To see this, let $H$ be a $c$-dihypergraph on ${}^\kappa\kappa$
and let
$I:=\proj_{\ddim, c}^{-1}(H){\restr}{}^\kappa\kappa$.
[Recall the notation $\proj_{\ddim,c}$ from the paragraph before Lemma <ref>.]
We claim that $\KLW^I_\kappa(X,Y)$ is equivalent to $\KLW_\kappa^H(X,Y)$.
\beginenumerate-(1)

* To see that the first options of the dichotomies are equivalent, note that
$y$ is an $H$-limit point of a set $A$ if and only if
$y$ is an $I$-limit point of $A$.

* If $g$ is as in the second option of $\KLW_\kappa^I(X,Y)$,
then $f:=g\restr{{}^\kappa c}$ is

a continuous homomorphism from $\dhH c$ to $H$ as in the proof of Lemma <ref> <ref>.
Moreover, $f$ reduces $(\RR^c_\kappa,\QQ^c_\kappa)$ to $(X,Y)$ since
$\RR^c_\kappa=\RR^d_\kappa\cap{{}^\kappa c}$
$\QQ^c_\kappa=\QQ^d_\kappa\cap{{}^\kappa c}$.
Now suppose that $f$ is as in the second option of $\KLW_\kappa^H(X,Y)$.
Fix a map $k: d\to c$ with $k\restr c = \id_c$ and $k(i)\neq 0$ for all $i\in d\,\setminus\, c$.
Define $g:{}^\kappa\ddim\to X$ by letting $g(x):=f(k\comp x)$.
Then $g$ is a continuous homomorphism from $\dhH d$ to $H$ as in the proof of Lemma <ref> <ref>.
This works for any $c\subsetneq \ddim$ with $0\in c$. But $0\in c$ is needed for both directions in both (1) and (2).
Since $x\in \RR^d_\kappa$ if and only if $k\circ x\in \RR^c_\kappa$ for all $x\in {}^\kappa\kappa$, $g$ reduces $(\RR^d_\kappa,\QQ^d_\kappa)$ to $(X,Y)$.
\endenumerate-(1)

§.§ The determinacy of Väänänen's perfect set game

In this subsection,
we obtain the determinacy of the games studied in <cit.>*Section 2

for all subsets of ${}^{\kappa}\kappa$
as an application of $\ODD\kappa\kappa({}^\kappa\kappa)$.
In fact, the result yields additional properties of the underlying set in the case that player $\pone$ has a winning strategy.
Similarly, $\ODD\kappa\kappa({}^\kappa\kappa,\defsetsk)$ implies the determinacy of  Väänänen's game for all definable subsets of ${}^\kappa\kappa$.

By Cantor-Bendixson analysis, any topological space can be decomposed into a scattered and a crowded[Recall that a topological space $X$ is scattered if each nonempty subspace contains an isolated point.
$X$ is crowded if it has no isolated points.]
subset by iteratively removing isolated points. If the space has a countable base, then the scattered part is countable.
Väänänen adapted this to subsets of generalized Baire spaces via
a game of length $\xi\leq\kappa$.

The following is a variant of Väänänen's game.
$\cbii(X)$ implies the determinacy of Väänänen's (original) perfect set game $\GV_\kappa(X,x)$ for all $x\in X$
$\cbi(X)$ implies the determinacy of $\GV_\kappa(X)$

Väänänen's perfect set game
of length $\xi\leq\kappa$ for a subset
$X$ of ${}^\kappa\kappa$
is played by two players $\pone$ and $\ptwo$ as follows.

$\pone$ plays a strictly increasing continuous sequence $\langle \gamma_\alpha : \alpha< \xi \rangle$ of ordinals below $\kappa$ with $\gamma_0=0$.
$\ptwo$ plays
an injective sequence
$\langle x_\alpha : \alpha< \xi \rangle$ of elements of $X$
with $x_\alpha \supseteq (x_\beta\restr \gamma_{\beta+1})$
for all $\beta<\alpha$.

$\pone$       $\gamma_0$                 $\gamma_1$                        …
$\gamma_{\alpha}$          …

$\ptwo$              $x_0$                   $x_1$                              …
$x_\alpha$                       …

Each player loses immediately if they do not follow these requirements.
If both follow the rules in all rounds, then $\ptwo$ wins.

Moreover, the game $\GV_\xi(X,x)$[In fact, Väänänen did not consider the full game $\GV_\xi(X)$, but only defined the local versions $\GV_\xi(X,x)$
for all $x\in {}^\kappa\kappa$ with the requirement $x_\alpha\in X$

for all $\alpha\geq 1$
Since we work with the games $\GV_\xi(X,x)$ for $x\in X$ only, we can amalgamate them into a single game.]
is defined just like $\GV_\xi(X)$ for any $x$ in $X$, except that $\ptwo$ must play $x_0=x$ in the first round.

We say that a player wins a game if they have a winning strategy.

Suppose that $\xi\leq\kappa$ and that $X$ is a subset of ${}^\kappa\kappa$.
\beginenumerate-(1)

The $\xi$-scattered part of $X$ is
$\Sc_\xi(X)= \{ x\in X :  \text{ \pone\ wins  $\GV_\xi(X,x)$} \}$.

The $\xi$-kernel of $X$ is
$\Ker_\xi(X)= \{ x\in X :  \text{ \ptwo\ wins  $\GV_\xi(X,x)$} \}$.
[Väänänen originally defined the $\xi$-kernel of $X$ as the set $\Ker^V_\xi(X)$ of all $x\in {}^\kappa\kappa$ such that $\ptwo$ wins $\GV_\xi(X,x)$.
Note that $\Ker_\xi^V(X)\cap X=\Ker_\xi(X)$ and that
$\bar{\Ker_\xi(X)}=\Ker^V_\xi(X)$ for all $\xi\geq\omega$.]
\endenumerate-(1)

$X$ is $\xi$-scattered if $X=\Sc_\xi(X)$ and
$\xi$-crowded if $X=\Ker_\xi(X)\neq\emptyset$.

These definitions extend well-know notions from topology:
$\omega$-scattered is equivalent to scattered and $\omega$-crowded is equivalent to crowded.
Moreover, any subset of ${}^\kappa\kappa$ whose closure is $\kappa$-perfect is a $\kappa$-crowded set.
In fact, a subset $X$ of ${}^\kappa\kappa$ is $\kappa$-crowded if and only if every $x\in X$ is contained in some subset $Y$ of $X$ whose closure is $\kappa$-perfect by Lemma <ref> below.

Note that $\Sc_\xi(X)$ is a $\xi$-scattered relatively open subset
and $\Ker_\xi(X)$ is a relatively closed subset of $X$.
$\Ker_\xi(X)$ is $\xi$-crowded if
this may fail when $\xi=\gamma+1$ where $\gamma$ is indecomposable
$\xi$ is an indecomposable ordinal.[I.e., $\alpha+\beta<\xi$ for all $\alpha,\beta<\xi$.]
\beginenumerate-(a)

* $X$ is $\xi$-scattered $\ \Longleftrightarrow\ $ $\pone$ wins $\GV_\xi(X)$.

* $\Ker_\xi(X)\neq\emptyset$ $\ \Longleftrightarrow\ $ $\ptwo$ wins $\GV_\xi(X)$.
\endenumerate-(a)

By the Gale-Stewart theorem, $\GV_\xi(X)$ is determined for all $\xi\leq\omega$.
This is no longer true for $\xi>\omega$.
For instance, there exists a closed subset $X$ of ${}^{\omega_1}\omega_1$ such that $\GV_{\omega+1}(X)$ is not determined,
This does not necessarily imply the non-determinacy of $\GV_\kappa(X)$
and in fact, it is consistent that there exists such a set $X$ of size $\omega_2$
[See <cit.> for variants of these results.]
The next property can be understood both as the determinacy of Väänänen's game of length $\kappa$ with an extra condition and as a variant of the Cantor-Bendixson analysis.

For any class $\mathcal C$,
$\cbi(\mathcal C)$ states that the following holds for all subsets $X\in\mathcal C$ of ${}^\kappa\kappa$:

$\cbi(X)$: Either
$|X|\leq\kappa$ and $\pone$ wins $\GV_\kappa(X)$,
$\ptwo$ wins $\GV_\kappa(X)$.

Equivalently, $\cbi(X)$ states that either $X$ is $\kappa$-scattered and of size ${\leq}\kappa$, or it contains a $\kappa$-crowded subset.
[If $\ptwo$ wins $\GV_\kappa(X)$, then $\Ker_\kappa(X)\neq\emptyset$ is a $\kappa$-crowded subset of $X$.]

This principle may fail even for closed sets.
In fact, it is consistent that
there exists an $\omega_1$-scattered closed subset of ${}^{\omega_1}\omega_1$ of size $\omega_2$ <cit.>*Theorem 3.[See <cit.> for variants of this result.]

We first show that
$\ODD\kappa\kappa({}^\kappa\kappa) \Longrightarrow \cbi(\pwrset({}^\kappa\kappa))$
$\ODD\kappa\kappa({}^\kappa\kappa,\defsets\kappa) \Longrightarrow \cbi(\defsetsk)$.
use the dihypergraph $\CC^X$
from Subsection <ref>.
This consists of all
$\kappa$-sequences $\bar x$ in ${}^\kappa \kappa$
with $\lim_{\alpha<\kappa}(\bar x)\in X \setminus \ran(\bar x)$.
What can we say about the complexity of $\CC^X$ if we know the complexity of $X$?

Let $X\subseteq{}^\kappa\kappa$.
If $\CC^X$ has a $\kappa$-coloring, then $|X|\leq\kappa$ and
$\pone$ wins $\GV_\kappa(X)$.

We have already shown $|X|\leq\kappa$ in
Corollary <ref>.
pagebreak inserted before lemma

For the second claim, recall that a subset $A$ of ${}^\kappa\kappa$ is $\CC^X$-independent if and only if $A'\cap X=\emptyset$,
where $A'$ denotes the set of limit points of $A$.
Take $\CC^X$-independent sets $A_\alpha$ with ${}^\kappa\kappa=\bigcup_{\alpha<\kappa} A_\alpha$.
We define the following strategy for player $\pone$.
In rounds of the form $\alpha+1$, $\pone$ chooses $\gamma_{\alpha+1}$ so that
\beginenumerate-(i)

x_{\beta}\restr\gamma_{\alpha+1}$ for all $\beta<\alpha$, and

$A_\alpha\cap N_{x_\alpha\restr\gamma_{\alpha+1}}\subseteq\{x_\alpha\}$.
\endenumerate-(i)
$\pone$ can achieve the first property because $\ptwo$ plays an injective sequence and the second property because $x_\alpha\in X$ cannot be a limit point of $A_\alpha$.
Suppose $\ptwo$ wins a run of the game where $\pone$ has used this strategy.
Then $y:=\bigcup_{\alpha<\kappa} x_\alpha\restr\gamma_{\alpha+1}$
this is where we use that $x_\alpha\supseteq x_\beta\restr\gamma_{\beta+1}$ for all $\beta<\alpha$
is an element of ${}^\kappa\kappa$,
so $y\in A_\beta$ for some $\beta<\kappa$.
Since $y\in N_{x_\beta\restr\gamma_{\beta+1}}$,
we must have $y=x_\beta$ by <ref>.
This contradicts <ref> for any $\alpha>\beta$.

The equivalence of <ref> and <ref> in the next lemma was shown in <cit.>*Proposition 2.69.
Our proof is essentially the same, but we include it due to the different notation.
[In <cit.>, a subset $Y$ of ${}^\kappa\kappa$ is called $\kappa$-strongly dense in itself
if $\closure Y$ is $\kappa$-perfect and $\Ker_\xi(X)$ denotes Väänänen's definition of the kernel.]

The following statements are equivalent for any $X\subseteq {}^{\kappa}\kappa$
and $x\in X$:
\beginenumerate-(1)

There exists $Y\subseteq X$ such that $x\in Y$ and $\closure Y$ is $\kappa$-perfect.

There is a nice reduction
$\Phi$ for $({}^\kappa\kappa, X)$ with $x\in[\Phi]\big(\QQ_\kappa)$.
The proof shows that $\Phi$ can equivalently be chosen to be continuous.
\endenumerate-(1)

<ref> $\Rightarrow$ <ref>: Suppose $\Phi$ is a nice reduction for $({}^\kappa\kappa, X)$.
Let $Y:=[\Phi]\big(\QQ_\kappa)$.
Then $x\in Y\subseteq X$.
$\closure{Y}=\ran([\Phi])$ is $\kappa$-perfect
by Lemma <ref>.

<ref> $\Rightarrow$ <ref>: Let
$Y$ be a subset of $X$ with $x\in Y$ such that $\bar Y$ is $\kappa$-perfect.
Then $T(Y)$ is a $\kappa$-perfect tree.
It is straightforward to construct a winning strategy $\tau$ for  $\ptwo$ in $\GV_\kappa(Y,x)$
using the fact that $T(Y)$ is cofinally splitting in successor rounds of the game and
the fact that $T(Y)$ is $\lle\kappa$-closed in limit rounds.
Since $Y\subseteq X$,
$\tau$ is also a winning strategy for  $\ptwo$ in $\GV_\kappa(X,x)$.

<ref> $\Rightarrow$ <ref>:
Suppose that $\tau$ is a winning strategy for  $\ptwo$ in $\GV_\kappa(X,x)$.
We construct a nice reduction for $({}^\kappa\kappa,X)$
by having  $\ptwo$ use $\tau$ repeatedly in response to
different partial plays of  $\pone$.
In detail, we construct
a $\perp$- and strict order preserving
function $\Phi: {}^{<\kappa}2\to T(X)$, a continuous
strict order preserving function
$\langle \gamma_s : s\in \SS_\kappa \rangle$
[Recall the definition of $\SS_\kappa$ from Subsection <ref>.]
of ordinals below $\kappa$
and a function $\langle x_s : s\in {}^{<\kappa}\kappa \rangle$ of elements of $X$ such that the following hold for all $s\in{}^{<\kappa}2$:
\beginenumerate-(i)

$\gamma_{\emptyset}=0$ and $x_{\emptyset}=x$.

$\Phi(s)=x_s\restr\gamma_{s\conc\langle 1\rangle}$.

$x_s=\tau\big(\langle\gamma_{t}:\, t\subseteq s,\, t\in\SS_\kappa\rangle\big).$
\endenumerate-(i)
For all $s\in{}^{<\kappa}\kappa$, let
$p_s:= \langle \gamma_t,\, x_t:\,t\subseteq s,\, t\in\SS_\kappa\rangle.$
<ref> states that $p_s$ is a partial run of $\GV_\kappa(X)$ where $\ptwo$ uses the strategy $\tau$.
Note that $x_s$ is constant in intervals disjoint from $\SS_\kappa$, but takes a new value at each $s\in \SS_\kappa$:
If $s\in\SS_\kappa$, then
$\bigcup \{p_u:\, u\subsetneq s,\,u\in\SS_\kappa\}$
is extended to $p_s$
by $\pone$ playing $\gamma_s$ as defined below
and $\ptwo$ choosing $x_s$ using $\tau$.

$s=t\conc \langle 0\rangle^{\alpha}$ for some $\alpha<\kappa$. Then $p_s=p_t$, so $x_s=x_t$.

We shall construct $x_s$, $\gamma_s$ and $\Phi(s)$
by recursion.
Regarding the order of the construction, let $\gamma_\emptyset=0$ and $x_\emptyset=x$.
If $s\in\SS_\kappa$ and $\lh(s)\in\Lim$, let
$\gamma_s:=\bigcup\{\gamma_{t}:\, t\subsetneq s,\,t\in\SS_\kappa\}$.
If $s\in{}^{<\kappa} 2$
$\gamma_{t}$ has been constructed for all $t\subseteq s$ with $t\in\SS_\kappa$, then
The recursion is ordered this way because if $s=t\conc\langle 1\rangle$, then $\gamma_s$ also has to be constructed at this point.
construct $x_s$, $\gamma_{s\conc\langle 1\rangle}$ and $\Phi(s)$ in the next step as follows.
Let $x_s$ be as in <ref>.

We aim to choose
$\gamma_{s\conc\langle 1\rangle}$ so that
$\Phi$ remains $\perp$- and strict order preserving for
$\Phi(s):=x_s\restr\gamma_{s\conc\langle 1\rangle}$.

The next claim shows that

extends $\Phi(t)$ for all $t\subsetneq s$ if $\gamma_{s\conc\langle 1\rangle}$ is chosen sufficiently large.

This will ensure that $\Phi$ is strict order preserving.

$\Phi(t)\subsetneq x_s$ for all $t\subsetneq s$.

If $s=t\conc\langle 0\rangle^\alpha$, then
$\Phi(t)\subseteq x_t=x_s$.
Otherwise, take the least $\alpha\geq \lh(t)$ with $s(\alpha)=1$.
Then $s\supseteq t\conc\langle 0\rangle^\alpha\conc\langle 1\rangle$.
By <ref>
applied to $t\conc\langle 0\rangle^\alpha$
and <ref>,
we have
\Phi(t)
\subseteq
\Phi(t\conc\langle 0\rangle^\alpha\conc\langle 1\rangle)
(x_{t\conc\langle 0\rangle^\alpha}\restr
\gamma_{t\conc\langle 0\rangle^\alpha\conc\langle 1\rangle})
\subseteq

It remains to show that $\Phi$ is $\bot$-preserving if $\gamma_{u^\smallfrown \langle 1\rangle}$ is chosen sufficiently large for all $u$ of successor length.
To see this, suppose that $s=t\conc\langle i\rangle$.
Since $\ptwo$ has to play an injective sequence in $\GV_\kappa(X,x)$,
we have
$x_{t\conc\langle 0\rangle}=x_t
\neq
x_{t\conc\langle 1\rangle}$
by <ref>.
Therefore $\Phi(t\conc\langle 0\rangle):= x_{t\conc\langle 0\rangle}\restr\gamma_{t\conc\langle 0,1\rangle}$
$\Phi(t\conc\langle 1\rangle):= x_{t\conc\langle 1\rangle}\restr\gamma_{t\conc\langle 1,1\rangle}$
are incompatible extensions of $\Phi(t)$, if
$\gamma_{t\conc\langle 0,1\rangle}$
$\gamma_{t\conc\langle 1,1\rangle}$
are sufficiently large.
This completes the construction.

Note that we can ensure that $\Phi$ is continuous by letting $\gamma_{s\conc\langle 1\rangle}:=
\bigcup_{t\subsetneq s}\gamma_{t\conc\langle 1\rangle}$ for $\lh(s)\in\Lim$.
Then $\Phi(s)=\bigcup_{t\subsetneq s}\Phi(t)$ by the previous claim and <ref>.

for all $u\in {}^{<\kappa}\kappa$.

We have
$[\Phi](\pi(u)\conc\langle0\rangle^\kappa) = \bigcup_{\alpha<\kappa} \Phi(\pi(u)\conc\langle0\rangle^\alpha)= x_{\pi(u)}$.
The first equality holds by the definition of $[\Phi]$.
The last one holds since $\langle \Phi(\pi(u)\conc\langle0\rangle^\alpha) : \alpha<\kappa \rangle$ is a strictly increasing chain of initial segments of $x_{\pi(u)}$ by <ref> and since $x_s=x_t$ if $s=t\conc \langle 0\rangle^{\alpha}$.

By the previous claim, $[\Phi](\pi(\emptyset)\conc\langle0\rangle^\kappa)=x_{\emptyset}=x$ and
\subseteq \{x_s:s\in{}^{<\kappa}\kappa\}\subseteq X$.

Since $\Phi$ is also $\perp$- and strict order preserving, it is as
required in <ref>.

The next theorem follows from the two previous lemmas and Theorem <ref>.

It also holds for $\CC^X\cap\dhIkappa$ instead of $\CC^X$ by Remark <ref>.
Question/idea: is it possible to get a set $X$ as in (1) whose closure is as in  <cit.>*Theorem 3?

Suppose $X$ is a subset of ${}^\kappa\kappa$.
\beginenumerate-(1)

If $\CC^X$ has a $\kappa$-coloring, then $|X|\leq\kappa$ and
$\pone$ wins $\GV_\kappa(X)$.

There exists a continuous homomorphism from $\dhH\kappa$ to $\CC^X$
if and only if
$\ptwo$ wins $\GV_\kappa(X)$.
\endenumerate-(1)
In particular,
each of
$\ODD\kappa{\CC^X}$ and $\KLW_\kappa({}^\kappa\kappa,X)$
[Note that $\ODD\kappa{\CC^X}$ and $\KLW_\kappa({}^\kappa\kappa,X)$ are equivalent by Theorem <ref>.]
implies $\cbi(X)$.

Suppose $X$ is a $\Gdelta(\kappa)$ subset of ${}^\kappa\kappa$.
the converse of <ref> holds by Corollary <ref>.
Hence $\cbi(X)\Longleftrightarrow\PSP_\kappa(X)$
by Corollary <ref> <ref>.

It is open whether
for all subsets $X$ of ${}^\kappa\kappa$,
or equivalently, whether
the converse of <ref> holds.
“NOT $\cbi(X)$ and $\Det(\GV_\kappa(X))$ for some closed set $X$” is consistent for $\kappa=\omega_1$ by <cit.>*Theorem 3 (which says that
it is consistent that
there exists an $\omega_1$-scattered closed subset $X$ of ${}^{\omega_1}\omega_1$ of size $\omega_2$).

Lemmas <ref> and <ref>
generalize the following observations
from closed subsets to arbitrary subsets of ${}^\kappa\kappa$:
\beginenumerate-(1)

* If $X$ is closed and
$|X|\leq\kappa$, then $\pone$ wins $\GV_\kappa(X)$ <cit.>*Proposition 3.
[Note that if $X$ is closed,
or more generally $\Gdelta(\kappa)$,
then $|X|\leq\kappa$ if and only if $\CC^X$ has a $\kappa$-coloring by Corollary <ref>.]

* If $X$ is closed, then
$\Ker_\kappa(X)$ is the union of all $\kappa$-perfect subsets of $X$ <cit.>.
\endenumerate-(1)
These observations already show $\cbi(X)\Longleftrightarrow\PSP_\kappa(X)$ for closed sets $X$.

holds for any $\Gdelta(\kappa)$ subset $X$ of ${}^\kappa\kappa$.
To see this, note that the direction from left to right follows immediately from the previous theorem and
Corollary <ref> <ref>.
For the converse direction,
it suffices to show that any $\Gdelta(\kappa)$ subset $X$ of ${}^\kappa\kappa$ such that
$\ptwo$ wins $\GV_\kappa(X)$
has a $\kappa$-perfect subset.
The proof is a slight variant of the construction in Lemma <ref> <ref> $\Rightarrow$ <ref>.
If $X=\bigcap_{\alpha<\kappa} U_\alpha$ with each $U_\alpha$ open,
we construct
$\Phi: {}^{<\kappa}2\to T(X)$,
$\langle \gamma_s : s\in \SS_\kappa \rangle$
and $\langle x_s : s\in {}^{<\kappa}\kappa \rangle$
such that in addition to the properties in that construction,
$N_{\Phi(s)}\subseteq U_{\lh(s)}$ for all $s\in{}^{<\kappa}2$.
This can be guaranteed since
$x_s\in X\subseteq U_{\lh(s)}$
and $U_{\lh(s)}$ is open.
Thus, $\ran([\Phi])$ is contained in
$X=\bigcap_{\alpha<\kappa} U_{\alpha}$.
Since $\Phi$ is $\perp$-preserving, this is a $\kappa$-perfect subset of $X$ by Lemma <ref> <ref>.

The next property is an analogue of Cantor-Bendixson analysis:
it ensures that a subset $X$ of ${}^\kappa\kappa$ can be turned into
a $\kappa$-crowded set
by first removing
a $\kappa$-scattered subset consisting of
up to $\kappa$ many points.

For any class $\mathcal C$,
$\cbii(\mathcal C)$ states that the following holds for all subsets $X\in\mathcal C$ of ${}^\kappa\kappa$:

\begin{equation*}
\cbii(X):\:
\text{ and }
\end{equation*}

Note that
$\cbii(X)$ implies that $\GV_\kappa(X,x)$ is determined for every $x\in X$,
[It is easy to see that this implies the determinacy of Väänänen's original games $\GV_\kappa(X,x)$ for all $x\in {}^\kappa\kappa$.]
and hence that $\GV_\kappa(X)$ is determined.

The next theorem shows that
$\ODD\kappa\kappa({}^\kappa\kappa) \Longrightarrow \cbii(\pwrset({}^\kappa\kappa))$ and
$\ODD\kappa\kappa({}^\kappa\kappa,\defsetsk) \Longrightarrow \cbii(\defsetsk)$.

Suppose that $\mathcal C$ is
a class.
\beginenumerate-(1)

$\cbii({\mathcal C}) \Longrightarrow \cbi({\mathcal C})$.

If $X\cap N_t\in{\mathcal C}$
for all subsets $X\in{\mathcal C}$ of ${}^\kappa\kappa$ and all $t\in{}^{<\kappa}\kappa$,

$$\cbi({\mathcal C}) \Longleftrightarrow \cbii({\mathcal C}).$$

In particular,

$\KLW_\kappa({}^\kappa\kappa,\mathcal C)$
[Equivalently, $\ODD\kappa{\CC^X}$ for all $X\in\mathcal C$, by Theorem <ref>.]

implies $\cbii(\mathcal C)$.
\endenumerate-(1)

For <ref>, let $X\subseteq{}^\kappa\kappa$ and assume
To see $\cbi(X)$, suppose that
$\ptwo$ does not win $\GV_\kappa(X)$.
By $\cbii(X)$,
$X$ is a $\kappa$-scattered set of size $\lleq\kappa$,
as required.

Now suppose that $\cbi({\mathcal C})$ holds for some class ${\mathcal C}$ as in <ref>, and take $X\in{\mathcal C}$. To show $\cbii(X)$, let
$S$ denote the set of those $s\in{}^{<\kappa}\kappa$ such that
$X\cap N_s$ is a $\kappa$-scattered set of size $\leq\kappa$.
$|\bigcup_{s\in S} (X\cap N_s)|\leq\kappa$
it suffices to show that

\[
\bigcup_{s\in S} (X\cap N_s)
\subseteq\Sc_\kappa(X).
\]

To see the first inclusion, suppose that $x\in X\oldsetminus\Ker_\kappa(X)$.
Then $\ptwo$ does not win $\GV_\kappa(X,y)$, so
there is some $\gamma$ such that
$\ptwo$ does not win $\GV_\kappa(X\cap N_{y\restr\gamma})$.
Since $X\cap N_{y\restr\gamma}\in{\mathcal C}$, we have
$\cbi(X\cap N_{y\restr\gamma})$.
Hence $|X\cap N_{y\restr\gamma}|\leq\kappa$ and
$\pone$ wins $X\cap N_{y\restr\gamma}$, so
$y\restr\gamma\in S$.

For the second inclusion,
let $y\in\bigcup_{s\in S} (X\cap N_s)$.
Then $y\restr\gamma\in S$ for some $\gamma<\kappa$.
$\pone$ has the following winning strategy in $\GV_\kappa(X,y)$:
$\pone$ first plays $\gamma_0=0$ and $\gamma_1=\gamma$.
In subsequent rounds, $\pone$ uses their
winning strategy in $\GV_\kappa(X\cap N_{y\restr\gamma})$
to determine their moves in $\GV_\kappa(X,y)$.
Thus, $y\in\Sc_\kappa(X)$.

The last assertion in <ref> now follows immediately from Theorem <ref>.

The next corollary follows from the results in this subsection and Theorem <ref>.

The following hold in all $\Col(\kappa,\lle\lambda)$-generic extensions of $V$:
\beginenumerate-(1)

if $\lambda>\kappa$ is inaccessible in $V$.

if $\lambda>\kappa$ is Mahlo in $V$.
\endenumerate-(1)

We can equivalently replace $\cbii(\mathcal C)$ by $\cbi(\mathcal C)$, where $\mathcal C$ denotes $\defsetsk$ and $\pwrset({}^\kappa\kappa)$, respectively.
In the model in <ref>, Väänänen's game is therefore determined for all subsets of ${}^\kappa\kappa$.
The previous corollary strengthens two results of Väänänen
<cit.>*Theorem 1 & Theorem 4.
He showed that both
$\cbiiomegaone(\closedsets(\omega_1))$ and
the statement $\cboomegaone$ defined right below
[$\cboomegaone$ follows from  $\cbiomegaone(\powerset({}^{\omega_1}\omega_1))$.]
hold in $\Col(\omega_1,\lle\lambda)$-generic extensions if $\lambda$ is measurable.
[In fact, Väänänen obtained these two statements from a
principle $I(\omega)$
which holds after Lévy-collapsing a measurable cardinal to $\omega_2$ <cit.>.
$I(\omega)$ is equiconsistent with a measurable cardinal since it implies the existence of a precipitous ideal on $\omega_2$.]

Galgon showed that
can be obtained by Lévy-collapsing an inaccessible cardinal to $\kappa^+$
is in fact equivalent to

<cit.>*Proposition 2.16.
Our proof of Theorem <ref> generalizes the proof of the latter result.

Using the previous corollary and <cit.>,
we can lower the consistency strength of
the dichotomy on the size of complete subhypergraphs (cliques)
of finite dimensional $\Gdelta(\kappa)$ dihypergraphs on $\analytic(\kappa)$ sets studied in <cit.>
and its generalization to families of dihypergraphs.
Given a family $\mathcal H$ of dihypergraphs on a set $X$,
we say that $Y\subseteq X$ is an $\mathcal H$-clique if
$\dhK{\ddim_H}Y\subseteq H$ for each $H\in\mathcal H$, where $\ddim_H$ denotes the arity of $H$.

If $\mathcal H$ is a set of $\kappa$ many finite dimensional $\Gdelta(\kappa)$

dihypergraphs on $X$
and there is an $\mathcal H$-clique of size $\kappa^+$,
then there is a $\kappa$-perfect $\mathcal H$-clique.
$\Gdelta(\kappa)$ can equivalently be replaced by open.
Moreover, $\DK_\kappa(X)$ is a statement about

certain product-open $\omega$-dihypergraphs,
as $Y$ is an $\mathcal H$-clique if and only if
$\dhK{\omega}Y\subseteq \bigcap_{H\in\mathcal H}\proj_{\omega, \ddim_H}^{-1}(H){\restr}X$. ]

$\DK_\kappa(X)$ extends a dichotomy of Doležal and Kubiś <cit.> to uncountable cardinals.${}$[They proved $\DK_\omega(\analytic)$, extending the
special cases for a single $\Gdelta$ graph <cit.> and
a single finite dimensional $\Gdelta$ hypergraph <cit.>.]
By <cit.>,${}$
[In <cit.>,
$\DK_\kappa(X)$, $\Diamondi\kappa$ and $\cbo$ are denoted by
$\mathrm{PIF}_\kappa(X)$, $\mathscr{DI}_\kappa$
$\DK_\kappa(X)$ is formulated as a dichotomy about the size of independent sets with respect to $\kappa$ many finite dimensional  $\Fsigma(\kappa)$ dihypergraphs.]
Footnote number size problem solved (twice) by adding ${}$.
$\DK_\kappa(\analytic(\kappa))$ follows from
$\Diamondi\kappa$ and

$\ptwo$ wins $\GV_\kappa(X)$ for all subsets $X$ of ${}^\kappa\kappa$ of size $\kappa^+$.

Since $\cbo$ follows from $\cbi(\powerset({}^\kappa\kappa))$,

the previous corollary yields:

holds in all $\Col(\kappa,\lle\lambda)$-generic extensions of $V$
if $\lambda>\kappa$ is a Mahlo cardinal in $V$.

This strengthens a joint result of Väänänen and the second-listed author <cit.> showing that $\DK_\kappa(\analytic(\kappa))$ holds in $\Col(\kappa,\lle\lambda)$-generic extensions for any measurable cardinal $\lambda>\kappa$.[They showed that the combination of $\Diamond_\kappa$ and a principle $I^-(\kappa)$ implies $\DK_\kappa(\analytic(\kappa))$. Note that $I^-(\omega_1)$ is the same as $I(\omega)$.
The consistency of $I^-(\kappa)$ can be proved similarly to that of $I(\omega)$ in <cit.>.]
It is open whether one can obtain $\DK_\kappa(\analytic)$ directly from $\ODD\kappa\kappa({}^\kappa\kappa)$ by constructing a suitable dihypergraph.
Moreover, we have not considered the case of $\lle\kappa$-dimensional dihypergraphs.

Suppose $X$ and $Y$ are subsets of ${}^\kappa\kappa$ and $G$ is a digraph on ${}^\kappa\kappa$.
We consider
a more general game $\GV^G_\kappa(X,Y)$ that is defined by modifying the winning condition in $\GV_\kappa(X)$.
$\ptwo$ wins a run if and only if
$(x_\alpha,x_{\alpha+1})\in G$ for all $\alpha<\kappa$
[If $G$ is an open digraph, then it is easy to see that we get an equivalent game if we require instead that $(x_\beta,x_\alpha)\in G$ for all $\beta<\alpha<\kappa$.]
$\bigcup_{\alpha<\kappa} x_\alpha\restr\gamma_{\alpha+1} \in Y$.
We claim that for any open digraph $G$ on ${}^\kappa\kappa$,
[Equivalently, $\ODD\kappa{\CC^{X,G}\restr Y}$, by Theorem <ref>.]
implies the following.

$\cbig G(X,Y)$:
Either $G\restr (X\cap Y)$ admits a $\kappa$-coloring and $\pone$ wins $\GV^G_\kappa(X,Y)$,
or $\ptwo$ wins $\GV^G_\kappa(X,Y)$.

This follows from <ref> and <ref> below.

For $Y={}^\kappa\kappa$ and $G=\gK {{}^\kappa\kappa}$,

$\cbig G(X,Y)$
is equivalent to $\cbi (X)$.
$\cbig G(X,Y)$ relates to $\ODD\kappa{G\restr(X\cap Y)}$ just like $\cbi (X)$ relates to $\PSP_\kappa(X)$.
In particular,
they are equivalent
if $X$ is a $\Gdelta(\kappa)$ subset of ${}^\kappa\kappa$
by <ref> below.
[The game $\GV^G_\kappa(X):=\GV^G_\kappa(X,{}^\kappa\kappa)$ for open graphs $G$
was studied in <cit.>*Section 3
with the aim of obtaining games that relate to $\OGD_\kappa(X)$ just like $\GV_\kappa(X)$ relates to $\PSP_\kappa(X)$ for closed sets $X$.
It is denoted by $\GV^1_\kappa(X,R)$ for $R=(X\times X)\oldsetminus G$ there.]

\beginenumerate-(1)

Suppose the first option in $\KLW^G_\kappa(Y,X)$ holds, i.e., $Y$ is the $\kappa$-union of sets $A_\alpha$

with no $G$-limit points[See Definition <ref>.] in $Y$.
We claim that
$G\restr(X\cap Y)$ admits a $\kappa$-coloring and $\pone$ wins $\GV^G_\kappa(X,Y)$.
The former follows from

Lemma <ref> <ref> $\Rightarrow$ <ref>.
The proof of the latter is similar to Lemma <ref>.
Consider the following strategy for $\pone$ in $\GV^G_\kappa(X,Y)$.
In rounds of the form $\alpha+1$, $\pone$ chooses $\gamma_{\alpha+1}$ so that
\beginenumerate-(i)

$(x_\beta,y)\in G$
for all $y\in N_{x_\alpha\restr\gamma_{\alpha+1}}$
if $\alpha=\beta+1$
$N_{x_\alpha\restr\gamma_{\alpha+1}}\subseteq G^{x_{\alpha-1}}$

$(x_\alpha,y)\notin G$
for all $y\in A_\alpha\cap N_{x_\alpha\restr\gamma_{\alpha+1}}$.
Equivalently: $G^{x_\alpha}\restr N_{x_\alpha\restr\gamma_{\alpha+1}}=\emptyset$

\endenumerate-(i)
$\pone$ can achieve <ref> since $G$ is open and $(x_\beta,x_\alpha)\in G$ if $\alpha=\beta+1$.
$\pone$ can achieve <ref> since

$x_\alpha\in X$ is not a $G$-limit point of $A_\alpha$.

Suppose $\ptwo$ wins a run of the game where $\pone$ has used this strategy and let
$y:=\bigcup_{\alpha<\kappa} x_\alpha\restr\gamma_{\alpha+1}$.
Then $y\in Y$, so
$y\in A_\alpha$ for some $\alpha<\kappa$.
$(x_\alpha,y)\notin G$ by <ref> since
$y\in N_{x_\alpha\restr\gamma_{\alpha+1}}$.
$(x_\alpha,y)\in G$ by <ref>
since $y\in N_{x_{\alpha+1}\restr\gamma_{\alpha+2}}$.

We claim that
the second option in $\KLW_\kappa^G(Y,X)$ holds if and only if
$\ptwo$ wins $\GV_\kappa^G(X,Y)$.
Recall that the former is equivalent to the existence of
an order homomorphism $\Phi$ for $({}^\kappa\kappa,G)$
such that $[\Phi]$ reduces $(\RR_\kappa,\QQ_\kappa)$ to $(Y,X)$
by Theorem <ref> <ref> <ref>$\Leftrightarrow$<ref>.
Suppose first that $\Phi$ is such an order homomorphism.
We define the following strategy for $\ptwo$ in $\GV_\kappa^G(X,Y)$.
As part of their strategy, $\ptwo$ also constructs a continuous strictly increasing sequence
$\langle s_\alpha : \alpha<\kappa \rangle$ in $\SS_\kappa$
$\Phi(s_\alpha\conc\langle 0\rangle^\kappa)=x_\alpha$
for all $\alpha<\kappa$,
and a sequence $\langle\delta_{\alpha}:\alpha<\kappa\rangle$  of ordinals below $\kappa$.

Suppose $\pone$ has played $\langle \gamma_\beta : \beta\leq\alpha \rangle$ and $\ptwo$ has played $\langle x_\beta : \beta<\alpha\rangle$ and has constructed
$s_\beta$ for all $\beta<\alpha$ and
$\delta_{\beta}$ for all $\beta$ with $\beta+1<\alpha$.

If $\alpha\in\Lim$,
$\ptwo$ lets $s_\alpha:=\bigcup_{\beta<\alpha}s_\beta$.

If $\alpha=\beta+1$,
$\ptwo$ picks $\delta_\beta<\kappa$ so that
$x_\beta\restr\gamma_\alpha\subseteq\Phi(s_\beta\conc\langle 0\rangle^{\delta_\beta})$
and lets $s_\alpha:=s_\beta\conc\langle 0\rangle^{\delta_\beta}\conc\langle 1\rangle$.

In both cases, $\ptwo$ plays $x_\alpha=[\Phi](s_\alpha\conc\langle 0\rangle^\kappa)$.
Note that $x_\alpha\in X$ since $[\Phi](\QQ_\kappa)\subseteq X$.

Suppose $\langle \gamma_\alpha, x_\alpha : \alpha<\kappa \rangle$ is a run according to this strategy and $\ptwo$ has constructed $s_\alpha$ and $\delta_\alpha$ for all $\alpha<\kappa$.
To see that $\ptwo$ wins,
it suffices to show that
\beginenumerate-(i)

$x_\beta\restr\gamma_{\beta+1}\subseteq x_\alpha$ for all $\beta<\alpha<\kappa$,

$(x_\beta,x_\alpha)\in G$ for all $\beta<\alpha<\kappa$,

$y:=\bigcup_{\alpha<\kappa} x_\alpha\restr\gamma_{\alpha+1}$ is in $Y$.
\endenumerate-(i)
In particular, note that <ref> implies that
$\langle x_\alpha:\alpha<\kappa\rangle$ is an injective sequence and $(x_\alpha,x_{\alpha+1})\in G$ for all $\alpha<\kappa$.

<ref> holds since
$x_\beta\restr\gamma_{\beta+1}\subseteq \Phi(s_\alpha)\subseteq
[\Phi](s_\alpha\conc\langle 0\rangle^\kappa)= x_\alpha$ for all $\beta<\alpha<\kappa$
by the construction.
For <ref>, suppose that $\beta<\alpha<\kappa$ and let
$u:=s_\beta\conc\langle 0\rangle^{\delta_\beta}$.
$\Phi(u\conc\langle 0\rangle)\subseteq x_\beta$

since $u\conc\langle 1\rangle = s_{\beta+1}\subseteq s_\alpha$, we also have
$\Phi(u\conc\langle 1\rangle)\subseteq \Phi(s_\alpha)\subseteq x_\alpha$.
Since $\Phi$ is an order homomorphism for $({}^\kappa\kappa,G)$, $(x_\beta,x_\alpha)\in N_{\Phi(u\conc\langle 0\rangle)}\times N_{\Phi(u\conc\langle 0\rangle)}\subseteq G$.
For <ref>,
let $z=\bigcup_{\alpha<\kappa} s_\alpha$. Then $z\in\RR_\kappa$ and
$[\Phi](z)=\bigcup_{\alpha<\kappa} \Phi(s_\alpha)=y$.
Hence $y\in [\Phi](\RR_\kappa)\subseteq Y$.

For the converse direction,
suppose that $\tau$ is a winning strategy for $\ptwo$ in $\GV^G_\kappa(X,Y)$.
Construct functions
$\Phi: {}^{<\kappa}2\to T(X)$,
$\langle \gamma_s : s\in \SS_\kappa \rangle$
and $\langle x_s : s\in {}^{<\kappa}\kappa \rangle$ as in the proof of
Lemma <ref> <ref> $\Rightarrow$ <ref>
with the additional property that
$N_{\Phi(s\conc \langle 0\rangle)}\times N_{\Phi(s\conc\langle 1\rangle)}\subseteq G$
for all $s\in{}^{<\kappa}\kappa$.
This can be guaranteed
since $G$ is open and
$(x_{s\conc\langle 0\rangle},x_{s\conc\langle 1\rangle})=
(x_s,x_{s\conc\langle 1\rangle})\in G$ for all $s\in{}^{<\kappa}\kappa$.
Then $\Phi$ is a nice reduction for $({}^\kappa\kappa,X)$
and moreover, it is an order homomorphism for $({}^\kappa\kappa,G)$ by the additional requirement.
It thus suffices to show that $[\Phi](\RR_\kappa)\subseteq Y$.
Let $z\in\RR_\kappa$.
It suffices to show that $[\Phi](z)$ is produced during a run of $\GV^G_\kappa(X,Y)$ where $\ptwo$ uses $\tau$.
$s\in \SS_\kappa$ for cofinally many initial segments $s$ of $z$ and
$\Phi(s)=x_s\restr\gamma_{s\conc\langle 1\rangle}$ for all $s\in{}^{<\kappa}\kappa$ by the definition of $\Phi$, we have

[\Phi](z)=\bigcup\{\Phi(s):s\subseteq z, s\in \SS_\kappa\}=
\bigcup\{x_s\restr\gamma_{s\conc\langle 1\rangle}
:s\subseteq z, s\in \SS_\kappa\}.

Thus, $[\Phi](z)$ is produced during the run
$\langle \gamma_s, x_s :s\subseteq z, s\in \SS_\kappa\rangle$
of $\GV^Y_\kappa(X)$ where $\ptwo$ uses $\tau$, as required.

If $X$ is a $\Gdelta(\kappa)$ subset of ${}^\kappa\kappa$,
$\cbig G(X,Y)\Longleftrightarrow\KLW_\kappa^G(Y,X)\Longleftrightarrow\ODD\kappa {G\restr X\cap Y}$.
In fact, it suffices that
$Y\setminus X$ can be separated from $X$ by a union of $\kappa$ many $G$-closed sets.
The first equivalence follows from
Lemma <ref> <ref> $\Rightarrow$ <ref>.
The second equivalence holds by
Corollary <ref>.

We claim that
$\ODD\kappa {G\restr X\cap Y}\Longleftrightarrow\cbig G(X,Y)$
for all $\Gdelta(\kappa)$ subsets $X$ of ${}^\kappa\kappa$ and all $Y\subseteq{}^\kappa\kappa$.
The direction from left to right follows from
Corollary <ref>, since
$\KLW_\kappa^G(Y,X)\Longrightarrow\cbig G(X,Y)$ by
<ref> and
For the converse direction, it suffices to show that
if $X$ is a $\Gdelta(\kappa)$ set and $\ptwo$ wins $\GV_\kappa^G(X,Y)$,
$G\restr (X\cap Y)$ has a $\kappa$-perfect complete subdihypergraph.
The proof of this is a slight variant of

the previous construction.
If $X=\bigcap_{\alpha<\kappa} U_\alpha$ with each $U_\alpha$ open,
we construct
$\Phi: {}^{<\kappa}2\to T(X)$,
$\langle \gamma_s : s\in \SS_\kappa \rangle$
and $\langle x_s : s\in {}^{<\kappa}\kappa \rangle$
with the additional property that
$N_{\Phi(s)}\subseteq U_{\lh(s)}$ for all $s\in{}^{<\kappa}2$.
This can be guaranteed since
$x_s\in X\subseteq U_{\lh(s)}$
and $U_{\lh(s)}$ is open.
Then $\ran([\Phi])$ is contained in
$X=\bigcap_{\alpha<\kappa} U_{\alpha}$
and hence
$[\Phi](\RR_\kappa)\subseteq Y$
as in the previous construction.
This set is a $G$-clique
since $\Phi$ is an order homomorphism for $({}^\kappa\kappa,G)$
$[\Phi](\RR_\kappa)\subseteq Y$,
any $\kappa$-perfect subset of it is as required.

\endenumerate-(1)

One can aim to generalize the preceding arguments to higher dimensions to show that $\KLW_\kappa^H(Y,X)$ implies an analogue of $\cbig G(X,Y)$
for any box-open dihypergraph $H$.
The special case for ${}^\kappa\kappa\times G$ is intended to show that  $\KLW_\kappa^{{}^\kappa\kappa\times G}(Y,X)$ implies an analogue of $\cbi$ with respect to a different variant

of Väänänen's game for open graphs $G$ that was studied in <cit.>*Section 3 for the special case of closed sets $X$ and $Y={}^\kappa\kappa$. It is denoted by $\GV^2_\kappa(X,R)$ for $R=(X\times X)\oldsetminus G$ there.

It is further likely that similar ideas as in the proof of Theorem <ref> show that $\cbig G(X,Y)$ is equivalent to an
[For any $x\in X$, define the game $\GV^G_\kappa(X,Y,x)$ just like $\GV^G_\kappa(X,Y)$ except that $\ptwo$ must play $x_0=x$ in the first round.
Let $\Sc^G_\kappa(X,Y)$, respectively $\Ker^G_\kappa(X,Y)$, denote the set of those $x\in X$ such that $\pone$, respectively $\ptwo$, wins $\GV^G_\kappa(X,Y,x)$.
The analogue of $\cbii$ would state that

$X$ can be decomposed as the union of $\Sc^G_\kappa(X,Y)$ and $\Ker^G_\kappa(X,Y)$
and $G\restr\Sc^G_\kappa(X,Y)$ has a $\kappa$-coloring.]
of $\cbii$ for $\GV^G_\kappa(X,Y)$ for any open graph $G$.
This analogue would then follow from $\OGD_\kappa(X\cap Y)$ if $X$ is a $\Gdelta(\kappa)$ set,
answering <cit.>*Question 3.57.

§.§ The asymmetric Baire property

def: kappa-meager
def: Baire property
A subset $X$ of ${}^\kappa\kappa$ is called $\kappa$-meager if it is the union of $\kappa$ many nowhere dense subsets of ${}^\kappa\kappa$ and $\kappa$-comeager if its complement is $\kappa$-meager.
Moreover, if $Y$ is a subset of ${}^\kappa\kappa$ then $X$ is called $\kappa$-meager in $Y$ if $X\cap Y$ is $\kappa$-meager[If $Y$ is open, then this is equivalent to $X\cap Y$ being a $\kappa$-meager subset of the space $Y$, but this is not true for arbitrary subspaces $Y$.]
and $\kappa$-comeager in $Y$ if $Y\setminus X$ is $\kappa$-meager.
$Y\setminus X= Y\cap({}^{\kappa}\kappa\setminus X)$ is $\kappa$-meager $\Leftrightarrow$ the complement of $X$ is $\kappa$-meager in $Y$

A subset $A$ of ${}^\kappa\kappa$ has the $\kappa$-Baire property if there exists an open subset $U$ of ${}^\kappa\kappa$ such that
$X\triangle U$ is $\kappa$-meager.
If $\kappa$ is uncountable, then the $\kappa$-Baire property always fails for some $\kappa$-analytic subset of ${}^\kappa\kappa$:
Halko and Shelah <cit.> observed that the club filter

$$ \mathrm{Club}_\kappa = \{  x \in {}^\kappa\kappa:  \{ \alpha<\kappa : 0<x(\alpha)\}  \text{ contains a club}  \} $$

does not have the $\kappa$-Baire property.
To see this,
suppose first that $ \mathrm{Club}_\kappa$ is $\kappa$-comeager in some basic open set $N_t$. Then there is a sequence $\langle U_\alpha : \alpha<\kappa\rangle$ of open dense subsets of $N_t$
such that
$\bigcap_{\alpha<\kappa}U_\alpha\subseteq \mathrm{Club}_\kappa$.
We construct a strictly increasing sequence $\langle t_\alpha : \alpha<\kappa \rangle$ in ${}^{<\kappa}\kappa$ with $t_0=t$ as follows.
If $\alpha=\beta+1$, choose $t_\alpha \supsetneq t_\beta$ with $N_{t_{\alpha}} \subseteq U_\beta$.
If $\alpha$ is a limit, let $t_{\alpha}:=\bigcup_{\beta<\alpha} t_\beta \cup \{ \langle \alpha, 0 \rangle \}$.
Let $x:=\bigcup_{\alpha<\kappa} t_\alpha$.
Then $x\in \bigcap_{\alpha<\kappa}U_\alpha$ and $x\notin \mathrm{Club}_\kappa$, a contradiction.
One can similarly see that $ \mathrm{Club}_\kappa $ is not $\kappa$-meager.

We consider a variant of the $\kappa$-Baire property from <cit.>.
Its definition uses the following types of strict order preserving maps.

Let $S\subseteq{}^{<\kappa}\kappa$, and  let $t\in{}^{<\kappa}\kappa$.
\beginenumerate-(a)

* $S$ is dense above $t$ if for any $u\supseteq t$
there exists $s\in S$ with $s\supseteq u$.
$S$ is nowhere dense if it is not dense above any $t\in{}^{<\kappa}\kappa$.

* A strict order preserving map $\iota:{}^{<\kappa}\kappa\to{}^{<\kappa}\kappa$ is
dense if for all $t\in{}^{<\kappa}\kappa$, the set
$\{\iota(t\conc\langle \alpha\rangle):\alpha<\kappa\}$
is dense above $\iota(t)$.
\endenumerate-(a)

The property of being $\kappa$-comeager
can be characterized via
such maps:
a subset $X$ of ${}^\kappa\kappa$ is $\kappa$-comeager in a basic open set $N_t$
if and only if there exists a continuous dense strict order preserving map
with $\iota(\emptyset)=t$ and
$\ran([\iota])\subseteq X$
<cit.>*Lemma 3.2.

A subset $X$ of  ${}^\kappa\kappa$ has the asymmetric $\kappa$-Baire property $\ABP_\kappa(X)$
[$\ABP_\kappa(X)$ was called the almost Baire property in <cit.>. Note that $X$ is $\kappa$-meager if and only if there exists a continuous dense strict order preserving map $\iota:{}^{<\kappa}\kappa\to{}^{<\kappa}\kappa$ with $\iota(\emptyset)=\emptyset$ and $\ran([\iota])\subseteq {}^\kappa\kappa\setminus X$, by <cit.>*Lemma 3.2.Therefore our definition of $\ABP_\kappa(X)$ is equivalent to the one given in <cit.>*Definition 3.3.]
either $X$ is $\kappa$-meager
or there exists a dense strict order preserving map
with $\ran([\iota])\subseteq X$.

For any class $\mathcal C$,
let $\ABP_\kappa(\mathcal C)$ state that $\ABP_\kappa(X)$ holds for all subsets $X\in\mathcal C$ of ${}^\kappa\kappa$.

property is equivalent to the determinacy of the Banach-Mazur game of length $\kappa$ <cit.>*Lemma 3.6.
Note that $\iota$ is not required to be continuous in the previous definition, so $\ABP_\kappa(\mathcal C)$
is weaker than the $\kappa$-Baire property
for subsets of ${}^\kappa\kappa$ in $\mathcal C$.
The version of $\ABP_\kappa(\mathcal C)$ with a continuous map $\iota$ is equivalent to the statement that each subset in $\mathcal C$ is either $\kappa$-comeager in some basic open set or $\kappa$-meager.
If $\mathcal C$
is closed under preimages of shift maps $\sigma_t: {}^\kappa\kappa\to {}^\kappa\kappa$ for each $t\in {}^{<\kappa}\kappa$, where $\sigma_t(x)=t^\smallfrown x$,
then the latter statement is equivalent to the Baire property for all sets in $\mathcal C$.[To see this, let $U$ be the union of all basic open sets $N_t$ such that $N_t\setminus X$ is $\kappa$-meager.
Then $U\triangle X$ is $\kappa$-meager.]
Since continuity is trivial for $\omega$,
$\ABP_\omega(\mathcal C)$ is equivalent to the Baire property for subsets of ${}^\omega\omega$ in $\mathcal C$.

We obtain the asymmetric $\kappa$-Baire property $\ABP_\kappa(X)$
as a special case of $\ODD\kappa\kappa(X,\defsetsk)$.
We use the dihypergraph $\dhD\kappa$ from Definition <ref>.
Note that $\dhD\kappa\in\defsetsk$ is a box-open dihypergraph on its domain ${}^\kappa\kappa$.

A subset $Y$ of ${}^\kappa\kappa$ is $\dhD\kappa$-independent if and only if
it is nowhere dense.

Suppose $Y$ is somewhere dense, i.e., $Y$ is dense in $N_t$ for some $t\in{}^{<\kappa}\kappa$.
Fix $y_u\in N_u\cap Y$ for all $u\supseteq t$.
Then any injective enumeration of the values $y_u$ is a
hyperedge of $\dhD\kappa\restr Y$.

Conversely, if $\dhD\kappa\restr Y$ is nonempty,
then $Y$ is dense is some basic open set $N_t$.

The next lemma shows that the property of being an order homomorphism for $(X,\dhD\kappa)$ in inherited to supersets of $X$.

Let $X\subseteq{}^\kappa\kappa$.
\beginenumerate-(1)

Let $\langle t_\alpha:\alpha<\kappa\rangle$ be a sequence of elements of  $T(X)$.
Then $\prod_{\alpha<\kappa} N_{t_\alpha}\cap X \subseteq \dhD \kappa$[Note that $\prod_{\alpha<\kappa} N_{t_\alpha}\cap X \neq\emptyset$, as $t_\alpha\in T(X)$ for all $\alpha<\kappa$.]
if and only if
$\{t_\alpha:\alpha<\kappa\}$ is dense above some $s\in{}^{<\kappa}\kappa$.
Thus, a subtree $T$ of ${}^{<\kappa}\kappa$ is $\dhD\kappa$-independent if and only if it's nowhere dense (i.e. not dense above any $s\in{}^{<\kappa}\kappa$)

Any order homomorphism $\iota$ for $(X,\dhD\kappa)$ is an order homomorphism for $\dhD\kappa$.
Note that $\domh{\dhD\kappa}={}^\kappa\kappa$
\endenumerate-(1)

We first show <ref>.
If $\{t_\alpha:\alpha<\kappa\}$ is dense above $s$,
then every element of
$\prod_{\alpha<\kappa} N_{t_\alpha}\cap X$ is dense in $N_s$.
Conversely, suppose that $\{t_\alpha:\alpha<\kappa\}$ is nowhere dense.
It suffices to construct a sequence
$\bar x\in\prod_{\alpha<\kappa} N_{t_\alpha}\cap X$ which is nowhere dense in ${}^\kappa\kappa$.
Let $I$ denote the set of $\alpha<\kappa$ such that $T(X)$ is dense above some $t\supseteq t_\alpha$, or equivalently, $\succ{t}\subseteq T(X)$ as $T(X)$ is downwards closed.
[Recall that $\succ{t}= \{ u\in {}^{<\kappa}\kappa : t\subsetneq u \}$.]
Then $\kappa\,\setminus\, I$ is precisely the set of those $\alpha<\kappa$ such that $({}^{<\kappa}\kappa)\setminus T(X)$ is dense above $t_\alpha$.

We first claim that any $\bar x\in\prod_{\alpha\in \kappa \setminus I} N_{t_\alpha}\cap X$ is nowhere dense in ${}^\kappa\kappa$.
Otherwise $\bar{x}$ is dense in $N_t$ for some $t\in {}^{<\kappa}\kappa$.
Pick some $\alpha\in \kappa\,\setminus\, I$ with $x_\alpha \in N_t$.
Since $\ran(x)\subseteq X$, $T(X)$ is dense above $t\cup t_\alpha$.
But then $\alpha\in I$.

It therefore suffices to find a sequence
$\bar y\in\prod_{\alpha\in I} N_{t_\alpha}\cap X$ that is nowhere dense in ${}^\kappa\kappa$.
To this end, we construct a sequence
$\langle u_\alpha:\alpha\in I\rangle$
such that the following hold for all $\alpha\in I$:
\beginenumerate-(i)

$u_\alpha\in T(X)$ and $t_\alpha\subsetneq u_\alpha$.

$u_\alpha\not\subseteq t_\gamma$ for all $\gamma<\kappa$.

For all $\beta\in I\cap \alpha$, either
$u_\alpha\perp u_\beta$ or $u_\alpha=u_\beta$.
\endenumerate-(i)
Suppose that $\alpha\in I$ and $u_\beta$ has been constructed for all $\beta\in I\cap \alpha$.
If there exists
$\beta\in I\cap \alpha$ with
$u_\beta\supsetneq t_\alpha$, then let $u_\alpha:=u_\beta$.
Otherwise, we have $t_\alpha\perp u_\beta$ for all $\beta<\alpha$ by <ref>.
As $\alpha\in I$, there exists $s\supseteq t_\alpha$ with $\succ{s}\subseteq T(X) $.
Since $\{t_\gamma:\gamma<\kappa\}$ is nowhere dense, there exists
$u_\alpha\supsetneq s$ in $T(X)$
such that $u_\alpha\not\subseteq t_\gamma$ for all $\gamma<\kappa$.
Then $u_\alpha\supsetneq t_\alpha$, so
$u_\alpha\perp u_\beta$ for all $\beta\in I\cap \alpha$.

Since $u_\alpha\in T(X)$ for all $\alpha<\kappa$, we may choose a sequence
$\bar y\in\prod_{\alpha\in I} N_{u_\alpha}\cap X \subseteq \prod_{\alpha\in I} N_{t_\alpha}\cap X$
such that $y_\alpha=y_\beta$ whenever $u_\alpha=u_\beta$.
Since the $u_\alpha$'s are pairwise incomparable,
$\bar y$ is nowhere dense in ${}^\kappa\kappa$.

<ref> shows that the property  $\prod_{\alpha<\kappa} N_{t_\alpha}\cap X \subseteq \dhD \kappa$ is inherited to supersets of $X$.
<ref> follows.

\beginenumerate-(1)

Any dense strict order preserving map $\iota:{}^{<\kappa}\kappa\to{}^{<\kappa}\kappa$
is an order homomorphism for $\dhD\kappa$.

If there exists an order homomorphism $\iota$ for $\dhD\kappa$, then there exists a dense strict order preserving map $\theta:{}^{<\kappa}\kappa\to{}^{<\kappa}\kappa$
with $\ran([\theta])\subseteq \ran([\iota])$.
Note that $\theta$ may not be continuous, even if $\iota$ was.
Uncomment for the example and proof

\endenumerate-(1)

<ref> is clear.${}$[This is the easy direction of Lemma <ref>.]
The symbols ${}$ were added just before the footnote to make the number appear in the right size.
For <ref>,
suppose there is an order homomorphism $\iota$ for $\dhD \kappa$.
By Lemma <ref> <ref> there exists a function $s: {}^{<\kappa}\kappa \to {}^{<\kappa}\kappa$ such that for all $t\in{}^{<\kappa}\kappa$,
is dense above $s(t)$.
Since $\iota$ is strict order preserving, $s(t)\supseteq\iota(t)$.

We define a
continuous strict order preserving map $e:{}^{<\kappa}\kappa\to{}^{<\kappa}\kappa$ such that
$\theta:=s\comp e$ is dense and strict order preserving.
We define $e(t)$ by recursion on $\lh(t)$.
Let $e(\emptyset):=\emptyset$. If $\lh(t)\in \Lim$, then let $e(t)=\bigcup_{u\subsetneq t} e(u)$.
Suppose $e(t)$ has been defined.
Let $\theta(t):=s(e(t))$.
Let $\langle e(t\conc \langle\beta\rangle):\beta<\kappa\rangle$ be an injective enumeration of
the set of those $e(t)\conc\langle\alpha\rangle$ such that
$\iota(e(t)\conc\langle\alpha\rangle)\supsetneq \theta(t)$.
It is clear from the construction that $\theta$ is dense and strict order preserving.

$\iota(e(t))\subseteq s(e(t))=\theta(t)\subseteq \iota\big(e(t\conc\langle\alpha\rangle)\big)$
for all $t\in{}^{<\kappa}\kappa$ and all $\alpha<\kappa$,
we have $[\theta]=[\iota]\comp[e]$.
Therefore $\ran([\theta])\subseteq\ran([\iota])$.

Suppose $X$ is a subset of ${}^\kappa\kappa$.
\beginenumerate-(1)

$\dhD \kappa{\restr}X$ admits a $\kappa$-coloring if and only if $X$ is $\kappa$-meager.
If $\dhD\kappa\restr X$ admits a $\kappa$-coloring, then there is a continuous homomorphism from $\dhH\kappa$ to $\dhD\kappa\restr({}^\kappa\kappa\setminus X)$. Is this true for any other dihypergraphs $H$? For any other ones from our applications?

Proof: Since $X$ is $\kappa$-meager, then there is a (continuous) dense strict order preserving map $\iota$ with $\ran([\iota])\subseteq {}^\kappa\kappa\setminus X$
(and $\iota(\emptyset)=\emptyset$).
Then $\iota$ is also an order homomorphism for $\dhD\kappa$ by the previous lemma and hence for $({}^\kappa\kappa\setminus X,\dhD\kappa)$.

There exists a continuous homomorphism from $\dhH\kappa$ to $\dhD \kappa{\restr}X$
if and only if there exists a dense strict order preserving map
$\iota:{}^{<\kappa}\kappa\to{}^{<\kappa}\kappa$ with $\ran([\iota])\subseteq X$.
\endenumerate-(1)
Thus, $\ABP_\kappa(X)$ is equivalent to $\ODD\kappa{\dhD\kappa{\restr}X}$.

<ref> holds since a subset of ${}^\kappa\kappa$ is  $\dhD \kappa$-independent if and only if it is nowhere dense by Lemma <ref>.

For <ref>, suppose there exists a continuous homomorphism from $\dhH\kappa$ to $\dhD \kappa{\restr}X$.
We obtain an order homomorphism $\iota$ for $(X,\dhD\kappa)$ by Lemma <ref>.
$\iota$ is also an order homomorphism for $\dhD\kappa$ by Lemma <ref>.
We then obtain the required map from Lemma <ref> <ref>.
The converse holds by Lemma <ref> <ref>.

In Theorem <ref>,  $\dhD\kappa$ can be equivalently replaced by the dihypergraph
$\dhD\kappa^-$ studied in Remark <ref>.
This follows from Corollary <ref>,
$\dhD \kappa^-\equivf \dhD \kappa$ and both are box-open on ${}^\kappa\kappa$.

Theorem <ref> and the previous theorem
yield a new proof of the following result from <cit.>.

holds in all $\Col(\kappa,\lle\lambda)$-generic extensions of $V$ if $\lambda>\kappa$ is inaccessible in $V$.

§.§ An analogue of the Jayne-Rogers theorem

We will derive an analogue of Jayne and Rogers' characterization of
${\bf \Delta}^0_2$-measurable functions

from the open dihypergraph dichotomy.
They proved the following principle for functions with analytic domain
in the countable setting.

Suppose throughout this subsection that $X$ is a subset of ${}^\kappa\kappa$
and $f:X\to {}^\kappa\kappa$ is a function.
We say that $f$ is
$\kappa$-continuous with closed pieces
if $X$ is the union of $\kappa$ many
relatively closed subsets $C$
such that $f\restr C$ is continuous.

Let $\JR_\kappa^f$ denote the statement that the following conditions are equivalent:
\beginenumerate-(1)

* $f$ is ${\bf \Delta}^0_2(\kappa)$-measurable.[See the end of Subsection <ref> for the definition. Equivalently, $f$ is ${\bf \Pi}^0_2(\kappa)$-measurable.]
uncomment for the proof of the claim in the footnote

* $f$ is $\kappa$-continuous with closed pieces.
\endenumerate-(1)
$\JR_\kappa(X)$ states that $\JR_\kappa^g$ holds for all $g:X\to{}^\kappa\kappa$.

$\JR_\kappa(\mathcal C)$ states that $\JR_\kappa(X)$ holds for all subsets $X\in\mathcal C$ of ${}^\kappa\kappa$, where $\mathcal C$ is any class.

We will obtain $\JR_\kappa^f$ as a special case of
$\ODD\kappa\kappa(\graph(f))$, where
$\graph(f)$ denotes the graph of $f$.
[I.e., $\graph(f)$ consists of those pairs $(x,y)$ in $X\times{}^\kappa\kappa$ such that $f(x)=y$.]
If $X=\dom(f)$ is in $\defsetsk$,
then $\ODD\kappa\kappa(\graph(f),\defsetsk)$ suffices.
Here, we mean the variant of  Definition <ref>
for the underlying space ${}^\kappa\kappa\times{}^\kappa\kappa$ instead of ${}^\kappa\kappa$.
Since there is a homeomorphism $k$ from
is equivalent to $\ODD\kappa\kappa(k(\graph(f)),\defsetsk)$.

Our proof is similar to that of <cit.>.
Recall from Subsection <ref> that $\CC^X$ denotes the set of all
$\kappa$-sequences $\bar x$ in ${}^\kappa\kappa$
with $\lim_{\alpha<\kappa}(\bar x)\in X\setminus \ran(\bar x)$.

We will work with the $\kappa$-dihypergraph $\JJ^f$ on
consisting of those sequences
$\langle (x_\alpha,y_\alpha):\alpha<\kappa\rangle$ such that
$\bar x=\langle x_\alpha:\alpha<\kappa\rangle \in \CC^{X}$ and $f(\lim_{\alpha<\kappa}(\bar x))$ is not in the closure of  ${\{y_\alpha:\alpha<\kappa\}}$.
$\JJ^f$ was defined in the proof of <cit.> for $\kappa=\omega$.
Note that $\JJ^f\restr\graph(f)$
consists of those sequences
that project to $\CC^X$ and witness discontinuity of $f$.

Clearly $\JJ^f$ is a box-open dihypergraph on ${}^\kappa\kappa\times{}^\kappa\kappa$,
and $\JJ^f\in\defsetsk$ when $f\in\defsetsk$.

Let $\proj_k$ denote the projection onto the $k^{\mathrm{th}}$ coordinate for $k\in\{0,1\}$.

Suppose $A\subseteq\graph(f)$ and $C$ is the closure of $\proj_0(A)$ in $X$.
Then $A$ is $\JJ^f$-independent if and only if
$f\restr C$ is continuous.

First, suppose that  $A$ is $\JJ^f$-independent.
Assuming $f\restr C$ is not continuous at $y\in C$,
take an injective sequence
$\bar x=\langle x_\alpha:\alpha<\kappa\rangle$ in $\proj_0(A)$ converging to $y$
such that $\langle f(x_\alpha):\alpha<\kappa\rangle$ does not converge to $f(y)$.
Then there exists a subsequence
$\langle x_{\alpha_i}:i<\kappa\rangle$ of $\bar x$ and some $\gamma<\kappa$
$f(x_{\alpha_i}) \notin N_{f(y)\restr\gamma}$
for all $i<\kappa$.
the sequence
$\langle (x_{\alpha_i}, f(x_{\alpha_i})):i<\kappa\rangle$
contradicts the fact that
$A$ is $\JJ^f$-independent.

Conversely, suppose that $f\restr C$ is continuous.
It suffices to show that for any sequence
$\bar x=\langle x_\alpha:\alpha<\kappa\rangle$ in $C$ with $\bar x \in \CC^X$,
the sequence
$\langle (x_\alpha,f(x_\alpha)):\alpha<\kappa\rangle$ is not in $\JJ^f$.
But this follows from the fact that
$f(\lim_{\alpha<\kappa} \bar x)=\lim_{\alpha<\kappa} (\langle f(x_\alpha):\alpha<\kappa\rangle)$
by the continuity of $f\restr C$.

The following lemmas
will be used to characterize the existence of continuous homomorphisms from $\dhH \kappa$ to $\JJ^f\restr\graph(f)$.
Recall from Subsection <ref> that for a strict order preserving map $\iota: {}^{<\kappa}\kappa\to {}^{<\kappa}\kappa$, $\Lim_t^{\iota}$ denotes the union of all
limit points of sets of the form
$\{[\iota](x_\alpha) : \alpha<\kappa\}$
$x_\alpha\in N_{t\conc\langle\alpha\rangle}\cap X$
for all $\alpha<\kappa$.
The next lemma shows that $\iota$ can be “thinned out”
to a map $\iota \circ e$ such that $\Lim_t^{\iota\circ e}$ contains at most one element for any $t\in {}^{<\kappa}\kappa$.

If $\iota:{}^{<\kappa}\kappa\to{}^{<\kappa}\kappa$
is strict order preserving, then there exists a
continuous strict $\wedge$-homomorphism
[See Definition <ref>. Note that a map $e: {}^{<\kappa}\kappa\to {}^{<\kappa}\kappa$ is a strict $\wedge$-homomorphism if and only if it is strict order preserving with $e(t\conc\langle\alpha\rangle) \wedge e(t\conc\langle\beta\rangle)= e(t)$ for all $t\in{}^{<\kappa}\kappa$ and all $\alpha<\beta<\kappa$.]
$|\Lim_t^{\iota\comp e}|\leq 1$ for all $t\in {}^{<\kappa}\kappa$.

We construct $e(t)$ by recursion on $t$.
Let $e(\emptyset)=\emptyset$ and if $t$ is a limit, let $e(t):=\bigcup_{s\subsetneq t} e(s)$.
Now suppose that $e(t)$ has been constructed.
If there exists no $\bar{x}=\langle x_\alpha : \alpha<\kappa\rangle$ in $\prod_{\alpha<\kappa} N_{ e(t)\conc\langle \alpha\rangle}$ such that $\langle [\iota](x_\alpha) : \alpha<\kappa\rangle$ has a convergent subsequence,
then let $e(t\conc \langle \alpha \rangle):=e(t)\conc \langle \alpha \rangle$ for all $\alpha<\kappa$.
Then $\Lim^{\iota\comp e}_t=\Lim^\iota_{e(t)}=\emptyset$.
Otherwise fix a strictly increasing sequence $\langle \alpha_i : i<\kappa\rangle$ of ordinals below $\kappa$ and $\langle x_i : i<\kappa \rangle$ in $\prod_{i<\kappa} N_{e(t)\conc\langle \alpha_i\rangle}$ such that $\langle [\iota](x_i) : i<\kappa\rangle$ converges to some $y$.
For each $i<\kappa$, take $e(t\conc \langle i \rangle)$ with
$e(t)\conc \langle \alpha_i \rangle \subsetneq e(t\conc \langle i \rangle) \subsetneq x_i$
and $\lh\big(e(t\conc \langle i \rangle)\big)\geq i$.
Then $\big\langle \iota\big(e(t\conc \langle i \rangle)\big) : i<\kappa \big\rangle$ converges to $y$,
[Equivalently, all sequences in
$\prod_{i<\kappa} N_{\iota(e(t)\conc\langle \alpha_i\rangle)}$ converge to $y$;
see the paragraph before Lemma <ref>.]

so $\Lim^{\iota\comp e}_t=\{y\}$.

We next extend the notion of order homomorphisms
to dihypergraphs on ${}^\kappa\kappa\times{}^\kappa\kappa$.
\bigcup_{\alpha<\kappa}({}^\alpha\kappa\times{}^\alpha\kappa)$.
Given a strict order preserving map
where $\ddim\leq\kappa$,
let $\theta_i:=\proj_i\comp\theta$
for $i\in\{0,1\}$ and
define $[\theta]: {}^\kappa\ddim\to {}^\kappa\kappa\times{}^\kappa\kappa$ by letting $[\theta](x):=([\theta_0](x),[\theta_1](x))$.
The definition of continuity for $\theta$ is analogous to Definition <ref> <ref>.
$\theta(t)=(\bigcup_{s\subsetneq t}\theta_0(s),$ $\bigcup_{s\subsetneq t}\theta_1(s))$ for all $t\in{}^\kappa\kappa$ with $\lh(t)\in\Lim$.]

In the following definition and lemma,
we assume that
$\ddim\geq 2$ is an ordinal,

$Y$ is a subset of
and $H$ is a
$\ddim$-dihypergraph on ${}^\kappa\kappa\times{}^\kappa\kappa$.

We say that a strict order preserving map
is an order homomorphism for $(Y,H)$
if $\ran([\theta])\subseteq Y$

for all $t\in {}^{<\kappa}\ddim$,

\[
\prod_{\alpha<\ddim}
(N_{\theta_0(t\conc\langle \alpha\rangle)}
\times
N_{\theta_1(t\conc\langle \alpha\rangle)}) \cap Y
\subseteq H.
\]

The next lemma is an analogoue of Lemma <ref> for dihypergraphs on ${}^\kappa\kappa\times {}^\kappa\kappa$.

\beginenumerate-(1)

If $\theta$ is an order homomorphism for $(Y,H)$, then
$[\theta]$ is a continuous homomorphism from $\dhH\ddim$ to $H\restr Y$.
This is a verbatim analogue of  Lemma <ref>. Weaker versions of both (1) and (2) would suffice, but it might be easier to read the verbatim analogue

If $H\restr Y$ is box-open on $Y$ and there exists
a continuous homomorphism $h$
from $\dhH\ddim$ to $H\restr Y$, then
there exists a continuous
order homomorphism $\theta$ for $(Y,H)$
and a continuous strict $\wedge$-homomorphism
with $[\theta]=h\comp [e]$.
\endenumerate-(1)

Let $k$ denote the natural coding of elements $(u,v)$ of
[We write
\cup
as elements
$k(u,v)$ of ${}^{\leq\kappa}\kappa$
such that
[That is, for all $\alpha<\lh(u)$, let
and let

$k$ induces the $\ddim$-dihypergraph $\bar H:=k^\ddim(H)$ on
Moreover, $\theta: {}^{<\kappa}\ddim\to {}^{<\kappa}\kappa\times {}^{<\kappa}\kappa$ is an order homomorphism for $(Y,H)$
if and only if $k\comp \theta$ is an
order homomorphism for $(k(Y),\bar H)$.
Use that $(x,y)$ is in
$N_{\theta_0(t\conc\langle \alpha\rangle)}
\times
N_{\theta_1(t\conc\langle \alpha\rangle)}$
if and only if
$k((x,y))$ is in $N_{(k\comp\theta)(t\conc\langle\alpha\rangle)}$.

For <ref>, suppose $\theta$ is an order homomorphism for $(Y,H)$.
Then $[k\comp\theta]=k\comp[\theta]$
is a continuous homomorphism from $\dhH\ddim$ to
$\bar H\restr k(Y)$

by Lemma <ref>.
Hence $[\theta]$ is a continuous homomorphism from $\dhH\ddim$ to
$H\restr Y$.

For <ref>, suppose $h$ is a continuous homomorphism from $\dhH\ddim$ to $H\restr Y$.
Then $k\comp h$ is a continuous homomorphism from $\dhH\ddim$ to $\bar H\restr k(Y)$.
Since $\bar H\restr k(Y)$ is a box-open dihypergraph on $k(Y)$,
there exists
a continuous order homomorphism $\iota$ for $(k(Y),\bar H)$
and a continuous strict $\wedge$-homomorphism
with $[\iota]=k\comp h\comp [e]$
by Lemma <ref>.
We can assume that $\lh(\iota(t))$ is even for all $t\in{}^\kappa\kappa$ by Remark <ref>.
Then $\theta:=k^{-1}\comp \iota$ is an order homomorphism for $(Y,H)$
with $[\theta]=k^{-1}\comp [\iota]=h\comp [e]$.
$\theta$ is continuous since both $k^{-1}$ and $\iota$ are continuous.

\beginenumerate-(1)

$\JJ^f\restr\graph(f)$  admits a $\kappa$-coloring if and only if $f$ is $\kappa$-continuous with closed pieces.

There exists a continuous homomorphism from $\dhH\kappa$ to
if and only if
there exists a continuous map $g:{}^\kappa 2\to X$ with the following properties:
\beginenumerate-(a)

$f\comp g$ is a reduction of $\RR_\kappa$ to a closed subset of ${}^\kappa\kappa$.[A reduction $h: {}^\kappa2\to {}^\kappa\kappa$ of a set $A$ to a set $B$ is a function with $A=h^{-1}(B)$.
$\RR_\kappa,\,\QQ_\kappa$ and $\SS_\kappa$ were defined in Subsection <ref>.]

$(f\circ g){\restr}\RR_\kappa$ is continuous.
\endenumerate-(a)
\endenumerate-(1)
In <ref>, we can equivalently take $g$ to be injective, or even a homeomorphism onto a closed subset of ${}^\kappa\kappa$.

For <ref>, suppose first that $\graph(f)=\bigcup_{\alpha<\kappa} A_\alpha$, where each $A_\alpha$ is a
$\JJ^f$-independent subset of $\graph(f)$.
For each $\alpha<\kappa$, let $C_\alpha$ be the closure of $\proj_0(A_\alpha)$ in $X$.
Then $f\restr C_\alpha$ is continuous by Lemma <ref>, and $X=\bigcup_{\alpha<\kappa} C_\alpha$,
so $f$ is $\kappa$-continuous with closed pieces.
Conversely, suppose that the sequence $\langle C_\alpha:\alpha<\kappa\rangle$ witnesses that $X$ is $\kappa$-continuous with closed pieces.
Since each $f\restr C_\alpha$ is continuous,
the sets
$A_\alpha:=\{(x,f(x)): x\in C_\alpha\}$ are $\JJ^f$-independent by  Lemma <ref>.
$X=\bigcup_{\alpha<\kappa} C_\alpha$,
we have $\graph(f)=\bigcup_{\alpha<\kappa} A_\alpha$.

For <ref>, suppose that $g:{}^\kappa 2\to X$ is a continuous map with <ref> and <ref>.
Recall the definition of $\pi$ in
Subsection <ref>.
Let $h:=(g\circ [\pi], f\circ g \circ [\pi]): {}^\kappa\kappa\to X\times {}^\kappa\kappa$.
Then $\ran(h)\subseteq\graph(f)$,
and $h$
is continuous
since $f\comp g$ is continuous on $\RR_\kappa=\ran([\pi])$.

It remains to show that $h$ is a homomorphism from $\dhH\kappa$ to $\JJ^f$.
To see this, suppose that $\langle y_\alpha : \alpha<\kappa\rangle$ is a hyperedge in $\dhH\kappa$ with $t^\smallfrown\langle\alpha\rangle\subseteq y_\alpha$ for all $\alpha<\kappa$.
Then $t^\smallfrown \langle 0\rangle^\alpha {}^\smallfrown\langle1\rangle \subseteq [\pi](y_\alpha)$ for all $\alpha<\kappa$,
so $\lim_{\alpha<\kappa} [\pi](y_\alpha) = \pi(t)^\smallfrown \langle 0\rangle^\kappa$.
Since $g$ is continuous, $\lim_{\alpha<\kappa} (g\circ[\pi])(y_\alpha)=x_t:=g(\pi(t)^\smallfrown\langle0\rangle^\kappa)$.
Since $f\circ g$ is a reduction of $\RR_\kappa$, $g(\RR_\kappa)$ and $g(\QQ_\kappa
)$ are disjoint.
Hence $(g\circ[\pi])(y_\alpha)\neq x_t$ for all $\alpha<\kappa$ and $\langle (g\circ[\pi])(y_\alpha): \alpha<\kappa\rangle \in \CC^X$.
Suppose that $f\circ g$ is a reduction of $\RR_\kappa$ to the closed set $C$.
Then $f(x_t)\notin\closure{\{z_\alpha : \alpha<\kappa \}}\subseteq C$, where $z_\alpha := (f\circ g \circ [\pi])(y_\alpha)$.
Thus $\langle h(y_\alpha) : \alpha<\kappa \rangle \in \JJ^f$ as required.

For the converse,
suppose that there is a continuous homomorphism from $\dhH\kappa$ to
Since $\JJ^f$ is box-open on ${}^\kappa\kappa\times{}^\kappa\kappa$,
there is an order homomorphism
$\theta=(\theta_0,\theta_1)$ for
with $\ran([\theta])\subseteq \graph(f)$
$\ran([\theta_0])\subseteq X$ and
by Lemma <ref>.
Then $\theta_0$ is an
order homomorphism for $({}^\kappa\kappa,\CC^X)$.
$\langle \theta_0(t\conc\langle \alpha\rangle):\alpha<\kappa\rangle$
converges to some
$x^{\theta_0}_t\in X$
for each $t\in{}^{<\kappa}\kappa$
by Lemma <ref>.

We now show that $\theta$ can be assumed to have some additional properties.

There exists an
order homomorphism $\psi$
with $\ran([\psi])\subseteq \graph(f)$
such that for all $t\in{}^{<\kappa}\kappa$,
$f(x^{\psi_0}_t)$ is not in the closure of $\ran([\psi_1])$.

Let $\theta$ be any order homomorphism
with $\ran([\theta])\subseteq \graph(f)$.
We will construct a
strict $\wedge$-homomorphism
$e:{}^{<\kappa}\kappa\to{}^{<\kappa}\kappa$ such that for $\psi:=\theta\comp e$ we have

$$\forall t\in{}^{<\kappa}\kappa\ \ f(x^{\psi_0}_{t})\notin\closure{\ran([\psi_1])}.$$

This suffices, since
$\psi$ will also be an order homomorphism for
$e$ is a $\wedge$-homomorphism and
$\JJ^f$ is closed under subsequences and permutations of hyperedges.

$[\theta_1]{\restr}N_{t}$ is constant for some $t\in {}^{<\kappa}\kappa$.

In this case, let $y$ denote the unique element of $[\theta_1](N_{t})$.
Let $e(u):=t\conc u$ for all $u\in {}^{<\kappa}\kappa$.
For $\psi:= \theta \circ e$, $\ran([\psi_1])$ has the unique element $y$.
It thus suffices to show $f(x_{u}^{\psi_0})\neq y$ for all $u\in {}^{<\kappa}\kappa$.
To see this, pick $y_\alpha$ with $u\conc\langle\alpha \rangle \subseteq y_\alpha$ for $\alpha<\kappa$.

$f(x^{\psi_0}_u)=f(x^{\theta_0}_{t\conc u})$
$f(x^{\psi_0}_u)=f(x^{\theta_0}_{t\conc u})$ and $[\psi_1](y_\alpha)=[\theta_1](t\conc y_\alpha)$ extends
$\theta_1(t\conc u\conc\langle\alpha\rangle)$ for each $\alpha$
is not in the closure $C$ of
$\{ [\psi_1](y_\alpha) : \alpha<\kappa \}=
\{[\theta_1](t\conc y_\alpha)\:\alpha<\kappa\}$, since $\theta$ is an order homomorphism
for $({{}^\kappa\kappa\times{}^\kappa\kappa},\,\JJ^f)$
But $C$ has the unique element $y$.

$|\{f(x_t^{\theta_0}) : s\subseteq t\in {}^{<\kappa}\kappa \} |<\kappa$ for some $s\in {}^{<\kappa}\kappa$
and Case <ref> does not occur.

We first claim that there exists some $u\in {}^{<\kappa}\kappa$ such that
$A_u:=\{ t\in{}^{<\kappa}\kappa : f(x_{u}^{\theta_0})=f(x_t^{\theta_0})\}$
is dense above some $v\in{}^{<\kappa}\kappa$.
[See Definition <ref>.]
Let $\bar{u}=\langle u_\alpha : \alpha<\gamma \rangle$ be a sequence in ${}^{<\kappa}\kappa$ such that $\gamma<\kappa$ and for any $t\in {}^{<\kappa}\kappa$ extending $s$, there exists some $\alpha<\gamma$ with $f(x_t^{\theta_0})=f(x_{u_\alpha}^{\theta_0})$.
Supposing each $A_{u_\alpha}$ is nowhere dense, we can construct a strictly increasing sequence $\langle s_\alpha : \alpha<\gamma \rangle$ in ${}^{<\kappa}\kappa$ with $s\subseteq s_0$ such that for each $\alpha<\gamma$, there exists no $t\supseteq s_\alpha$ with $f(x_t^{\theta_0})=f(x_{u_\alpha}^{\theta_0})$.
Then for $t= \bigcup_{\alpha<\gamma} s_\alpha$, we have $f(x_t^{\theta_0})\neq f(x_{u_\alpha}^{\theta_0})$ for all $\alpha<\gamma$.
But this contradicts the choice of $\bar{u}$.

Since $[\theta_1]\restr N_{v}$ is not constant,
there exists some $y\in N_v$ with $[\theta_1](y)\neq f(x^{\theta_0}_u)$.
Take $w\subsetneq y$ with $v\subseteq w$ and $\theta_1(w)\perp f(x^{\theta_0}_u)$.
Since $A_u$ is dense above $w$, we can
construct a strict $\wedge$-homomorphism
$e: {}^{<\kappa}\kappa \to {}^{<\kappa}\kappa$
with $\ran(e)\subseteq A_u$
and $w\subseteq e(\emptyset)$.
Let $\psi:=\theta\circ e$.
for any $t\in {}^{<\kappa}\kappa$
since $e$ is a $\wedge$-homomorphism, so

since $\ran(e)\subseteq A_u$.

since $\ran([\psi_1])$ is a subset of $N_{\theta_1(w)}$.

Neither Case <ref> nor Case <ref> occurs.

By Lemma <ref> for $\theta_1$,
there exists a strict $\wedge$-homomorphism
such that $|\Lim^{\theta_1\comp \epsilon}_t|\leq 1$ for all $t\in{}^{<\kappa}\kappa$.
Since $\JJ^f$ is closed under subsequences and permutations of hyperedges,
$\theta\comp \epsilon$
is an order homomorphism for
we may assume that $|\Lim^{\theta_1}_t|\leq 1$ for all $t\in{}^{<\kappa}\kappa$,
by replacing $\theta$ with $\theta\comp \epsilon$ if necessary.

Fix an enumeration $\langle t_\alpha : \alpha<\kappa \rangle$ of ${}^{<\kappa}\kappa$ such that
$t_\alpha\subsetneq t_\beta$ implies $\alpha<\beta$.

There exists a strict $\wedge$-homomorphism
$e: {}^{<\kappa}\kappa \to {}^{<\kappa}\kappa$ such that $\psi:=\theta\comp e$ has the following properties for all $\alpha,\beta<\kappa$:
\beginenumerate-(i)

$f(x_{t_\beta}^{\psi_0}) \notin \Lim_{t_\alpha}^{\psi_1}$ if $\alpha\leq\beta$.

$f(x_{t_\alpha}^{\psi_0}) \notin N_{\psi_1(t_\beta)}$ if $\alpha<\beta$.
\endenumerate-(i)

Note that <ref> already holds for $\theta_1$ and $\alpha=\beta$.
To see this, observe that $f(x^{\theta_0}_{t}) \notin \Lim_{t}^{\theta_1}$ for all $t\in {}^{<\kappa}\kappa$,
since $\langle\theta_0(t\conc\langle\alpha\rangle):\alpha<\kappa\rangle$ converges to
and $\theta$ is an order homomorphism for
Since this property remains true for $\psi_1=\theta_1\comp e$ for any choice of $e$,
we only need to ensure <ref> in the case $\alpha<\beta$.

We first show how to arrange <ref>.
We define $e(t_\beta)$ by recursion on $\beta$.
Let $B_\beta:=\bigcup_{\alpha<\beta} \Lim_{e(t_\alpha)}^{\theta_1}$.
Since $|\Lim_{t_\alpha}^{\theta_1}|\leq 1$ for all $\alpha<\kappa$, we have $|B_\beta|<\kappa$.
$u:=e(t_\gamma)\conc\langle\eta\rangle$ if $t_\beta=t_\gamma\conc\langle\eta\rangle$ for some $\gamma<\beta$ and $\eta<\kappa$.
If $t_\beta$ has limit length, then let
$u:=\bigcup_{s\subsetneq t_\beta} e(s)$.
[This is defined since each $s\subsetneq t_\beta$ is equal to $t_\alpha$ for some $\alpha<\beta$.]
Since $|\{f(x^{\theta_0}_t) : u\subseteq t\in{}^{<\kappa}\kappa \} |\geq \kappa$ by the case assumption,
we can find some $t$ extending

with $f(x^{\theta_0}_t)\notin B_\beta$.
Then $e(t_\beta):=t$ satisfies <ref>.
Proof: $f(x^{\psi_0}_{t_\beta})=f(x^{\theta_0}_t)\notin B_\beta\supseteq\bigcup_{\alpha<\kappa}\Lim_{t_\alpha}^{\psi_1}$
Note that $e$ is a strict $\wedge$-homomorphism.

Assume $\theta$ satisfies <ref>.
We next show how to arrange <ref>.
Again, we define $e(t_\beta)$ by recursion on $\beta$.
We let $e$ be continuous at nodes $t_\beta$ of limit length.
Suppose that $t_\beta$ is a direct successor of $t_\gamma$ with $t_\beta=t_\gamma\conc\langle \eta \rangle$.
Since $[\theta_1](N_{e(t_\gamma)\conc\langle \eta \rangle})$ has at least $2$ elements by the case assumption, we can pick some proper extension $u_0$ of $e(t_\gamma)\conc\langle \eta \rangle$ such that $\theta_1(u_0)$ is incompatible with $f(x^{\theta_0}_{e(t_0)})$.
Continuing in the same fashion, we construct a strictly increasing sequence $\langle u_\alpha : \alpha<\beta\rangle$ such that $\theta_1(u_\alpha)$ is incompatible with $f(x^{\theta_0}_{e(t_\alpha)})$ for all $\alpha<\beta$.
Let  $e(t_\beta) := \bigcup_{\alpha<\beta}u_\alpha$.
Then <ref> holds for $t_\beta$.
Proof: $\psi_1(t_\beta)=\theta_1(e(t_\beta))$ is incomatible with $f(x^{\psi_0}_{t_\alpha})=f(x^{\theta_0}_{e(t_\alpha)})$ for all $\alpha<\beta$.
Again, $e$ is a strict $\wedge$-homomorphism.

Let $\psi:=\theta\comp e$, where $e$ is as in the previous subclaim.
We show that
$f(x_{t}^{\psi_0})\notin\closure{\ran([\psi_1])}$ for all $t\in{}^{<\kappa}\kappa$.
Suppose that this fails for some $t\in{}^{<\kappa}\kappa$.
Suppose that $t=t_\alpha$.

We first show that $f(x_{t}^{\psi_0})\notin\ran([\psi_1])$. Otherwise,
suppose that  $y\in{}^\kappa\kappa$ with
$f(x_{t}^{\psi_0})=[\psi_1](y)=\bigcup_{u\subsetneq y}\psi_1(u)$.
Take any $\beta$ with $\alpha<\beta<\kappa$ and $t_\beta\subsetneq y$.
$f(x_{t}^{\psi_0})\in N_{\psi_1(t_\beta)}$, contradicting <ref>.
We next show $f(x_{t}^{\psi_0})\notin\closure{\ran([\psi_1])}$.
Suppose otherwise.
Since $\closure{\ran([\psi_1])}=
\ran([\psi_1])\cup\bigcup_{\beta<\kappa}\Lim^{\psi_1}_{t_\beta}$ by Lemma <ref> <ref>,
we have $f(x_{t}^{\psi_0})\in \Lim^{\psi_1}_{t_\beta}$
for some $\beta<\kappa$.
Then $\alpha<\beta$ by <ref>.
\closure{\bigcup_{\eta<\kappa} N_{\psi_1(t_\beta\conc\langle\eta\rangle)}}
\subseteq N_{\psi_1(t_\beta)}$,
we have
$f(x_{t}^{\psi_0})\in N_{\psi_1(t_\beta)}$.
But this contradicts <ref>.

Let $\psi$ be an order homomorphism for
as in the previous claim.
Since $\psi_0$ is an
order homomorphism for

there exists
a nice reduction $\Phi:{}^{<\kappa}2\to{}^{<\kappa}\kappa$ for
and a strict $\wedge$-homomorphism
[Note that a strict order preserving map $\epsilon:\SS_\kappa\to{}^{<\kappa}\kappa$  is a strict $\wedge$-homomorphism if and only if $\epsilon(s\conc\langle 0\rangle^\alpha\conc\langle 1\rangle) \wedge \epsilon(s\conc\langle 0\rangle^\beta\conc\langle 1\rangle)=\epsilon(s)$ for all $s\in\SS_\kappa$ and $\alpha<\beta<\kappa$.]
such that
$\psi_0\comp \epsilon=\Phi\restr\SS_\kappa$
by Lemma <ref>.

We will show $g:=[\Phi]:{}^\kappa 2\to {}^\kappa \kappa$ is as required in

Since $\Phi$ is $\perp$- and strict order preserving,
$g$ is a homeomorphism onto a closed subset of ${}^\kappa\kappa$
this is needed because of the last sentence in the theorem

by Lemma <ref> <ref>.
We next show that $\ran(g)\subseteq X$ and that $(f\circ g){\restr}\RR_\kappa$ is continuous.
It is clear that $g(\QQ_\kappa)\subseteq X$
and $g(\RR_\kappa)\subseteq \ran([\psi_0])$.
Since $\ran([\psi])\subseteq\graph(f)$,
we have
$\ran([\psi_0])\subseteq X$
and hence $\ran(g)\subseteq X$.
Moreover, note that
$f\comp[\psi_0]=[\psi_1]$ since
Thus $(f\comp g)\restr\RR_\kappa = [\psi_1 \comp \epsilon]$
and hence it is continuous.

We now show that $f\comp g$ is a reduction of $\RR_\kappa$ to a closed subset of ${}^\kappa\kappa$.
Since $(f\comp g)(\RR_\kappa)\subseteq\ran([\psi_1])$,
it suffices to show that
$(f\comp g)(\QQ_\kappa)$ and $\closure{\ran([\psi_1])}$ are disjoint.
Suppose that $z\in\QQ_\kappa$. Then $z=s\conc\langle 0\rangle^\kappa$ for some $s\in\SS_\kappa$.
For each $\alpha<\kappa$, let $t_\alpha:=s\conc\langle 0\rangle^\alpha\conc \langle 1\rangle$.
We claim that
This suffices, since
$f(x^{\psi_0}_{\epsilon(s)})\notin\closure{\ran([\psi_1])}$ by the choice of $\psi$.

Since $\epsilon$ is a $\wedge$-homomorphism from $\SS_\kappa$ to ${}^{<\kappa}\kappa$,

there exists an injective sequence
$\langle \eta_\alpha:\alpha<\kappa\rangle$ of ordinals below $\kappa$ such that
$\epsilon(s)\conc\langle \eta_\alpha\rangle\subseteq \epsilon(t_\alpha)$
hence $\psi_0(\epsilon(s)\conc\langle \eta_\alpha\rangle)\subseteq \Phi(t_\alpha)$.

On the one hand,
$\langle\Phi(t_\alpha):\alpha<\kappa\rangle$ converges to
$g(z)=[\Phi](s\conc\langle 0\rangle^\kappa)$.
On the other hand,
$\psi_0(\epsilon(s)\conc\langle \alpha\rangle)$
converges to $x^{\psi_0}_{\epsilon(s)}$
and therefore so does
$\psi_0(\epsilon(s)\conc\langle \eta_\alpha\rangle)$.
$\psi_0(\epsilon(s)\conc\langle \eta_\alpha\rangle)\subseteq \Phi(t_\alpha)$,
we must have

We next obtain an analogue of Jayne and Rogers' characterization of ${\bf \Delta}^0_2$-measurable functions <cit.>
from the open dihypergraph dichotomy.

implies $\JR_\kappa^f$.

Note that any $\kappa$-continuous function with closed pieces is ${\bf\Delta}^0_2(\kappa)$-measurable.
Therefore it suffices to show that the condition in Theorem <ref> <ref> fails for any ${\bf\Delta}^0_2(\kappa)$-measurable function $f: X\to {}^\kappa\kappa$.
To see this,
suppose that there exists a continuous map $g:{}^\kappa 2\to X$ such that $f\comp g$ is a reduction of $\RR_\kappa$ to a closed subset of ${}^\kappa\kappa$.
Then $\RR_\kappa$ would be a ${\bf \Sigma}^0_2(\kappa)$ subset of ${}^\kappa\kappa$.

Let $\JJ^f_i$ denote the variant of $\JJ^f$ where $\CC^X$ is replaced with the $\kappa$-dihypergraph $\CC_i^X:=\CC^X\cap \dhIkappa$ of all injective sequences in $\CC^X$.

In the previous theorem and corollary,
$\JJ^f$ can be equivalently replaced by $\JJ_i^f$.
This follows from Corollary <ref>
since $\JJ_i^f$
is box-open on ${}^\kappa\kappa$,
and any hyperedge of $\JJ^f$ has a subsequence in $\JJ^f_i$ and hence $\JJ^f_i\equivf\JJ^f$.

\beginenumerate-(1)

\JR_\kappa^f$
for any $X\in\defsetsk$ and any $f:X\to{}^\kappa\kappa$.

\endenumerate-(1)

We prove <ref> and <ref> simultaneously.
Suppose that $X\in\defsetsk$ and $f:X\to{}^\kappa\kappa$.
It suffices to show that $\JR^f_\kappa$ follows from each of
and $\ODD\kappa\kappa(\defsetsk,\defsetsk)$.
First suppose that $f\in\defsetsk$.
Then $\JJ^f$ is in $\defsetsk$
and we thus have
by the previous corollary.
Since $\graph(f)$ is then in $\defsetsk$, we further have
If on the other hand $f\notin\defsetsk$, then $\JR^f_\kappa$ holds trivially since
all ${\bf\Delta}^0_2(\kappa)$-measurable functions on $X$ and all $\kappa$-continuous functions on $X$ with closed pieces are in $\defsetsk$.

Theorem <ref> and the previous corollary yield the following result.

$\JR_\kappa(\defsetsk)$ holds
in all $\Col(\kappa,\lle\lambda)$-generic extensions
if $\lambda>\kappa$ is inaccessible in $V$.

§.§ Some implications

In this subsection,
we collect implications and separations
between some of the applications of the open dihypergraph dichotomy from the previous subsections such as
the Kechris-Louveau-Woodin dichotomy, its generalization to arbitrary sets,
the $\kappa$-perfect set property, versions of the Hurewicz dichotomy
and analogues to the Jayne-Rogers theorem.
The known information on implications between the above principles is summarized in the following Figures <ref> and <ref>.
[Blue arrows are clickable and lead to the corresponding results.]
Throughout this subsection,
$\mathcal C$ and $\mathcal D$ denote classes of subsets of ${}^\kappa\kappa$ and
$\KLWdis_\kappa(\mathcal C,\mathcal D)$ denotes the statement that $\KLW_\kappa(X,Y)$ holds for all disjoint subsets $X\in\mathcal C$ and $Y\in\mathcal D$ of ${}^\kappa\kappa$.
[Recall that $\KLW_\kappa(\mathcal C,\mathcal D)$ states that $\KLW_\kappa(X,Y)$ holds for arbitrary subsets $X\in\mathcal C,\,Y\in\mathcal D$ of ${}^\kappa\kappa$.]
Thus, $\KLWdis_\kappa(\mathcal C):=\KLWdis(\mathcal C,\pwrset({}^\kappa\kappa))$
is a direct analogue of the Kechris-Louveau-Woodin dichotomy
for classes $\mathcal C$ of subsets of the $\kappa$-Baire space.

theory1=[draw,rounded corners,scale=.900, minimum width=8.0mm, minimum height=8.0mm,line width=,VeryDarkBlueNode]
theory2=[draw,rounded corners,scale=.900, minimum width=8.0mm, minimum height=8.0mm,line width=,GreyBlueNode]
info=[rounded corners, inner sep= 2pt, align=left,scale=0.82]
label=[rounded corners, inner sep= 2pt, align=left,scale=0.95]

medarrow=[>=Stealth, line width=, scale=1.8, VeryDarkBlueNode]

(11) at (0,3.6) [theory1] [c][.6cm]2.1cm$\,\KLW_\kappa(\mathcal{C},\mathcal D)$;
(13) at (8,3.6) [theory1] [c][.6cm]1.5cm$\,\PSP_\kappa(\mathcal{C})$;
(21) at (0,1.8) [theory1] [c][.6cm]2.3cm$\,\KLWdis_\kappa(\mathcal{C},\mathcal D)$
(22) at (4,1.8) [theory1] [c][.6cm]2.15cm
$\,\THD_\kappa(\mathcal C,\mathcal D)$
(23) at (8,1.8) [theory1] [c][.6cm]1.6cm$\,\THD_\kappa(\mathcal C)$;
(24) at (11.6,1.8) [theory1] [c][.6cm]1.38cm$\,\HD_\kappa(\mathcal C)$;
(33) at (8,0.0) [theory1] [c][.6cm]1.25cm$\,\JR_\kappa(\mathcal C)$;

(pointA1) at (6.15,4.10);
(pointA2) at (6.15,3.05);
(pointB1) at (6.15,2.70);
(pointB2) at (6.15,2.15);
(pointC1) at (6.15,1.85);
(pointC2) at (6.15,1.35);
(pointD1) at (6.15,0.42);
(pointD2) at (6.15,-0.45);
[medarrow, GreyBlueLine]
([yshift=0mm]pointA1) edge [dashed]
[medarrow, GreyBlueLine]
([yshift=0mm]pointB1) edge [dashed]
[medarrow, GreyBlueLine]
([yshift=0mm]pointC1) edge [dashed]
[medarrow, GreyBlueLine]
([yshift=0mm]pointD1) edge [dashed]

[medarrow,  VeryDarkBlueNode!80]
([yshift=0mm]22.east) edge [->]
node[midway,above]${}^\kappa\kappa\in \mathcal D$

[medarrow, BlueArrow]
([yshift=0mm]11.east) edge [->]
node [connection btw KLW, KLWdis and PSP]xxx
node[midway,above]${}^\kappa\kappa\in \mathcal D$

([yshift=0mm]11.south) edge [->>]
node sep: KLW and KLWdisxxx

([yshift=0.7mm]21.east) edge [->]
node [cor: KLW implies THD weakly compact]xxx
node[midway, above]$\mathcal C, \mathcal D$ closed

([yshift=-0.7mm]21.east) edge [<->]
node [cor: KLW implies THD weakly compact]xxx

node[below, inner sep=5 pt][c]1.70 cm
$\mathcal C\, {\subseteq}\,$eventually bounded

([yshift=0mm]23.east) edge [<->]
node [proposition: HD and THD options]xxx

([yshift=-1.85mm,xshift=-0.165mm]22.east) edge [->]
node [cor: THD implies JR weakly compact]xxx

node[midway,below,sloped] $\Gdelta(\kappa){\subseteq}\mathcal C{\subseteq}\mathcal D$
node[midway,above,sloped]$\mathcal C$ closed

[medarrow, BlueArrow]
([yshift=0mm]23.north) edge [->]

node [sep: PSP and THD]xxx
[medarrow, BlueArrow](4.37,1.47) edge (4.52,1.47);

[medarrow, BlueArrow]

([yshift=-1.85mm,xshift=0.15 mm]13.west) edge [->]
node [sep: not PSP to THD omega]xxx
node[midway,above, sloped]$\kappa=\omega$
([yshift=1.85mm,xshift=-0.15 mm]22.east);
[medarrow, BlueArrow](3.37,1.545) edge (3.49,1.465);

[medarrow, BlueArrow]
([yshift=-1.85mm,xshift=-0.15 mm]13.east) edge [<-]
node [sep: PSP and HD]xxx
([yshift=1.85mm,xshift=0.145 mm]24.west);
[medarrow, BlueArrow](5.41,1.45) edge (5.53,1.53);

Implications in the case that $\kappa$ is weakly compact or $\omega$

theory1=[draw,rounded corners,scale=.900, minimum width=8.0mm, minimum height=8.0mm,line width=,VeryDarkBlueNode]
theory2=[draw,rounded corners,scale=.900, minimum width=8.0mm, minimum height=8.0mm,line width=,GreyBlueNode]
info=[rounded corners, inner sep= 2pt, align=left,scale=0.82]
label=[rounded corners, inner sep= 2pt, align=left,scale=0.95]

medarrow=[>=Stealth, line width=, scale=1.8, VeryDarkBlueNode]

(11) at (0,3.6) [theory1] [c][.6cm]2.1cm$\,\KLW_\kappa(\mathcal{C},\mathcal D)$;
(13) at (8,3.6) [theory1] [c][.6cm]1.5cm$\,\PSP_\kappa(\mathcal{C})$;
(21) at (0,1.8) [theory1] [c][.6cm]2.3cm$\,\KLWdis_\kappa(\mathcal{C},\mathcal D)$
(22) at (4,1.8) [theory1] [c][.6cm]2.15cm
$\,\THD_\kappa(\mathcal C,\mathcal D)$
(23) at (8,1.8) [theory1] [c][.6cm]1.6cm$\,\THD_\kappa(\mathcal C)$;
(24) at (11.6,1.8) [theory1] [c][.6cm]1.38cm$\,\HD_\kappa(\mathcal C)$;
(33) at (8,0.0) [theory2] [c][.6cm]1.25cm$\,\JR_\kappa(\mathcal C)$;

(pointA1) at (6.15,4.08);
(pointA2) at (6.15,2.15);
(pointB1) at (6.15,1.88);
(pointB2) at (6.15,-0.45);
[medarrow, GreyBlueLine]
([yshift=0mm]pointA1) edge [dashed]
[medarrow, GreyBlueLine]
([yshift=0mm]pointB1) edge [dashed]

[medarrow, VeryDarkBlueNode!80]

([yshift=0mm]22.east) edge [->]
node[midway,above]${}^\kappa\kappa\in \mathcal D$

[medarrow, BlueArrow]

([yshift=0mm]11.east) edge [->]
node [connection btw KLW, KLWdis and PSP]xxx
node[midway,above]${}^\kappa\kappa\in \mathcal D$

([yshift=0mm]11.south) edge [->>]
node sep: KLW and KLWdisxxx

([yshift=-1.85mm,xshift=0.165 mm]13.west) edge [->>]

node [impl and sep: PSP and THD non weakly compact]xxx

([yshift=1.85mm,xshift=-0.145 mm]22.east)

[medarrow, BlueArrow]
([yshift=-1.85mm,xshift=-0.15 mm]13.east) edge [<-]
node[midway,above, sloped]$\kappa$ inaccessible
node [sep: PSP and HD]xxx
([yshift=1.85mm,xshift=0.145 mm]24.west);
[medarrow, BlueArrow](5.41,1.45) edge (5.53,1.53);

Implications in the case that $\kappa$ is not weakly compact.

A double arrowhead denotes strict implication in both diagrams.
Some of the implications in the diagrams use additional assumptions
such as the statements that
$\mathcal C$ or $\mathcal D$ are closed under continuous preimages or
that $\mathcal C$ consists solely of eventually
bounded subsets of ${}^\kappa \kappa$.
These conditions are abbreviated in the diagrams.
For all implications in Figure <ref>, we further assume that
$\mathcal C$, $\mathcal D$ are closed under finite intersections
and $\mathcal D$ is closed under complements,
but note that
the former is relevant only for the implication $\THD_\kappa(\mathcal C,\mathcal D)\Longrightarrow\JR_\kappa(\mathcal C)$
and the latter only for the implications between $\KLWdis_\kappa(\mathcal C,\mathcal D)$, $\THD_\kappa(\mathcal C, \mathcal D)$ and $\JR_\kappa(\mathcal C)$.

We now proceed with proving the implications in the diagrams.
Note that
$\KLW_\kappa(\mathcal C,\mathcal D)$ implies $\KLWdis_\kappa(\mathcal C,\mathcal D)$.
The next proposition discusses connections with the $\kappa$-perfect set property.

that for all $X\in\mathcal C$, there exists $Y\in\mathcal D$ with $X\subseteq Y$.
[For example, this holds if ${}^\kappa\kappa\in\mathcal D$.
More generally, it would suffice to assume in the proposition that
all elements of $\mathcal C$ can be obtained as the intersection
of sets $X\in\mathcal C$ and $Y\in\mathcal D$ such that $X\setminus Y$ can be separated from $Y$ by a $\Fsigma(\kappa)$ set.]
\beginenumerate-(1)

$\KLW_\kappa(\mathcal C,\mathcal D)\Longrightarrow \PSP_\kappa(\mathcal C)$.

If $\mathcal C$ is closed under intersections with elements of $\mathcal D$ and their complements, then

$$\KLW_\kappa(\mathcal C,\mathcal D)\Longleftrightarrow\KLWdis_\kappa(\mathcal C,\mathcal D)\land\PSP_\kappa(\mathcal C).$$

\endenumerate-(1)

follows immediately from Corollary <ref> <ref>.
For <ref>,
the direction from left to right follows from <ref>.
The converse follows from Corollary <ref> <ref>.

In particular,
$\KLW_\kappa(\mathcal C,\mathcal C)\Longleftrightarrow\KLWdis_\kappa(\mathcal C,\mathcal C)$
if $\kappa$ is weakly compact or $\omega$
and $\PSP_\kappa(\mathcal C)$ holds
and $\mathcal C$ is the class of definable or $\kappa$-Borel subsets of ${}^\kappa\kappa$.
sep: KLW and KLWdis
the two dichotomies can be separated for the class of $\Fsigma(\kappa)$ sets: $\KLWdis_\kappa(\Fsigma(\kappa))$ is trivially true,

$\KLW_\kappa(\Fsigma(\kappa))$ implies
$\PSP_\kappa(\Fsigma(\kappa))$ by Corollary <ref> <ref>.

In the countable setting,
neither the perfect set property $\PSP_\omega(\mathcal C)$
nor the open graph dichotomy $\OGD_\omega(\mathcal C)$
implies the Kechris-Louveau-Woodin dichotomy $\KLWdis_\omega(\mathcal C,\Fsigma)$
for the class
$\mathcal C:=L(\RR)[U]$, where $L(\RR)$ is a Solovay model and $U$ is a selective ultrafilter on $\omega$ that is generic over $L(\RR)$.
This follows from results of Di Prisco and Todorčević

[$\OGD_\omega(X)$ is denoted by $\OCA(X)$ and $\KLW_\omega(X,Y)$ is called the Hurewicz separation property in <cit.>.]
In particular,
<cit.>*Theorem 5.1 shows that $\OGD_\omega(\mathcal{C})$ holds, while
<cit.>*Lemma 6.1 shows that $\KLW_\omega(X,Y)$ fails for the subset $X$ of the Cantor space induced by $U$ and $Y=\QQ_\omega$.
Dorottya's note to self: Every countable dense subset of the Cantor space is homeomorphic to the rationals and its complement is homeomorphic to ${}^\omega\omega$, by <cit.>*Exercises 7.12 & 7.13. Hence the Hurewicz separation property as defined in <cit.> for disjoint sets $X$ and $Y$ is equivalent to $\KLW_\omega(X,Y)$.

Turning to
versions of the Hurewicz dichotomy,
recall that $\THD_\kappa(\mathcal C,\mathcal D)$ states that $\THD_\kappa(X,Y)$ holds for all sets $X\subseteq Y$ with $X\in\mathcal C$ and $Y\in\mathcal D$.
$\THD_\kappa(\mathcal C):=\THD_\kappa(\mathcal C,{}^\kappa\kappa)$
is equivalent to $\HD_\kappa(\mathcal C)$ at weakly compact cardinals $\kappa$ and $\kappa=\omega$
by Remark <ref>.

The connections between $\THD_\kappa(\mathcal C,\mathcal D)$ and
$\KLWdis_\kappa(\mathcal C,\mathcal D)$ for weakly compact cardinals and $\omega$
in Figure <ref>
will follow from the
implications in Figure <ref>.
Note that these hold at all infinite cardinals $\kappa$ with $\kappa^{<\kappa}=\kappa$.

theory1=[draw,rounded corners,scale=.5, minimum width=10mm, minimum height=6.5mm,scale=1.8,line width=,VeryDarkBlueNode]
info=[rounded corners, inner sep= 2pt, align=left,scale=0.82]

medarrow=[>=Stealth, line width=, scale=1.8]

(1) at (25,-2) [theory1] [c][1cm]7.5cm$X$ contains a relatively closed subset of $Y$ which is homeomorphic to ${}^\kappa\kappa$.
(2) at (25,-11) [theory1] [c][1.5cm]7.5cmThere is a homeomorphism
from ${}^\kappa 2$ onto a closed subset of ${}^\kappa\kappa$
that reduces
$(\RR_\kappa,\QQ_\kappa)$ to $(X,{}^\kappa \kappa\setminus Y)$.
(4) at (0,-2) [theory1] [c][1cm]7.5cm$X$ is contained in a $K_\kappa$ subset of $Y$.

(5) at (0,-11) [theory1] [c][1.5cm]7.5cm$X$ is contained in a $\Fsigma(\kappa)$ subset $A$ of ${}^\kappa\kappa$ with $A\subseteq Y$.

[medarrow, BlueArrow]
([xshift=85mm]1.south west) edge[<->]
node [KLW and THD for precompact X]xxx
node[midway,right] [c]1.50 cm$X$ contained in a $K_\kappa$ set
([xshift=85mm]2.north west)

([xshift=85mm]4.south west) edge[<->]
node [KLW and THD for precompact X]xxx
node[midway,right] [c]1.50 cm$X$ contained in a $K_\kappa$ set
([xshift=85mm]5.north west);

[medarrow, VeryDarkBlueNode]
([xshift=45mm]2.north west) edge[->>] ([xshift=45mm]1.south west)
([xshift=45mm]4.south west) edge[->>] ([xshift=45mm]5.north west);

The options in $\THD_\kappa(X,Y)$ and $\KLW_\kappa(X,{}^\kappa\kappa\setminus Y)$ for $X\subseteq Y$

It is clear that the implications $\ths$ $\Longrightarrow$ $\klws$
and $\klwl$ $\Longrightarrow$ $\thl$ hold
for all subsets $X\subseteq Y$ of ${}^\kappa\kappa$.
The converse implications hold  for subsets $X$ of $K_\kappa$ sets by the next proposition.
However, they clearly fail
for any closed homeomorphic copy $X$ of ${}^\kappa\kappa$,
and in particular, for $X={}^\kappa 2$ if $\kappa$ is not weakly compact by <cit.>.

If $X\subseteq Y\subseteq{}^\kappa\kappa$ and $X$
is contained in a $K_\kappa$ subset of ${}^\kappa\kappa$,

then the corresponding options in Figure <ref> are equivalent.
In particular,
$\THD_\kappa(X,Y)\Longleftrightarrow\KLW_\kappa(X,{}^\kappa\kappa\setminus Y)$.

Let $\HH:=\dhth Y\restr X$ and $\CC:=\CC^{{}^\kappa\kappa\setminus Y}_i\restr X=(\CC^{{}^\kappa\kappa\setminus Y}\cap \dhIkappa)\restr X$.
[Recall the definitions of $\dhth Y$ and $\CC^Y$ from
Definition <ref> and Subsection <ref>, respectively.]
Write $X$ as the $\kappa$-union of $\kappa$-precompact sets $X_\alpha$.

We show that $\CC\restr X_\alpha\equivf\HH\restr X_\alpha$
[See Definition <ref>.]
for each $\alpha<\kappa$.
This implies $\HH\equivf\CC$ and hence
the corresponding options in $\ODD\kappa\HH$ and $\ODD\kappa\CC$ are equivalent by Corollary <ref>.
The claim then follows from
Theorems <ref>, <ref> and Remark <ref>.
Since $\CC\subseteq \HH$,
it suffices to see that
any hyperedge $\bar x$ of $\HH\restr X_\alpha$ contains a subsequence in $\CC$.
Note that $\bar x$ is injective.
Since $\bar x$ is
contained in a $\kappa$-precompact set,
it must contain a convergent subsequence $\bar y$.
As $\bar x\in \HH$, the limit of $\bar y$ cannot be an element of $Y\cup\ran(\bar y)$
and hence $\bar y\in \CC$, as required.

Suppose $\kappa$ is weakly compact or $\kappa=\omega$ and $\mathcal D$ is closed under complements.
\beginenumerate-(1)

Suppose that $\mathcal C$ is a collection of eventually bounded subsets of ${}^\kappa\kappa$. Then

\KLWdis_\kappa(\mathcal C,\mathcal D)
\Longleftrightarrow
\THD_\kappa(\mathcal C,\mathcal D).$$

Suppose that $\mathcal C$ and $\mathcal D$ are closed under homeomorphic images.

$$\KLWdis_\kappa(\mathcal C,\mathcal D)
\Longrightarrow
\THD_\kappa(\mathcal C,\mathcal D).$$

\endenumerate-(1)

<ref> follows immediately from the previous proposition.
For <ref>,
let $h$ be a homeomorphism from ${}^\kappa\kappa$ onto a subset of ${}^\kappa 2$.
[E.g., let $h:=[\pi]$ where $\pi$ is the map defined in Subsection <ref>.]
Take subsets $X\in\mathcal C$ and $Y\in\mathcal D$ of ${}^\kappa\kappa$.
Then $\KLW_\kappa(h(X),{{}^\kappa\kappa\setminus h(Y)})$ is equivalent to $\THD_\kappa(h(X),h(Y))$ by the previous proposition.
The latter is equivalent to  $\THD_\kappa(X,Y)$.

It is open whether the previous corollary holds without the assumption of weak compactness.
The next corollary
follows from <ref> and
Proposition <ref> <ref>.

Suppose that $\mathcal D$ is closed under complements, and $\mathcal C$  is  a collection of  $\kappa$-precompact sets closed under intersections with elements of $\mathcal D$.

$$\KLW_\kappa(\mathcal C,\mathcal D)\Longleftrightarrow\THD_\kappa(\mathcal C,\mathcal D)\land \PSP_\kappa(\mathcal C).$$

In particular,
$\KLW_\kappa(\mathcal C,\mathcal C)\Longleftrightarrow\THD_\kappa(\mathcal C,\mathcal C)$
if $\kappa$ is weakly compact or $\omega$, $\mathcal C$ is the class of definable or $\kappa$-Borel subsets of ${}^\kappa 2$ and $\PSP_\kappa(\mathcal C)$ holds.

The connections between the $\kappa$-perfect set property and the versions of the Hurewicz dichotomy follow from results in <cit.> and <cit.>
which are summarized in the next remark.

\beginenumerate-(1)

It is consistent that $\THD_\kappa(\analytic(\kappa))$
pagebreak inserted
holds and the $\kappa$-perfect set property for closed sets
fails simultaneously at all regular uncountable cardinals $\kappa$ <cit.>*Corollary 5.4.
Note that these statements are consistent with the existence of weakly compact cardinals, since the iteration $\PP_{<\kappa}$ used in the proof of <cit.>*Corollary 5.4 preserves the $(\kappa+2)$-strong unfoldability of any cardinal $\kappa$ by <cit.>*Lemmas 4.3 and 4.7 and thus these cardinals, which may exist in the ground model $L$, remain weakly compact in the extension.

A similar separation between $\HD_\kappa(\analytic(\kappa))$ and the $\kappa$-perfect set property for closed sets is possible for all inaccessible cardinals $\kappa$ by
the arguments in <cit.>*Section 3.2.
In particular, this follows from
<cit.>*Proposition 3.19 and Lemma 3.23 (4) $\Rightarrow$ (1)
and the observation that for inaccessible cardinals $\kappa$,
the first option in $\HD_\kappa(X)$ holds if and only if $X$ is eventually bounded.

It is consistent that
$\THD_\kappa(\analytic(\kappa),\pwrset({}^\kappa\kappa))$ holds but
the $\kappa$-perfect set property for closed sets fails
at all uncountable non-weakly compact cardinals,
since the model in <cit.>*Corollary 5.4 satisfies the
strong Hurewicz dichotomy for $\kappa$-analytic sets
[See Remark <ref> for the definition of the strong Hurewicz dichotomy.]
at all such cardinals.
To see this, note that the argument in the proof of <cit.>*Theorem 1.7 works for the construction in <cit.>*Theorem 1.9 too.

$\PSP_\kappa(X)$ implies the strong Hurewicz dichotomy
for $X$ <cit.>*Proposition 2.7
and thereby

at all uncountable non-weakly compact cardinals.

However, this implication fails in the countable setting, since
$\PSP_\omega(\mathcal C)$ does not imply
$\THD_\omega(\mathcal C,\Gdelta)$
for the class $\mathcal C:=L(\RR)[U]$, where $L(\RR)$ is a Solovay model and $U$ is a selective ultrafilter on $\omega$ that is generic over $L(\RR)$,
by the results of  Di Prisco and Todorčević <cit.> discussed in
Remark <ref>
and by Proposition <ref>.
\endenumerate-(1)

All other implications between the $\kappa$-perfect set property and versions of the Hurewicz dichotomy are open.
For instance, implications from $\PSP_\kappa(\defsets\kappa)$ to $\THD_\kappa(\defsets\kappa)$, $\HD_\kappa(\defsets\kappa)$, $\THD_\kappa(\defsets\kappa,\pwrset({}^\kappa\kappa))$ and the converse are open for weakly compact cardinals and $\omega$.
Regarding <ref> and <ref>

Similarly, it is open whether $\PSP_\omega(\defsets\omega)$ implies $\THD_\omega(\defsets\omega)$ or conversely. For weakly compact cardinals, it is open whether $\PSP_\kappa(X)$ implies $\THD_\kappa(X)$, $\HD_\kappa(X)$ or $\THD_\kappa(X,\pwrset({}^\kappa\kappa))$.

Concerning larger classes, it is open whether $\THD_\kappa(\defsetsk)$ or $\HD_\kappa(\defsetsk)$ imply $\PSP_\kappa(\defsetsk)$ for all regular uncountable cardinals $\kappa$.

In <ref>, it is open whether one can construct a model where $\HD_\kappa(\analytic(\kappa))$ holds, but the $\kappa$-perfect set property for closed sets for successor cardinals $\kappa$.
It is also open whether  <ref> works for successor cardinals.
Does  $\THD_\kappa(\defsets\kappa,\pwrset({}^\kappa\kappa))$ imply $\HD_\kappa(\defsets\kappa)$ for non-weakly compact cardinals?

We now turn to the implications between the Jayne-Rogers theorem and other applications.
The next proposition shows that
for functions $f$ whose range
is $\kappa$-precompact,
$\JR_\kappa^f$ follows from
Kechris-Louveau-Woodin dichotomy.
Moreover, it follows
from the
two-parameter topological Hurewicz dichotomy
if the domain of $f$ is also $\kappa$-precompact.
Recall that $\JR_\kappa^f$ follows from $\ODD\kappa\JJ$ for the $\kappa$-dihypergraph $\JJ$ in the next proposition
by Theorem <ref> and
Remark <ref>.

Suppose $X\subseteq{}^\kappa\kappa$ and $f:X\to{}^\kappa\kappa$.
Let $\JJ:=\JJ^f_i$,
and let
$A:=k(\graph(f))$ and
$B:=k\big((X\times {}^\kappa\kappa)\setminus\! \graph(f)\big)=
k\big(X\times{}^\kappa\kappa\big)\setminus A$,
$k$  is a homeomorphism from ${}^\kappa\kappa\times{}^\kappa\kappa$ onto ${}^\kappa\kappa$.

\beginenumerate-(1)

$\ODD\kappa\JJ\Longleftrightarrow\KLW_\kappa(A,B)$ if $\ran(f)$
is contained in a $K_\kappa$ subset of ${}^\kappa\kappa$.

(1) also follows from (2) in the case that $X$ is also contained in a $K_\kappa$ set by Proposition <ref>

$\ODD\kappa\JJ\Longleftrightarrow\THD_\kappa(A,{}^\kappa\kappa\setminus  B)$ if both $X$ and $\ran(f)$ are
contained in a $K_\kappa$ subsets of ${}^\kappa\kappa$.

\endenumerate-(1)

Let $h:=k^{-1}$, and
let $\CC:=
%{h}^\ddim(\CC_i^{{}^\kappa\kappa\setminus B}\restr{A})=
$\HH:=(h)^{\ddim}(\dhth{{}^\kappa\kappa\setminus B}\restr A)=(h)^{\ddim}(\dhth{{}^\kappa\kappa\setminus B})\restr \graph(f)$.

The next claim show that
$\JJ\equivf\CC$ if $\ran(f)$ is $\kappa$-precompact
and $\JJ\equivf\HH$ if both $X$ and $\ran(f)$ are $\kappa$-precompact.

\beginenumerate-(i)

$\JJ\leqf\CC$ if $\ran(f)$ is $\kappa$-precompact.

$\HH\leqf\CC$ if $X$ and $\ran(f)$ are both $\kappa$-precompact.
\endenumerate-(i)

For <ref>,
note that $\CC$ consists of those injective convergent sequences $\bar a$ in $\graph(f)$ with
$(x^{\bar a},y^{\bar a}):=\lim_{\alpha<\kappa}(\bar a)$ in $(X\times{}^\kappa\kappa)\setminus\! \graph(f)$.
It suffices to show that any such sequence $\bar a=\langle (x_\alpha,f(x_\alpha)):\alpha<\kappa\rangle$ has a subsequence in $\JJ$.
Note that
$x^{\bar a}$ is in $X$ and
$f(x^{\bar a})\neq y^{\bar a}=\lim_{\alpha<\kappa}f(x_\alpha)$.
By passing to a subsequence of $\bar a$, we may assume that $x\neq x_\alpha$ and $f(x)\neq f(x_\alpha)$ for all $\alpha<\kappa$.
Then $\langle x_\alpha:\alpha<\kappa\rangle$ is in $\CC^X_i$
If $\langle x_\alpha:\alpha<\kappa\rangle$ weren't injective, then $\bar a$ would not be either since $f$ is well defined.
and $f(x)\notin\closure{\{f(x_\alpha):\alpha<\kappa\}}$.
This sequence is in $\JJ$, as required.

We prove <ref>
and <ref> simultaneously.
Note that $\HH$ consists of those injective $\kappa$-sequences
in $\graph(f)$ whose limit points are all contained in $(X\times{}^\kappa\kappa)\setminus\!\graph(f)$.
Suppose $\bar a=\langle (x_\alpha,f(x_\alpha):\alpha<\kappa\rangle$ is a hyperedge of $\JJ$.
Since $\bar x:=\langle x_\alpha:\alpha<\kappa\rangle$ is in $\CC_i^X$, it must converge to some $x\in X$.
Hence all limit points of $\bar a$ are of the form $(x,y)$ for some $y\in{}^\kappa\kappa$.
Moreover, $(x,f(x))$ is not a limit point of $\bar a$
since $f(x)$ is not in $\closure{\{f(x_\alpha):\alpha<\kappa\}}$.
Since $\bar a$ is injective, it is a hyperedge of $\HH$
Now, assume that $\ran(f)$ is $\kappa$-precompact. Then $\langle f(x_\alpha):\alpha<\kappa\rangle$ and therefore $\bar a$ has a convergent subsequence,
This subsequence is in $\CC$,
since its limit is in $(X\times{}^\kappa\kappa)\setminus\!\graph(f)$ and it is injective.

For <ref>, note that if $X$ and $\ran(f)$ are both $\kappa$-precompact,
then so is $\graph(f)$.
As in the proof of Proposition <ref>, it suffices to see that
any hyperedge $\bar a$
of $\HH$ contains a subsequence in $\CC$.
Since $\bar a$ is
contained in the $\kappa$-precompact set $\graph(f)$,
it must contain a convergent subsequence $\bar b$.
As $\bar a\in \HH$, the limit of $\bar b$
is an element of $(X\times{}^\kappa\kappa)\setminus\!\graph(f)$.
Hence $\bar b\in \CC$, as required.

For <ref>,
suppose that $\ran(f)$ is contained in a $K_\kappa$ set $\bigcup_{\alpha<\kappa} C_\alpha$, where each $C_\alpha$ is $\kappa$-compact.

Let $G_\alpha:=\graph(f)\cap(X\times C_\alpha)$ for each $\alpha<\kappa$.
$\graph(f)=\bigcup_{\alpha<\kappa}G_\alpha$, and
$\JJ\restr G_\alpha\equivf\CC\restr G_\alpha$ for each $\alpha<\kappa$ by the previous claim.
Hence $\JJ\equivf\CC$,
and thus
$\ODD\kappa\JJ$ and $\ODD\kappa\CC$ are equivalent
by Corollary <ref>.
The latter is further equivalent to $\KLW_\kappa(A,B)$ by
Theorem <ref> and Remark <ref>.

For <ref>,
note that if both $X$ and $\ran(f)$ are contained in $K_\kappa$ sets, then so is $\graph(f)$.
Write $\graph(f)$ as the $\kappa$-union of $\kappa$-precompact sets $G_\alpha$.
Then $\JJ\restr G_\alpha\equivf\HH\restr G_\alpha$ for each $\alpha<\kappa$ by the previous claim.
Hence $\JJ\equivf\HH$, and therefore
$\ODD\kappa\JJ$ is equivalent to $\ODD\kappa\HH$
by Corollary <ref>,
which is further equivalent to $\THD_\kappa(A,{}^\kappa\kappa\setminus  B)$
by Theorem <ref>.

Suppose that $\kappa$ is weakly compact or $\kappa=\omega$.
If the classes $\mathcal C\subseteq \mathcal D$ contain all $\Gdelta(\kappa)$ sets,
$\mathcal C$ is closed under continuous preimages,
$\mathcal D$ under complements and
both are closed under finite intersections,
$\THD_\kappa(\mathcal C,\mathcal D)\Longrightarrow\JR_\kappa(\mathcal C)$.

Suppose that $\THD_\kappa(\mathcal C,\mathcal D)$ holds.
It suffices to show $\JR_\kappa^g$ for all
subsets $Y\in\mathcal C$ of ${}^\kappa\kappa$
and all $\DeltaZeroTwo(\kappa)$-measurable functions $g:Y\to{}^\kappa\kappa$.
Let $h$ be a homeomorphism from ${}^\kappa\kappa$ onto a subset of ${}^\kappa 2$

$f:=h\comp g\comp h^{-1}$.
Since $X:=\dom(f)=h(Y)$ and $\ran(f)=h(\ran(g))$ are both
$\THD_\kappa(A,{}^\kappa\kappa\setminus B)$
is equivalent to $\ODD\kappa\JJ$,
$\JJ$, $k$, $A:=k(\graph(f))$ and
$B:=k(X\times{}^\kappa\kappa)\setminus A$
are defined as in the previous proposition.
We will show that $A\in\mathcal C$ and ${}^\kappa\kappa\setminus B\in\mathcal D$.
This suffices, since
$\JR_\kappa^f$ and thus $\JR_\kappa^g$ follow from
$\THD_\kappa(A,{}^\kappa\kappa\setminus B)$
by Corollary <ref> and Remark <ref>

Since $f$ is $\DeltaZeroTwo(\kappa)$-measurable, $\graph(f)$ is a $\Gdelta(\kappa)$ subset of ${}^\kappa\kappa \times{}^\kappa\kappa$.
To see this, note that

(X \times ({}^\kappa\kappa\setminus N_t))\cup
\big],$$

since for all $(x,y)\in X\times{}^\kappa\kappa$ we have $f(x)=y$ $\Longleftrightarrow$ $\forall t\in{}^{<\kappa}\kappa
\big(y\in N_t\Rightarrow f(x)\in N_t\big)$.

for some $\Gdelta(\kappa)$ subset $G$ of ${}^\kappa\kappa \times{}^\kappa\kappa$.

Since $f$ is also $\DeltaZeroTwo(\kappa)$-measurable,
fix a $\DeltaZeroTwo(\kappa)$ set $U_t$ with
$f^{-1}(N_t)=U_t\cap X$
for all $t\in{}^{<\kappa}\kappa$.

({}^\kappa\kappa \times N^c_t)\cup
\big),$$

where $N^c_t:={}^\kappa\kappa\setminus N_t$.
Then $G$ is a $\Gdelta(\kappa)$ set since
${}^\kappa\kappa \times N^c_t$ is clopen and
$U_t\times{}^\kappa\kappa$ is $\DeltaZeroTwo(\kappa)$ for each
We claim that $\graph(f)=G\cap(X\times{}^\kappa\kappa)$.
To see this, let $(x,y)\in X\times{}^\kappa\kappa$.

\begin{multline*}
\Longleftrightarrow
\Longleftrightarrow
\forall t\in{}^{<\kappa}\kappa
\big(y\in N_t\Rightarrow f(x)\in N_t\big)
\Longleftrightarrow
\\
\forall t\in{}^{<\kappa}\kappa
\big(y\notin N_t \lor f(x)\in N_t\big)
\Longleftrightarrow
(x,y)\in G.
\end{multline*}

The last equivalence holds since $x\in X$ and hence
$f(x)\in N_t\Leftrightarrow x\in U_t$.

Note that
$X\in \mathcal C$.
Since $X\times{}^\kappa\kappa$ is the preimage of $X$ under a projection and $k$ is a homeomorphism, we have $k(X\times{}^\kappa\kappa)\in \mathcal C$.
Since $A:=k(\graph(f))$ is a $\Gdelta(\kappa)$ subset of $k(X\times {}^\kappa\kappa)$, we have $A\in \mathcal C$.
Therefore ${}^\kappa\kappa\setminus B= A\cup k(({}^\kappa\kappa\setminus X)\times{}^\kappa\kappa)\in \mathcal D$.

§ OUTLOOK

As above, $\kappa$ always denotes an infinite cardinal with $\kappa^{<\kappa}=\kappa$.
With regards to the main results, we ask for a global version of Theorem <ref> <ref> for all regular cardinals and arbitrary box-open dihypergraphs.
The difficulty with proving such a result is that in an iterated forcing extension as in the proof of Theorem <ref>, new box-open dihypergraphs are added by the tail forcings and the box-open dihypergraph dichotomy does not automatically hold for them.

Assume there is a proper class of Mahlo cardinals.
Is there a tame class forcing extension of $V$ where $\ODD\kappa\kappa(\defsetsk)$ holds for all

The consistency strength of $\ODD\kappa\kappa(\defsets\kappa)$ is open.
In particular, we do not know if it implies that there is an inner model with a Mahlo cardinal.
We further ask about the consistency strength of each of its variants and applications.

Do the following need a Mahlo cardinal for some uncountable $\kappa$?
\beginenumerate-(1)

\beginenumerate-(a)

\endenumerate-(a)

\beginenumerate-(a)

*  $\KLW^\ddim_\kappa(\defsetsk)$, where $2\leq\ddim<\kappa$.
[See Subsection <ref>.

Note that $\KLW^\kappa_\kappa(\defsetsk)$
is equivalent to $\ODD\kappa\kappa(\defsetsk)$ by Corollary <ref>.]
I.e., $\KLW^\ddim_\kappa(X,Y)$ for all $X\in\defsetsk$ and ALL $Y\subseteq{}^\kappa\kappa$. If $\ddim<\kappa$, this is stronger than $\ODD\kappa{\ddim}(\defsetsk)$,
which is equivalent to: $\KLW_\kappa^\ddim(X,Y)$ for all $\Gdelta(\kappa)$ sets $Y$ and all $X\in\defsetsk$. The latter is equiconsistent with an inaccessible.

*  $\KLW_\kappa(\defsetsk)$.

*  $\KLW_\kappa({}^\kappa\kappa)$.
\endenumerate-(a)

*  $\KLWdis_\kappa(\defsetsk)$.
[Recall that $\KLWdis_\kappa(\defsetsk)$ denotes the statement that $\KLW_\kappa(X,Y)$ holds for all disjoint subsets $X\in\defsetsk$ and $Y$ of ${}^\kappa\kappa$, while $\KLW_\kappa(\defsetsk)$ is the version where $X$ and $Y$ are not necessarily disjoint.]

\beginenumerate-(a)

$\cbo$.[$\cbo$ and $\DK_\kappa(X)$ were defined in the paragraphs preceding Corollary <ref>.]

The determinacy of $\GV_\kappa(X)$ for all subsets $X$ of ${}^\kappa\kappa$.
\endenumerate-(a)

\endenumerate-(1)

In <ref><ref>, one can equivalently consider
Note that

<ref><ref> $+$ $\Diamondi\kappa$
[These implications follow from the results in Subsections <ref> and <ref>.]

The statements
are each equiconsistent with the existence of an inaccessible cardinal, since they each imply $\PSP_\kappa(X)$ for all closed sets $X$.
For non-weakly compact $\kappa$,
$\THD_{\kappa}(\defsets{\kappa},\powerset({}^{\kappa}\kappa))$ is consistent relative to an inaccessible cardinal,
since it follows from $\PSP_{\kappa}(\defsets{\kappa})$
<cit.>*Proposition 2.7,
and hence so is $\THD_{\kappa}(\defsets{\kappa},\defsets{\kappa})$.
For weakly compact $\kappa$, $\THD_{\kappa}(\defsets{\kappa},\powerset({}^{\kappa}\kappa))$ follows from $\KLWdis_\kappa(\defsets{\kappa})$ by Corollary <ref>.

Do the following need an inaccessible cardinal for some uncountable $\kappa$?
\beginenumerate-(1)

* \beginenumerate-(a)

* $\HD_\kappa(\defsets\kappa)$.

* $\THD_\kappa(\defsets\kappa)$.
[Recall that this is the same statement as

* $\THD_\kappa(\defsets\kappa,\defsets\kappa)$.
\endenumerate-(a)

* can we use the failure in $L$ of “every proper $\Fsigma(\kappa)$ set is $\Fsigma(\kappa)$-complete to show that $\omega_1$ is inaccessible in $L$?

* $\ABP_\kappa(\defsets\kappa)$.

* $\JR_\kappa(\defsetsk)$. this is equivalent to: “$\JR_\kappa^f$ for all $f\in\defsetsk$”. Also, this is the version which we've stated in Corollary <ref>

* $\OGD_\kappa({}^\kappa\kappa)$.
\endenumerate-(1)

Note that it is consistent relative to $\ZFC$ that
$\THD_\kappa(\analytic(\kappa))$ holds
for all regular uncountable cardinals $\kappa$ <cit.>*Theorem 1.13,
$\THD_\kappa(\analytic(\kappa),\pwrset({}^\kappa\kappa))$ holds for all uncountable non-weakly compact cardinals $\kappa$,
[To see this, note that the model constructed in <cit.>*Theorem 1.13
satisfies the strong Hurewicz dichotomy for $\kappa$-analytic sets at all non-weakly compact cardinals by the proof of  <cit.>*Theorem 1.7.]
and $\HD_\kappa(\analytic(\kappa))$ holds for all inaccessible cardinals $\kappa$
<cit.>*Section 3.2.
[See Remark <ref> <ref> for a detailed explanation.]
It is open whether the latter holds for uncountable successor cardinals $\kappa$.
In the countable setting,
$\ABP_\omega(\defsets\omega)$ is equivalent to the Baire property for all sets of reals in $\defsets\omega$.
Hence it is consistent relative to $\ZFC$ by a well-known result of Shelah <cit.>.
$\OGD_\kappa({}^\kappa\kappa)$ is not provable in $\ZFC$ for $\kappa=\omega_2$.
To see this, note that in $L$,
there exists a $\kappa$-Kurepa tree $T$ that is a continuous image of ${}^\kappa\kappa$
for any successor cardinal $\kappa=\mu^+$, where $\mu$ has uncountable cofinality
Then $\PSP_\kappa([T])$ fails in $L$, so $\OGD_\kappa({}^\kappa\kappa)$ fails as well.[Note that $\OGD_\kappa(X)$ is preserved under taking continuous images by Lemma <ref> and Corollary <ref>.]
In the countable setting, it is open whether an inaccessible cardinal is required to obtain $\THD_\omega(\defsets\omega)$,
and while  <cit.> claimed a solution, Jörg Brendle pointed out to us that this argument does not work.

We expect the open dihypergraph dichotomy to have further applications.
In current work, we have shown that $\ODD\omega\omega(X)$ is equivalent to the closed-sets covering property $\mathsf{CCP}(X)$
introduced by Louveau <cit.> and further studied by Petruska <cit.>, Solecki <cit.>,
Di Prisco and Todorčević <cit.>.
We have also obtained a variant of this result in the uncountable case.
It is likely that some further applications of $\mathsf{CCP}(X)$
in <cit.> lift to the uncountable setting.
Moreover, we expect that
the level-two results of Lecomte-Zeleny <cit.>,
and more generally,
the higher dimensional version of the Kechris-Louveau-Woodin dichotomy
in <cit.>*Section 4, can be extended to the uncountable setting.
Carroy and Miller showed in unpublished work that their basis results for the classes of non-Baire class $1$ functions and functions that are not $\sigma$-continuous with closed pieces <cit.> follow from the Kechris-Louveau-Woodin dichotomy.
Does this hold in the uncountable setting?

Since the Baire property for subsets of ${}^\omega\omega$ follows from instances of $\ODD\omega\omega(X)$, we ask whether a similar result holds for Lebesgue measurability.
Note that there is no analogous proof, since the ideal of Lebesgue null sets is not generated by closed sets and it thus cannot be generated by the independent sets with respect to a box-open dihypergraph on the Baire space.

implications and separations
between applications of the open dihypergraph dichotomy are open,
for instance all missing implications in Figures <ref>
and <ref>.
In particular, these diagrams suggest
to ask about implications between the $\kappa$-perfect set property $\PSP_\kappa(\mathcal C)$ and
the Kechris-Louveau-Woodin dichotomy $\KLWdis_\kappa(\mathcal C)$ for
uncountable $\kappa$ and
sufficiently closed classes $\mathcal C$.
Note that the former does not imply the latter in the countable case
the results of Di Prisco and Todorcevic <cit.> described in
Remark <ref>.
[In fact, these results shows that the open graph dichotomy for $\mathcal C$ does not imply $\KLW_\omega(\mathcal C,\Fsigma)$.]
Their results further show that
neither the perfect set property
nor the open graph dichotomy
implies the Baire property,

since the subset of the Cantor space induced by
a nonprincipal ultrafilter on $\omega$ does not have the Baire property.
Moreover, the Baire property does not imply the perfect set property for $\defsets\omega$ in the countable setting <cit.>.

Can this be done in the uncountable setting as well?

Suppose that $\kappa$ is uncountable and $\mathcal C$ is closed under continuous preimages.
Do any of the statements
$\PSP_\kappa(\mathcal{C})$, $\ABP_\kappa(\mathcal{C})$ or
$\KLWdis_\kappa(\mathcal{C})$ imply any of the other ones?

$\KLW_\kappa(\mathcal C,\mathcal C)$ is equivalent to $\KLWdis_\kappa(\mathcal C,\mathcal C)\land\PSP_\kappa(\mathcal C)$
if $\mathcal C$ is closed under intersections and complements,
[See Proposition <ref> <ref>.]
this is related to the problem whether
$\PSP_\kappa(\mathcal C)$ implies $\KLW_\kappa(\mathcal C,\mathcal C)$ or even $\KLW_\kappa(\mathcal C)$.
Since $\KLW_\kappa({}^\kappa\kappa,\mathcal C)$ implies $\cbi(\mathcal C)$,
one can similarly ask whether $\PSP_\kappa(\mathcal C)$ implies $\cbi(\mathcal C)$.
${}$[See Theorem <ref>. If $\mathcal C$ is closed under intersections with basic open sets, then
$\cbi(\mathcal C)$ is equivalent to $\cbii(\mathcal C)$
by Theorem <ref>.]
Footnote size problem solved in the usual way
It is equivalent to ask whether
$\GV_\kappa(X)$ is determined for every
$X\in\mathcal C$ of size $\kappa$.

The reverse implication

is equivalent to the statement that every $\kappa$-crowded set
in $\mathcal C$
of size at least $\kappa^+$
I wrote “at least $\kappa^+$” insted of just “$\kappa^+$” since I'm not sure every $\kappa$-croweded set of size at least $\kappa^+$ has a $\kappa$-crowded set of size exactly $\kappa^+$.
has a $\kappa$-perfect subset.
Note that these statements are equivalent for the class $\Gdelta(\kappa)$ and moreover, they are equivalent to $\KLW_\kappa({}^\kappa\kappa,\Gdelta(\kappa))$.
[See Corollary <ref> <ref> and
Remark <ref>.]

We proved that $\ODD\kappa\kappa(\mathcal C)$ and $\KLW^\kappa_\kappa(\mathcal C)$ are equivalent.
[See Corollary <ref>.]
Are some other applications of $\ODD\kappa\kappa(\defsetsk)$ equivalent to it?
For lower dimensions, we ask whether
$\KLW^\ddim_\kappa(\mathcal C)$ implies
$\ODD\kappa\ddim(\mathcal C)$.
Note that
$\ODD\kappa\ddim(\mathcal C)$ is equivalent to $\KLW^\ddim_\kappa(\mathcal C,\Gdelta(\kappa))$
if $\mathcal C$ is closed under intersections with $\Gdelta(\kappa)$ sets.
[See Corollary <ref> <ref>.]

Note that one can prove the consistency of $\ODD\kappa\ddim(\defsetsk)$ for $\ddim<\kappa$ more easily than in this paper along the lines of the proof of the consistency of $\PSP_\kappa(\defsets\kappa)$ in <cit.>*Section 2.2.
This is how we proved it originally.
The fact that the proofs are very similar suggests that
$\ODD\kappa\ddim(\defsetsk)$ and $\PSP_\kappa(\defsetsk)$ may be equivalent if $\ddim<\kappa$.

Does $\PSP_\kappa(\mathcal C)$ imply $\OGD_\kappa(\mathcal C)$ if $\mathcal C$ is closed under continuous preimages?
[Recall that $\OGD_\kappa(X)$ and $\ODD\kappa 2(X)$ are equivalent for all subsets $X$ of ${}^\kappa\kappa$ by Corollary <ref>.]

We further ask which of the open dihypergraph dichotomies for different dimensions $2\leq\ddim\leq\kappa$ can be separated, and in particular:

Do the following implications hold if $3\leq\ddim<\kappa$ and $\mathcal C$ is closed under continuous preimages?
\beginenumerate-(1)

$\OGD_\kappa(\mathcal C)\Longrightarrow\ODD\kappa\ddim(\mathcal C)$.

$\ODD\kappa\ddim(\mathcal C)\Longrightarrow\ODD\kappa\kappa(\mathcal C)$.
\endenumerate-(1)

The version of the implication in <ref> for definable $\kappa$-dihypergraphs is also open.
Note that
$\OGD_\omega(\mathcal{C})$ does not imply $\ODD\omega\omega(\mathcal{C},\defsets\omega)$ by
the results of Di Prisco and Todorcevic <cit.> described in
Remark <ref>.
We ask if this implication fails for regular uncountable cardinals $\kappa$, too.

The restriction $\ODD\kappa\kappa({}^\kappa\kappa)$ to the underlying set ${}^\kappa\kappa$ has interesting consequences such as
$\ODD\kappa\kappa({\bf \Sigma}^1_1(\kappa))$.
[See Theorem <ref>.]
This motivates the next problem and its restriction to definable dihypergraphs.
Similar questions are of interest for  $\KLW^\ddim_\kappa({}^\kappa\kappa)$ and other applications.

Does $\ODD\kappa\kappa({}^\kappa\kappa)$ imply $\ODD\kappa\kappa(\defsetsk)$?

Summarizing and expanding some of the above questions,
we ask which of the implications in
Figure <ref>
can be reversed.
[The second and third implication in each row follows from Lemmas <ref> and <ref>, and Corollary <ref>, respectively.
The equivalences in the figure follow from Lemma <ref>.
Blue arrows are clickable and lead to the corresponding results.]
Note that $\OGD_\kappa(\defsets\kappa)$ is equivalent to $\ODD\kappa 2(\defsets\kappa)$ in the figure[See Corollary <ref>.]
and the same holds for other classes.
The first two items in the third row are equivalent to $\ODD\kappa\kappa({}^\kappa\kappa)$ and $\ODD\kappa\kappa({}^\kappa\kappa,\defsetsk)$, respectively.
“intextsep” and “abovecaptionskip” were set locally in the figure

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([yshift=0mm]oddAD.south) edge [<->]
node [ODD for continuous images]xxx

([yshift=0mm]ogdA.south) edge [<->]
node [ODD for continuous images]xxx

([yshift=0mm]oddCD.east) edge [->]

node ODD^kappa implies definable ODD^ddimxxx

([yshift=0mm]ogdC.east) edge [->]
node [cor: PSP from ODD]xxx

Implications between $\mathsf{ODD}$ for dimensions $2\leq\ddim<\kappa$ and $\PSP$

Separating the various dichotomies and applications discussed above is difficult, since the models
where they are known to hold are in most cases obtained by Lévy collapses.
One can try to force over a model of $\ODD\kappa\kappa(\defsets\kappa)$ to preserve some applications but break others as in <cit.> in the countable setting.

Figure <ref> in Subsection <ref> summarizes
implications and separations
between several variants of the open dihypergraph dichotomy with additional requirements on the homomorphisms.
Which of these implications can be reversed?
A striking difference between the countable and uncountable settings is that one can choose the continuous homomorphisms to be injective in the latter case, assuming $\Diamondi\kappa$ holds.[See Propositions <ref> and <ref> and Theorem <ref>.]
We ask whether this assumption can be eliminated.

Does $\ODD{\kappa}H \Longrightarrow \ODDI{\kappa}H$ hold
for all uncountable $\kappa$ and all relatively box-open $\kappa$-dihypergraphs $H$?
$\,$[It is equivalent to ask whether
the existence of a continuous homomorphism from $\dhH\kappa$ to a relatively box-open $\kappa$-dihypergraph $H$ implies the existence of an injective map with the same properties.]
footnote size problem solved in the usual way

This implication holds for $\ddim$-dihypergraphs for $\ddim<\kappa$.
[In fact, the homomorphism can be chosen to be a homeomorphism onto a closed set; see Theorem <ref> and Corollary <ref>.]
Since $\Diamondi\kappa$ holds if $\kappa$ is inaccessible or a successor cardinal ${\geq}\omega_2$ <cit.>,[See Lemma <ref>.]
the remaining cases are $\omega_1$
and weakly inaccessible, but not inaccessible, cardinals.
To separate $\ODD{\omega_1}H$ from $\ODDI{\omega_1}H$ for some relatively box-open $H$, one would need to consider models where $\CH$ holds but $\Diamond_{\omega_1}$ fails.
To our knowledge, the first such model was constructed by Jensen <cit.>.
However, a simpler construction is possible using recent work of Aspero and Mota <cit.>.
Do these models separate the above two principles for some $H$?
If one can show that $\ODDI\kappa\kappa(\defsets\kappa)$ implies $\Diamondi\kappa$ for all $\kappa$, then any model of $\CH + \neg\Diamond_{\omega_1} + \ODD{\omega_1}{\omega_1}(\defsets\kappa)$
is as required.
A related question is whether it is consistent that
$\ODDI\kappa H\Longrightarrow\ODDH\kappa H$ holds for all relatively box-open $\kappa$-dihypergraphs $H$.

The questions studied in this paper make sense for
other spaces
such as the strong $\kappa$-Choquet spaces studied in <cit.>
and <cit.>.
Is it consistent that
the open dihypergraph dichotomy holds for all such spaces and does it have similar applications?
Are there analogues to the results in this paper for

singular cardinals $\kappa$ with $2^{<\kappa}=\kappa$?

Omer Ben-Neria,
Sakae Fuchino, István Juhász, Justin Moore and Teruyuki Yorioka asked about the following problem on higher analogues to the open graph axiom.
Let $\OGA_\kappa$ denote the statement that $\OGA_\kappa(X)$
[See the paragraph before Definition <ref>.]
holds for all subsets $X$ of ${}^\kappa\kappa$.

Is it consistent that $\OGA_\kappa$ holds for some regular uncountable cardinal $\kappa$ with $\kappa^{<\kappa}=\kappa$?

Our results provide a choiceless model of this statement, since
Theorem <ref> shows that the model $L({}^\kappa\Ord)$ in a generic extension obtained by the Lévy-collapse of an inaccessible cardinal to $\kappa^+$ satisfies $\OGA_\kappa$.
This suggests to approach Problem <ref> by forcing over a choiceless model of $\OGA_\kappa$.
It may be useful to know whether $\OGA_\kappa$ implies
$\OGD_\kappa(\analytic(\kappa))$ or the same for $\defsets\kappa$. This would motivate the strategy to force over  a choiceless model of $\OGD$ to obtain $\OGA$.

§ APPENDIX

§.§ Preliminaries

Complete embeddings and projections.

The concept of projections (defined below) will not need to be used in our main proof in Section 4 (or anywhere else). It is enough to use complete embeddings/reductions/the quotient forcing associated to a complete embedding.

The lemma below gives a connection between projections and complete embeddings (and reductions).
Note that we will not need to use this lemma in the proofs in this paper either.

(see <cit.>, <cit.>)
Definition 1.17 from <cit.>
Suppose that $\PP$ and $\QQ$ are forcings.
\beginenumerate-(a)

A projection $\pi: \QQ\to \PP$ is a homomorphism with respect to $\leq$ such that
\beginenumerate-(i)

$\pi[\QQ]$ is dense in $\PP$, and

for all $q\in \QQ$
and all $p\leq \pi(q)$, there is a condition $\bar{q}\leq q$ with $\pi(\bar{q})\leq p$)
(i.e., $\pi[\pred q\QQ]$
is dense below $\pi(q)$ for all $q\in\QQ$).
\endenumerate-(i)

Suppose that $\pi: \QQ\to \PP$ is a projection and $G$ is a $\PP$-generic filter over $V$.
The quotient forcing $\QQ/G$ for $G$ in $\QQ$ relative to $\pi$ is defined as the subforcing

$$\QQ/G=\{q\in \QQ: \pi(q)\in G\}$$

of $\QQ$.
Moreover, we fix a $\PP$-name $(\QQ/\PP)^{\pi}$ for the quotient forcing for $\dot{G}$ in $\QQ$ relative to $\pi$, where $\dot{G}$ is a $\PP$-name for the $\PP$-generic filter, and will refer to this as (a name for) the quotient forcing for $\PP$ in $\QQ$ relative to $\pi$.
\endenumerate-(a)

Suppose that $\PP$ and $\QQ$ have weakest elements $\one_\PP$ and $\one_\QQ$, respectively. It is clear that in
Definition <ref> <ref>,
if we assume the condition <ref>,
then the condition <ref>
is equivalent to the condition that $\pi(\one_\QQ)=\one_\PP$.

Suppose that $\PP,\QQ$ are forcings,
$i:\PP\to\QQ$ is a homomorphism w.r.t. $\leq,\perp,\one$,
and $\pi:\QQ\to \PP$ is a homomorphism w.r.t. $\leq,\one$.
Suppose that $\pi\comp i=\id_{\PP}$.
\beginenumerate-(1)

The following are equivalent:
\beginenumerate-(a)

For all $q\in\QQ$, $\pi(q)$ is a reduction of $q$ to $\PP$.

$\pi$ is a projection, and $i(\pi(q))\compat q$ for all $q\in \QQ$.
Is “$i(\pi(q))\compat q$ for all $q\in \QQ$” needed? Or does it already follow from the assumption that $\pi$ is a projection (and the other assumptions of the lemma)?
\endenumerate-(a)

Suppose <ref> <ref> holds. Then:
\beginenumerate-(i)

For all $q\in\QQ$ and $p\in\PP$, we have $p\compat\pi(q)$ $\Leftrightarrow$ $i(p)\compat q$.equivalently: $p\perp\pi(q)$ $\Leftrightarrow$ $i(p)\perp q$

If $\PP$ is separative, then $\pi(q)$ is the largest reduction of $q$ to $\PP$.

for all $\PP$ generic filters $G$, we have $\QQ/G=(\QQ/G)^{\pi}$.
\endenumerate-(i)
\endenumerate-(1)

Note that <ref><ref> implies that $i$ is a complete embedding.

suppose $\PP,\QQ$ are Boolean algebras and $i:\PP\to\QQ$ is a complete embedding
$\pi_i$ is the retraction associated to $i$.
$\pi_i\comp i=\id_{\PP}$ and <ref><ref> holds (see the proof of Lemma <ref>).
Therefore $\pi_i$ is a projection.
The conclusion of Lemma <ref>
is equivalent to <ref> and <ref> in <ref> for $\pi=\pi_i$,
and therefore also follows from the above lemma.
The above lemma and its proof can be seen as a more general version of Lemma <ref>.

We first show <ref>. First, assume <ref>.
For a given $q\in\QQ$, <ref> and the fact that $\pi(q)\leq\pi(q)$ imply that $i(\pi(q))\compat q$.
To see that $\pi$ is a projection, let $q\in\QQ$, and let $p\leq\pi(q)$.
Then by <ref>, we have $i(p)\compat q$.
Let $\bar q\in\QQ$ be such that $\bar q\leq i(p)$ and $\bar q\leq q$.
By the assumptions that $\pi$ preserves $\leq$ and that $\pi\comp i=\id_\PP$,
$\bar q\leq i(p)$ implies that
$\pi(\bar q)\leq \pi(i(p))=p$.
This shows that $\pi$ is a projection.

Now, assume <ref>. Let $q\in\QQ$. To show that $\pi(q)$ is a reduction of $q$, let $p\leq\pi(q)$.
Since $\pi$ is a projection, there exists $\bar q\leq q$ such that $\pi(\bar q)\leq p$.
Since $i$ preserves $\leq$, the second statement implies that $i(\pi(\bar q))\leq i(p)$.
The facts that $\bar q\leq q$ and $i(\pi(\bar q)\leq i(p)$ and the assumption that
$\bar q\compat i(\pi(q))$ imply that $q\compat i(p)$.
This shows that $\pi(q)$ is a reduction of $q$.

We now show <ref>. Suppose that <ref> holds.
To show <ref>, let $q\in\QQ$ and let $p\in\PP$.
Using the fact that $\pi(q)$ is a reduction of $q$, it is easy to see that
$p\compat\pi(q)$ implies that $i(p)\compat q$.
To see the other direction, suppose $i(p)\compat q$, and let $s\leq q$ be such that $s\leq i(p)$. Since $\perp$ preserves $\leq$,
this implies that $\pi(s)\leq \pi(q)$ and that
Thus, $p\compat \pi(q)$.

To show <ref>, suppose that $G$ is $\PP$-generic and that $q\in\QQ$.
Then by the definition of $\QQ/G$ and by <ref>, we have
if and only if
$p\compat \pi(q)$ for all $p\in G$.
Since $G$ is $\PP$-generic, this last statement holds
$\Leftrightarrow$ $\pi(q)\in G$,
or equivalently,
$\Leftrightarrow$ $q\in(\QQ/G)^{\pi}$.

To show that <ref> holds, suppose $\PP$ is separative, and let $q\in\QQ$.
By the assumption that <ref> holds, $\pi(q)$ is a reduction of $q$.
To see that $\pi(q)$ is the largest reduction of $q$,
suppose $p\in\PP$ and $p\not\leq\pi(q)$.
Then by the separativity of $\PP$, there exists $r\in\PP$ such that $r\leq p$ and $r\perp\pi(q)$. Such an $r$ witnesses that $p$ is not a reduction of $q$.

§.§ Section <ref>

$ $

Remarks about Lemma <ref>.

These do not need to be included in the paper, since they are all easy arguments.

<ref>: $H'\subseteq\dhK\ddim X$ by its definition.

follows from the fact that
$\proj_{\ddim,\eta}\restr({}^\ddim X)$ is a continuous and open map from ${}^\ddim X$ to ${}^\eta X$ (assuming either that  ${}^\ddim X$ and ${}^\eta X$ are equipped with the box-topology or that they are equipped with the product topology).

We have $\proj_{\ddim,\eta}(x)=y$ $\Leftrightarrow$ $x\restr\eta=y$ $\Leftrightarrow$
This assertion is definable by a first order formula with parameter $\eta$.

\[
H=\proj_{\ddim,\eta}[H']=\{y:\exists x (x\in H'\wedge\proj_{\ddim,\eta}(x)=y)\}.\]

Thus, if $H'\in\defsets\kappa$, then $H\in\defsets\kappa$.

\[
\{x:\: x\in{}^\ddim X\,\wedge\,(\exists\alpha<\beta<\ddim)(x(\alpha)\neq x(\beta))\,\wedge\,
\exists y (y\in H\wedge\proj_{\ddim,\eta}(x)=y)\}.\]

Thus, if $H\in\defsets\kappa$ and $X\in\defsets\kappa$, then $H'\in\defsets\kappa.$

Let $Y\subseteq X$.
First, note that if $\bar y\in{}^\eta Y$ and $\bar x\in{}^\ddim Y$ and $\proj_{\ddim,\eta}(\bar x)=\bar y$, then $\bar x\in H'$ $\Leftrightarrow$ $\bar y$ in $H$, by the definition of $H'$.
Also, for any $\bar y\in{}^\eta Y$, there exists $\bar x\in{}^\ddim Y$ such that $\proj_{\ddim,\eta}(\bar x)=\bar y$.
These facts imply firstly that

\[
{}^\eta Y\cap H=\emptyset
\quad\text{$\Leftrightarrow$}\quad
{}^\ddim Y\cap H=\emptyset,
\]

which is equivalent to <ref>.
Secondly, they imply that

\[
\dhK\eta Y\not\subseteq H
\quad\text{$\Leftrightarrow$}\quad
\dhK\ddim Y\not\subseteq H',
\]

which is equivalent to <ref>.

In <ref>, $f$ is a homomorphism from $\dhH\eta$ to $H$ if and only if $f'$ is a homomorphism from $\dhH\ddim$ to $H'$.
To see the direction not shown in the proof above, suppose $f$ is a homomorphism from $\dhH\eta$ to $H$ and let $\bar x\in\dhH\ddim$.
Then $\bar y=\proj_{\ddim,\eta}(\bar x)\in\dhH\eta$, and therefore
$f(\bar y)\in H$.
This fact and the fact that
$\proj_{\ddim,\eta}(f'(\bar x))=f(\bar y)$
implies that
$f'(\bar x)\in H'$.

But the converse of <ref> does not hold: essentially because if $f':{}^\kappa\ddim\to X$ and $f=f'\restr{}^\kappa\eta$, then the continuity of $f$ does not imply the continuity of $f'$.