AD and simultaneous well-orderability principle

Question

Is the axiom of determinacy (AD) consistent with the following choice principle, and if yes, does it hold in $L(ℝ)$ under AD:

Simultaneous well-orderability: For every function $f:P(Ord)→\text{Wellorderable}$, there is $g$ such that for all $x$, $g(x)$ is a well-ordering of $f(x)$.

Notes:
- $\text{Wellorderable}$ is the class of all well-orderable sets. $P(Ord)$ is the class of all sets of ordinals. Domains of functions are sets.
- Without the $P(Ord)$ restriction, the principle would be equivalent to the axiom of choice for collections of well-orderable sets (i.e. choice for $V→\text{Wellorderable}$), which does not prove AC, but contradicts AD.

The principle looks strong, but I do not see a contradiction with AD. Under AD, AC fails for functions $ω_1→ℝ$, but $ℝ$ cannot be well-ordered. Also, under AD, there is no function that for every Turing degree gives a real of that degree (even though the set of such reals is countable). However, Turing degrees are not sets of ordinals; while Turing degrees are encodable by reals, the encoding is non-unique.

Under $V=\mathrm{HOD}(P(Ord))$, it suffices to consider OD $f$ in the principle. Also, under $V=\mathrm{HOD}(P(Ord))$, a natural strengthening of the principle is that for every $s∈P(Ord)$, every well-orderable non-empty $\mathrm{OD}(s)$ set has an $\mathrm{OD}(s)$ element (equivalently, $\mathrm{OD}(s)$ well-ordering as we can extract elements one by one here). Under $V=\mathrm{HOD}(2^λ)$ for an ordinal $λ$ (and thus for $V=L(ℝ)$), the unstrengthened principle is equivalent to existence of a set of ordinals $t$ such that every $s∈P(Ord)$, every well-orderable non-empty $\mathrm{OD}(s)$ set has an $\mathrm{OD}(s,t)$ element (equivalently, $\mathrm{OD}(s,t)$ well-ordering).

Other choice principles under AD:
- ZF + AD proves DC in $L(ℝ)$.
- Under AD, axiom of choice for sets parameterized by reals (i.e. choice for $f:ℝ→V$) implies, and under $V=L(P(ℝ))$, is equivalent to $\text{AD}_ℝ$ + "$Θ$ is regular".
- It is open whether under AD, every set of reals must be in $\mathrm{HOD}(P(Ord))$ (under $\text{AD}^+$, $P(ℝ)∈L(Ord^{<Θ})$).

Answer 1

The following is something quite close. Perhaps with a bit more work one could get it in the exact form above. Assume $V = L(\mathbb{R})$ although much of this will work in $V = L(\mathscr{P}(\mathbb{R}))$ and $\mathsf{AD}^+$.

Woodin's dichotomy theorem states that under $\mathsf{AD}^+$ if $E$ is an equivalence relation on $\mathbb{R}$, then either $\mathbb{R}$ injects into $\mathbb{R} / E$ or $\mathbb{R} / E$ is wellorderable. If you look at the proof of the result, in the case that $\mathbb{R} / E$ is wellorderable, the wellordering is produced uniformly from the $\infty$-Borel code for $E$ (using an ultrapower by the Martin measure).

In $V = L(\mathbb{R})$, if $A \subseteq \mathbb{R}$, $S$ is a set of ordinals, and $A$ is $OD_S$, then $A$ has an $OD_S$ $\infty$-Borel code.

Let $\delta$ be an ordinal, $X$ is a set that $\mathbb{R}$ surjects onto, and $\Phi : \mathscr{P}(\delta) \rightarrow \mathscr{P}(X)$ so that for all $A \in \mathscr{P}(\delta)$, $\Phi(A)$ is a wellorderable subset of $X$. Let $\pi : \mathbb{R} \rightarrow X$ be a surjection. (All sets in $L(\mathbb{R})$ are ordinal definable from a real.) Without loss of generality, say $\pi$ is $OD$ and $\Phi$ is $OD$. For each $A \in \mathscr{P}(\delta)$, let $E_A$ be the equivalence relation on $\mathbb{R}$ defined by $x \ E_A \ y$ if and only if [$\pi(x) = \pi(y)$ and $\pi(x) \in \Phi(A)$] or $(\pi(x) \notin \Phi(A) \wedge \pi(y) \notin \Phi(A))$. Note that $E_A$ is $OD_A$ and $\mathbb{R} / E_A$ is essentially in bijection with $\Phi(A)$, except for one extra class. By the observation above, $E_A$ has an $OD_A$ $\infty$-Borel code. Let $S_A$ be the $OD_A$-least such $\infty$-Borel code using the canonical ordering of $HOD_A$. Apply the uniform procedure from the Woodin dichotomy theorem using this $\infty$-Borel code $S_A$ to uniformly obtain a wellordering of $\mathbb{R} / E_A$ and hence $\Phi(A)$.

If $X$ is not a surjective image of $\mathbb{R}$, then it can be decomposed uniformly into pieces which are images of $\mathbb{R}$ so perhaps one could extends this result subset of such $X$ (now having many equivalence relations to patch together). So possibly the result holds when $\Phi$ is a set and not class function.

The equivalence relation $E_A$ (or many equivalence relations) needs to be obtained uniformly. If $\Phi : \mathscr{P}(Ord) \rightarrow V$ and the image do not sit inside a common set, then it is immediately clear to how to obtain these equivalence relations.

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See Theorem 3.2 and Fact 3.3 of "CARDINALITY OF WELLORDERED DISJOINT UNIONS OF QUOTIENTS OF SMOOTH EQUIVALENCE RELATIONS" by Chan and Jackson.