quotient set $X/E$ to $Y/F$? Is it at least true for some class of equivalence relations? This answer mentions that:

In natural situations, in particular in models of determinacy, we can replace "Borel reducibility", that is, "Borel cardinality" via Borel injections by actual cardinality.

so I guess the answer to my question is yes. But the best I can do myself is this: suppose there is an injective set mapping $f:X/E\rightarrow Y/F$, and consider $\Gamma=\{(x,y)\in X\times Y:f([x])=[y]\}$. Under strong determinacy all sets can be uniformized, so we have a reduction $g:X\rightarrow Y$, which is at least Baire measurable, and therefore Borel measurable on a comeager set. Can this be improved further?

Edit: I realized that basically I'm looking for some $E$ that is Baire reducible but not Borel reducible to $F$. We want to find such $E,F$ under $\mathsf{AD}_\mathbb{R}$ instead of $\mathsf{ZFC}$; see the comments below.

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  • $\begingroup$ Is "strong determinacy" different from $\mathsf{AD}$? I've not heard that term before. $\endgroup$ Commented Apr 30, 2023 at 20:08
  • $\begingroup$ @NoahSchweber I mean any determinacy axiom that is strong...Apparently $\mathsf{AD}$ alone doesn't imply $\varphi=$ "all sets can be uniformized", and $\mathsf{AD}+\varphi$ is also known as $\mathsf{AD}_\mathbb{R}$ $\endgroup$ Commented Apr 30, 2023 at 20:14
  • $\begingroup$ @EdwardH I don't fully understand your question. My logic is that under $\mathsf{AD}_\mathbb{R}$, a counterexample exists iff there exists $E$ Baire but not Borel reducible to $F$. I don't know whether Baire reducibility is absolute in a suitable sense. Also, if you could pardon my ignorance in dst, what is a reference for the free shift example being Baire but not Borel reducible? $\endgroup$ Commented May 23, 2023 at 14:37

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