# Surjective (strong) reducibility of Borel equivalence relations

Suppose $E$ and $F$ are Borel equivalence relations on Polish spaces $X$, $Y$, resp. Say that $E$ is surjectively Borel reducible to $F$ iff there is a Borel surjection $f:X \to Y$ such that $xEy$ iff $f(x) F f(y)$. This is (at least on the surface) stronger than normal Borel reducibility, weaker than Borel isomorphism.

What do we know about this reducibility? If $E$ is effectively Borel with only countably many classes, is $E$ effectively surjectively reducible to the trivial equivalence relation on $\omega$?

edit: Probably this is the version I am interested in: Say $E$ is strongly Borel reducible to $F$ iff $E$ reduces to $F$ via a Borel $f$ such that the range of $f$ is Borel. When $Y=\omega$ and $E$ has infinitely many classes, "$E$ is effectively stongly Borel reducible to the relation $F$ on $Y$" exactly means "$E$ is effectively surjectively Borel reducible to $F$".

edit again: Actually, the arXiv version of Fokina-Friedman-T¨ornquist "The Effective Theory of Borel Equivalence Relations" probably has a (minor) mistake. On page 9, the equivalence relation $E^*$ is $\Pi^1_1$, not Hyp. The $\Sigma^1_1$ line
$$(\exists x_0,x_1) ( x_0 \in H(d_0) \wedge x_1 \in H(d_1) \wedge x_0 E x_1)$$ fails when $H(d_0) = \emptyset$. The proof of Proposition 2 can be easily repaired, but the issue of empty set is worth attention when we seriously require the range being $\Delta^1_1$.

• If a Borel equivalence relation $E$ has only countably many equivalence classes, isn't each class Borel? – Noah Schweber Mar 15 '15 at 16:37
• Yes. But the range of the reduction is $\Sigma^1_1$, if $E$ is effective Borel. I wonder if it is possible to make the range $\Delta^1_1$. – Yizheng Zhu Mar 15 '15 at 16:57

The answer is no. The question is basically asking for a $\Delta^1_1$ transversal for a $\Delta^1_1$ Borel equivalence relation with only countably many classes. Let $A$ be a closed subset of $\mathbb{R}\times \omega$ whose projection to the second coordinate is not $\Delta^1_1$. Put $(x,m)E(y,n)$ iff either $(x,m)\notin A \wedge (y,n)\notin A$ or $(x,m)\in A \wedge (y,n)\in A\wedge m=n$. Then $E$ does not have a $\Delta^1_1$ transversal.