and Lemma <ref> yield the required properties of $\nna$.

Here is the detailed proof.
First, note that for all sufficiently large $\gamma\in[\nu,\lambda)$, we have that
$\na\in V[G_\gamma]$ is a $\PP_\gamma$-name.
This holds
because $\PP_\lambda$ has the $\lambda$-c.c. and $\na\in V[G_\nu]$ is a $\PP_\lambda$-name for an element of ${}^\kappa\kappa$.

Let $\gamma$ be sufficiently large with $\gamma=\xi+1$ for some ordinal
$\xi<\lambda$ with $\xi^{<\kappa}=\xi$.
Then $\PP_\gamma$
has size $\kappa$ in $V[G_\gamma]$.
Since $\PP_\gamma$ is also non-atomic and $\lle\kappa$-closed,
it is equivalent in $V[G_\gamma]$ to $\Add(\kappa,1)$,
and in fact there exists a dense embedding $\iota:\Addd(\kappa,1)\to\PP_\gamma$ in $V[G_\gamma]$ by Lemma <ref>.
[See Definition <ref>.]
This is an $\Addd(\kappa,1)$-name and thus an $\Add(\kappa,1)$-name.

We first show that in $V[G_\gamma]$,
$\nna$ is a name for a new element of ${}^\kappa\kappa$ and
<ref> holds.
Let $J$ be $\Add(\kappa,1)$-generic over $V[G_\gamma]$,
and let $x:=\nna^J$. It suffices to prove the following claim.

$x\notin V[G_\gamma]$ and
$\one_{\PP_\lambda}\forces^{V[G_\gamma][J]}\check x\in\Xphia.$

Let $K$ be the upwards closure of $\iota[J\cap \Addd(\kappa,1)]$ in $\PP_\gamma$.
Then $K$ is a $\PP_\gamma$-generic filter over $V[G_\gamma]$ such that

Then $x\notin V[G_\nu]$
Lemma <ref> <ref>.
Since $K$ and $G_{[\nu,\gamma)}$ are mutually generic over $V[G_\nu]$,
we have $V[K]\cap V[G_{\gamma}]=V[G_\nu]$.
Therefore $x\notin V[G_\gamma]$.

We also have
$\one_{\PP_\lambda} \forces^{V[G_\nu]}{\na}\in\Xphia$
by Lemma <ref> <ref>.
Since $\PP_\lambda\simeq \PP_\gamma\times\PP^\gamma$,

$$\one_{\PP^\gamma} \forces^{V[G_\nu][K]} \check{x}\in\Xphia.$$

$\PP^\gamma\simeq\PP_{[\nu,\gamma)}\times \PP_\lambda$
[See Corollary <ref>.]
and $V[G_\gamma][K]$ is a $\PP_{[\nu,\gamma)}$-generic extension of $V[G_\nu][ K]$,

$$\one_{\PP_\lambda} \forces^{V[G_\gamma][K]} \check{x}\in\Xphia.$$

The claim holds since

To see <ref>, let $q\in\Add(\kappa,1)$ and
Since $\nna\in V[G_\gamma]$, we have
$T\in\mathcal T_\gamma$
by Lemma <ref> <ref>.

$T$ is not $H$-independent.[I.e., $T\notin\indtrees\gamma$.]

Choose some $q^*\in \Addd(\kappa,1)$ with $q^*\leq q$,
and let
By Lemma <ref>,
$T^*\in V[G_\nu]\subseteq V[G_\gamma]$ and
$T^*$ is not $H$-independent.

It thus suffices to show that $T^*\subseteq T$.
To prove this, work in $V[G_\gamma]$ and
let $v\in T^*$.
Choose $p\in\PP_\lambda$ with $p\leq \iota(q^*)$ and
$p\forces_{\PP_\lambda} v\subseteq\na$.
Here and in (ii) below, it's better to keep track of the forcing in $\forces$
Let $p\restr\gamma:=p\restr((\gamma\times\kappa)\cap\dom(p))$.

$p\restr\gamma$ is the retraction of $p$ to $\PP_\gamma$,
i.e., $p\restr\gamma=\pi_{\id}(p)$ for the identity map $\id:\PP_\gamma\to\PP_\lambda$ and
the map
$\pi_{\id}$ from Definition <ref>.
Then $p\restr\gamma\in\PP_\gamma$ and
since $\iota(q^*)\in\PP_\gamma$.
As the formula “$x\subseteq y$” is absolute,
$p\restr\gamma\forces_{\PP_\gamma} v\subseteq\na$.
By the density of $\ran(\iota)$ in $\PP_\gamma$,
there exists $r\in\Addd(\kappa,1)$
such that
$\iota(r)\leq p\restr\gamma$. Then
\beginenumerate-(i)

$\iota(r)\leq\iota(q^*)$ and

$\iota(r)\forces_{\PP_\gamma} v\subseteq\na$.
\endenumerate-(i)
Since $\iota$ is a dense embedding,
<ref> implies $r\leq q^*$ as $\Addd(\kappa,1)$ is separative.
Because $\nna=\na^{\iota}$,
<ref> implies $r\forces v\subseteq \nna$.
Thus, $r$ witnesses that
$v\in T$

as required.

This completes the proof of Lemma <ref>.

We now define the notion of an $H$-witness function for $\nna$ and $\QQ$.
The existence of such a function
is an explicit
and equivalent way of stating that
the trees $T_\QQ^{\nna,q}$ are not $H$-independent for all $q\in\QQ$.

Suppose $\QQ$ is a forcing and $\nna$ is a $\QQ$-name for an
element of ${}^\kappa\kappa$.
An $H$-witness function for $\nna$ and $\QQ$
is a function
such that for all $p\in\QQ$ and $\alpha<\ddim$
\beginenumerate-(1)

* $w(p,\alpha)\in T_\QQ^{\nna,p}$ and

* $\prod_{\beta<\ddim}N_{w(p,\beta)}\subseteq H$.
\endenumerate-(1)

Such functions exist for the names
$\nna$ from
Lemma <ref> and $\QQ=\Add(\kappa,1)$.
[Note that $H$-witness functions need not exist.
For instance, if $\nna=\check x$ for some $x$ in the ground model, then
for any $q\in\QQ$.
Therefore $T_\QQ^{\nna,q}$ is
$H$-independent and
there is no $H$-witness function for $\check x$.]

A step function for a given forcing $\QQ$ is a function $s:\QQ\times\ddim\to \QQ$
such that
$s(q,\alpha)\leq q$
for all $q\in\QQ$ and $\alpha<\ddim$.

We now show that if there is an $H$-witness function, then there is an $H$-witness function with some nice properties and a corresponding step function.

Suppose $\QQ$ is a forcing and $\nna$ is a $\QQ$-name for an element of ${}^\kappa\kappa$.
\beginenumerate-(1)

If there exists an $H$-witness function for $\nna$ and $\QQ$, then there exists an $H$-witness
function $w$ such that

\[\nna_{[p]}\subsetneq w(p,\alpha)\,
\footnote{Recall the notation $\nna_{[p]}$ from Definition~\ref{sigma[q] def}. Note that this requirement is equivalent to: if $p\forces v\subseteq\nna$, then $v\subsetneq w(p,\alpha)$.}
\]

for all $p\in\QQ$ and $\alpha<\ddim$.

If $w$ is an $H$-witness function for $\nna$ and $\QQ$, then
there exists a step function $s$ for $\QQ$ such that

$$s(p,\alpha)\forces w(p,\alpha)\subseteq\nna.$$

\endenumerate-(1)

To see <ref>, suppose $u:\QQ\times\ddim\to{}^{<\kappa}\kappa$
is an $H$-witness function for $\nna$ and $\QQ$.
For any $p\in \QQ$, we have $\nna_{[p]}\cup u(p,\alpha)\in T_\QQ^{\nna,p}$ for all $\alpha<\ddim$, since $\nna_{[p]}$ equals the stem of $T_\QQ^{\nna,p}$
by Lemma <ref> <ref>.
It is easy to see that $T_\QQ^{\nna,p}$ has no terminal nodes.
Hence there is a sequence
$\langle w_p^\alpha:\alpha<\ddim\rangle$ in $ T_\QQ^{\nna,p}$
such that $\nna_{[p]}\cup u(p,\alpha)\subsetneq w_p^\alpha$ for all $\alpha<\ddim$.
by letting
$w(p,\alpha):=w_p^\alpha$ for all $p\in\QQ$ and $\alpha<\ddim$. It is clear that $w$ is as required.

To see <ref>, let $w:\QQ\times\ddim\to{}^{<\kappa}\kappa$ be an $H$-witness function for $\nna$ and $\QQ$.
By the definition of $T_\QQ^{\nna,q}$,
there exists $s_q^\alpha$
such that $s_q^\alpha\leq p$ and
$s_q^\alpha\forces w(p,\alpha)\subseteq\nna$
for all $q\in\QQ$ and $\alpha<\ddim$.
Define $s:\QQ\times\ddim\to\QQ$ by letting $s(q,\alpha):=s^\alpha_q$
for all $q\in\QQ$ and $\alpha<\ddim$.
Then $s$ is as required.

§.§.§ Construction of continuous homomorphisms

In our proof of Theorem <ref>,
we will define a continuous homomorphism $f$ from $\dhH\ddim$ to $H\restr X$.
The values of the homomorphism come from the evaluations of the $\Add(\kappa,1)$-name $\nna$ from the previous subsection
by a $\kappa$-perfect set of generic filters.
The difficulty is to ensure that $\ran(f)\subseteq X$.
The next lemma, the Nice Quotient Lemma for $\Col(\kappa,\lle\lambda)$, does this by controlling quotient forcings:
It will guarantee the existence
in $V[G]$
of a continuous function
whose range consists of elements with well-behaved quotient forcings.
This function will induce the required homomorphism $f$ via $\nna$.

We will construct
this function
along an appropriately chosen step function.
We will need the next definition.

Suppose $s$ is a step function for a given forcing $\QQ$.
A strict order reversing map
$h:{}^{<\kappa}\ddim\to\QQ$ is called an $s$-function
if for all $t\in{}^{<\kappa}\ddim$ and all $\alpha<\ddim$:

\begin{equation}
\label{eq: s-function}
s\big(h(t), \alpha\big).
\end{equation}

Note that it is necessary to require that $h$ is strict order reversing in the previous definition
since this does not follow from (<ref>).
(<ref>) does not guarantee that $h$ is order reversing at limits
However, (<ref>) does imply that
$h(u)<h(t)$ when $u$ is a direct successor of $t$.

if $\QQ=\Add(\kappa,1)$
and $h$ is an $s$-function, then
$[h]:{}^\kappa\ddim\to{}^\kappa\kappa$ is continuous and

$$[h](x)=\bigcup_{\alpha<\kappa}s\big( h(x\restr\alpha), x(\alpha)\big)$$

holds for all $x\in{}^\kappa\ddim$.

(Nice Quotient Lemma for $\Col(\kappa,\lle\lambda)$)

The following holds in $V[G]$.
Suppose that $\gamma<\lambda$ and
$s$ is a step-function for $\Add(\kappa,1)$.
Then there exists an $s$-function $h$
such that for all $x\in \ran([h])$:
\beginenumerate-(1)

$x$ is $\Add(\kappa,1)$-generic over $V[G_\gamma]$.
[Here, we identify $x$ with
$\kkppred{x}=\{t\in\Add(\kappa,1): t\subsetneq x\}$.

$V[G]$ is a
extension of $V[G_\gamma][x]$.
\endenumerate-(1)

We shall prove two cases of the Nice Quotient Lemma separately: the countable case in
Subsection <ref> and the uncountable case in
Subsections <ref> and <ref>.

In the rest of this subsection, we aim to construct a continuous homomorphism
from $\dhH\ddim$ to $H\restr X$.
To this end,
fix an ordinal $\gamma$ and an $\Add(\kappa,1)$-name $\nna$ in $V[G_\gamma]$
as in Lemma <ref>.
Then there exists an $H$-witness function for $\Add(\kappa,1)$ and $\nna$.
Thus, we may also fix
an $H$-witness function $w$ for $\nna$ and step function $s$ for $\nna$ and $w$ as in Lemma <ref>.
Lastly, let
$h$ be an $s$-function as in the previous lemma.

We argue in $V[G]$.
We define a function $f:{}^\kappa\ddim\to{}^\kappa\kappa$ by letting

\[
f(z):= \nna^{[h](z)}
\]

for all $z\in{}^{\kappa}\ddim$.

$\ran(f)\subseteq X$.

Let $z\in({}^{\kappa}\ddim)^{V[G]}$, and let $x:=[h](z)$.
Then $f(z)=\nna^x$.
Since $\nna$ was chosen so that
Lemma <ref> <ref>
holds in $V[G_\gamma]$
and $x$ is $\Add(\kappa,1)$-generic
over $V[G_\gamma]$
by Lemma <ref> <ref>,

\[
\qquad\qquad
\one_{\PP_\lambda}\forces f(z)\in\Xphia.\]

holds in $V[G_\gamma][x]$.
by Lemma <ref> <ref>.

It now suffices to prove the following.

$f$ is a continuous homomorphism from $\dhH\ddim$ to $H$.

We will define
a map
and show that it is an order homomorphism for
with $f=[\iota]$.
This is sufficient by Lemma <ref> <ref>.

Let $\iota(\emptyset):=\emptyset.$
If $\lh(v)\in\Succ$ and $v=u\conc\langle\alpha\rangle$, then let
$\iota(v):=w\big(h(u), \alpha\big)$.
If $\lh(v)\in\Lim$,
let $\iota(v):=\bigcup_{u\subsetneq v}\iota(u)$.

For all $v\in {}^{<\kappa}\ddim$, $h(v)\forces_{\Add(\kappa,1)}^{V[G_\gamma]} \iota(v)\subseteq \tau$.

This is proved by induction on $\lh(v)$.
It is clear for $v=\emptyset$.
For the successor case, note that

$$s(h(u),\alpha) \forces_{\Add(\kappa,1)} w(h(u),\alpha)\subseteq \tau$$

holds in $V[G]$
by Lemma <ref> <ref>.
Therefore this also holds in $V[G_\gamma]$ since the formula
“$p\forces \theta\subseteq \eta$” is absolute between transitive models of $\ZFC$.

can be defined by a recursion which
uses only absolute concepts; see <cit.>.

Since $h$ is an $s$-function, we have $s(h(u),\alpha)\subseteq h(u\conc\langle \alpha\rangle)$.
Moreover, $\iota(u\conc\langle\alpha\rangle)=w(h(u),\alpha)$ by the definition of $\iota$.

$$h(u\conc\langle\alpha\rangle) \forces_{\Add(\kappa,1)}^{V[G_\gamma]} \iota(u\conc\langle\alpha\rangle)\subseteq \tau.$$

If $\lh(v)\in\Lim$, then $h(v)\supseteq \bigcup_{u\subsetneq v} h(u)$ and $\iota(v)= \bigcup_{u\subsetneq v} \iota(u)$.
Thus $h(v)\forces_{\Add(\kappa,1)}^{V[G_\gamma]} \iota(v)\subseteq\tau$ by the inductive hypothesis.

Now $\iota$ is strict order preserving, since

$$ \iota(u)\subseteq \tau_{[h(u)]}  \subsetneq  w(h(u),\alpha)= \iota(u\conc \langle\alpha\rangle)$$

by the previous claim, Lemma <ref> <ref> and the definition of $\iota$.

To see that $\iota$ is an order homomorphism for $({}^\kappa\kappa,H)$, note that

\[
\prod_{\alpha<\ddim} N_{\iota(u\conc\langle\alpha\rangle)}=
\prod_{\alpha<\ddim} N_{w(h(u),\alpha)}
\subseteq H
\]

by the definition of $\iota$ and
since $w$ is an $H$-witness function.
To see that
$f=[\iota]$, recall that $f(z)=\tau^{[h](z)}$ by the definition of $f$.
Since $h(z\restr \alpha) \forces_{\Add(\kappa,1)}^{V[G_\gamma]} \iota(z\restr \alpha)\subseteq \tau$, we have $\iota(z\restr \alpha)\subseteq f(z)$ for all $\alpha<\kappa$.

This completes the proof of Theorem <ref> except for the proof of the
Nice Quotient Lemma for $\Col(\kappa,\lle\lambda)$.

§.§ The countable case

In this section, we complete the proof of the open dihypergraph dichotomy for definable sets in the countable case (Theorem <ref> for $\kappa=\omega$).
The missing step is the Nice Quotient Lemma for $\Col(\omega,\lle\lambda)$.
We make use of the next lemma:

(Solovay <cit.>, see <cit.>)

For any $x\in{({}^\omega\Ord)}^{V[G]}$, there exists a $\PP_\lambda$-generic
filter[For $\kappa=\omega$, $\PP_\lambda$ equals $\Col(\omega,\lle\lambda)$, where $\lambda$ is inaccessible. ]
$K$ over $V[x]$ such that $V[G]=V[x][K]$.

The proof idea for Solovay's lemma is that for any regular uncountable cardinal $\mu$, $\Col(\omega,\mu)$ is the unique forcing of size $\mu$ that makes $\mu$ countable <cit.>.
The phrase “quotient forcing of $V[x]$” makes sense by Proposition 10.10 in Kanamori's book. See Proposition 10.21 therein for a proof of Solovay's lemma.
Using this, one shows that the
quotient forcing of $V[x]$ in $V[G]$ is equivalent to $\Col(\omega,\lle\lambda)$.

We use the notation from the previous section for $\kappa=\omega$.
It remains to prove the following.

(Nice Quotient Lemma for $\Col(\omega,\lle\lambda)$)

The following holds in $V[G]$.
Suppose that $\gamma<\lambda$ and
$s$ is a step-function for $\Add(\omega,1)$.
Then there exists an $s$-function $h$
such that for all $x\in \ran([h])$:
\beginenumerate-(1)

$x$ is $\Add(\omega,1)$-generic over $V[G_\gamma]$.

$V[G]$ is a $\PP_\lambda$-generic extension of $V[G_\gamma][x]$.
\endenumerate-(1)

By Solovay's Lemma <ref> for $V[G_\gamma]$, it suffices to construct an $s$-function $h$ such that <ref> holds.

Let $\langle D_n:n<\omega\rangle\in V[G]$ enumerate all dense subsets of $\Add(\omega,1)$ that are elements of $V[G_\gamma]$.
This is possible since $({2^\omega})^{V[G_\gamma]}$ is countable in $V[G]$.
We construct $h:{}^{<\omega}\ddim\to\Add(\omega,1)$ with the following properties for all $u,v\in{}^{<\omega}\ddim$ and $\alpha<\kappa$:
\beginenumerate-(i)

If $u\subsetneq v$, then $h(v)< h(u)$.

s(h(u), \alpha)$.

*  $h(u)\in D_{\lh(u)}$.
\endenumerate-(i)
We construct $h(u)$ by recursion on $\lh(u)$.
Let $h(\emptyset)\in D_0$.
Now fix $u\in{}^{<\omega}\ddim$, and suppose that $h(u)$ has been constructed.
For each $\alpha<\ddim$,
choose $h(u\conc\langle\alpha\rangle)< s(h(u),\alpha)$ in $D_{\lh(u)+1}$.
Then $h$ satisfies
By <ref>, $h$ is order preserving, and so by <ref>, it is an $s$-function.
By <ref>, each $x\in\ran([h])$ is $\Add(\omega,1)$-generic over $V[G_\gamma]$.

While $\OGD^\omega_\omega(\analytic)$ is provable in $\ZFC$ <cit.>, $\OGD^\omega_\omega(\defsets\omega)$ has the consistency strength of an inaccessible cardinal, since $\OGD^\omega_\omega(\bf{\Pi}^1_1)$ implies that $\PSP_\omega(\bf{\Pi}^1_1)$ holds and thus there is an inaccessible cardinal in $L$
(see <cit.>.)

§.§ The uncountable case

In this section, we complete the proof of the open dihypergraph dichotomy for definable sets in the uncountable case (Theorem <ref> for $\kappa>\omega$).
We that assume $\kappa$ is uncountable

throughout the section.

§.§.§ kappa-analytic sets

We begin with the special case of Theorem <ref> for

The proof is much more direct than in the general case,
since it does not need the Nice Quotient Lemma for $\Col(\kappa,\lle\lambda)$.
It adapts an argument from <cit.>*Section 3.1
for the open graph dichotomy $\OGD_\kappa(\analytic(\kappa))$.

We first give a proof
for closed sets.

In $V[G]$, suppose $X$ is a closed subset of ${}^\kappa\kappa$. Then $\ODD\kappa {H\restr X}$ holds.

Recall our Assumptions <ref>.
In particular, we assume that $H\restr X$ does not admit a $\kappa$-coloring. It suffices to show that
there is a continuous homomorphism $f:{}^{\kappa}\ddim\to X$ from $\dhH\ddim$ to $H\restr X$.

In $V[G]$, let $S$ be a subtree of ${}^{<\kappa}\kappa$ such that $[S]=X=\Xphia$.
There is some $\alpha<\lambda$ such that $S\in V[G_\alpha]$, since $\PP_\lambda$ has the $\lambda$-c.c.
Now let $\gamma>\alpha$ and let $\nna$ be an $\Add(\kappa,1)$-name as in Lemma <ref>.

$\one\forces\nna\notin\check V$ is not needed in this proof. (It is needed for the proof of the general case.)
In $V[G_\gamma]$,
$\one_{\Add(\kappa,1)}$ forces that
$\one_{\PP_\lambda}\forces\check{\nna}\in [S]$ by
Lemma <ref> <ref>.
Since the formula “$x\in [S]$” is absolute between transitive models of $\ZFC$, we have

\begin{equation}
\label{an eq 1}
\one_{\Add(\kappa,1)} \forces^{V[G_\gamma]} \nna\in[S].
\end{equation}

In $V[G]$,
we construct functions
with the following properties
for all $u,v\in{}^{<\kappa}\ddim$:
\beginenumerate-(i)

If $u\subsetneq v$, then $t(u)\subsetneq t(v)$ and $h(v)< h(u)$.

*  $h(u)\forces^{V[G_\gamma]} t(u)\subseteq \nna$.

\prod_{\alpha<\ddim}N_{t({u\conc\langle\alpha\rangle})}\subseteq H
\endenumerate-(i)

Since the tree $T^{\tau,q}_{\Add(\kappa,1)}$ is not $H$-independent for all $q\in\Add(\kappa,1)$
by Lemma <ref> <ref>,
there exists an $H$-witness function for $\nna$.
Thus, we may fix an $H$-witness function $w$ and a step function $s$ for $\nna$ as in Lemma <ref>.

Construct $h(v)$ and $t(v)$ by recursion on $\lh(v)$.
Let $h(\emptyset):=\one_{\Add(\kappa,1)}$
and $t(\emptyset):=\emptyset$.
Now suppose that $v\in{}^{<\kappa}\ddim$ and
$h(u),\,t(u)$ have been constructed for all $u\subsetneq v$.
If $\lh(v)\in\Lim$,
let $t(v):=\bigcup_{u\subsetneq v}t(u)$.
$\Add(\kappa,1)$ is $\lle\kappa$-closed, there is some $h(v)< h(u)$ for all $u\subsetneq v$.
If $\lh(v)\in\Succ$ and $v=u\conc\langle\alpha\rangle$,
let $t(v):=w(h(u),\alpha)$
and $h(v)<s(h(u),\alpha)$.

Now suppose that
$t:{}^{<\kappa}\ddim\to S$
have been constructed.
Note that $t$ is strict order preserving by <ref>.
Hence $f=[t]:{}^\kappa\ddim\to X$ is defined.

By <ref>, $t$ is an order homomorphism for $({}^\kappa\kappa,H)$, so
$f$ is a continuous homomorphism from $\dhH\ddim$ to $H$.

It thus suffices that
$\ran(f)\subseteq [S]=X$.

This holds since
(<ref>) and <ref> imply that
$\ran(t)\subseteq S$.

The following alternative version of the previous proof can be understood as a special case of our proof of Theorem <ref>.
Let $S$, $\gamma$ and $\nna$ be as in the previous proof. Then
$\one_{\Add(\kappa,1)} \forces^{V[G_\gamma]} \nna\in[S]$.
Choose an $H$-witness function $w$ and a step function $s$ as in Subsection <ref>.
A similar construction as in

the Nice Quotient Lemma for $\Col(\omega,\lle\lambda)$
[See the proof of Lemma <ref>.]
shows that there exists an $s$-function
such that $[h](x)$ is $\Add(\kappa,1)$-generic over $V[G_\gamma]$ for all $x\in{}^\kappa\ddim$.[This is a weaker version of the Nice Quotient Lemma <ref>
obtained by omitting <ref>.]
For $f: {}^\kappa\ddim\to {}^\kappa\kappa$ with $f(x)=\nna^{[h](x)}$,
it follows that $\ran(f)\subseteq [S]$ by the absoluteness of “$x\in[S]$”.
One can show that $f$ is a homomorphism from $\dhH\ddim$ to $H$ as in
Lemma <ref>.
Thus, $f$ is a continuous homomorphism from $\dhH\ddim$ to $H\restr[S]=H\restr X$.

By the previous lemma, Theorem <ref> holds for all closed subsets of the $\kappa$-Baire space.
Since every

subset of ${}^\kappa\kappa$ is the continuous image of a closed subset of ${}^\kappa\kappa$,
we can extend this

sets by Lemma <ref>.

Theorem <ref> holds for all

subsets of ${}^\kappa\kappa$.

§.§.§ Definable sets

We prove the Nice Quotient Lemma for $\Col(\kappa,\lle\lambda)$ and thus Theorem <ref> for uncountable $\kappa$.
The main difficulty is that quotient forcings are not well behaved in the uncountable setting:

Solovay's Lemma <ref> does not extend to uncountable regular cardinals $\kappa$ by the following well-known counterexample.
Let $V[J*K]$ be an $\Add(\kappa,1)*\dot{\QQ}$-generic extension of $V$, where $\Add(\kappa,1)$ adds a stationary set $S$ and $\dot{\QQ}$ shoots a club through its complement.
Thus $\dot{\QQ}^J$ is not stationary set preserving in $V[J]$.
Hence  $V[J*K]$ is not a generic extension of $V[J]$ by a ${<}\kappa$-closed forcing.
This is discussed in more detail in <cit.>.

Instead of Solovay's lemma, we will use the following Nice Quotient Lemma for $\Add(\kappa,1)$.
Its proof is postponed to
the next subsection.

(Nice Quotient Lemma for $\Add(\mu,1)$)

Let $M$ be a transitive class model of $\ZFC$.[I.e., $M$ can be a proper class or a set.]
In $M$, suppose that $\mu$ is an uncountable cardinal with $\mu^{<\mu}=\mu$,
$2\leq\ddimq\leq\mu$ and
$s:\Add(\mu,1)\times\ddimq\to\Add(\mu,1)$ is a step function for $\Add(\mu,1)$.

For any $\Add(\mu,1)$-generic filter $K$ over $M$,
there exists in $M[K]$ an $s$-function

such that for all
$x\in \ran([h])$:

\beginenumerate-(1)

$x$ is $\Add(\mu,1)$-generic over $M$.

$M[K]$ is an $\Add(\mu,1)$-generic extension of $M[x]$.
\endenumerate-(1)
In fact, in <ref>,

for all distinct $y,z\in{}^\mu\ddimq$,
$[h](y)$ and $[h](z)$ are mutually $\Add(\mu,1)$-generic over $M$.

(Nice Quotient Lemma  for $\Col(\kappa,\lle\lambda)$)

The following holds in $V[G]$.
Suppose that $\gamma<\lambda$ and
$s$ is a step-function for $\Add(\kappa,1)$.
Then there exists an $s$-function

such that for all $x\in \ran([h])$:
\beginenumerate-(1)

$x$ is $\Add(\kappa,1)$-generic over $V[G_\gamma]$.

$V[G]$ is a

extension of $V[G_\gamma][x]$.
\endenumerate-(1)
In fact in <ref>, for all distinct $y,z\in{}^\kappa\ddim$,
$[h](y)$ and $[h](z)$ are mutually $\Add(\kappa,1)$-generic over $V[G_\gamma]$.

We rearrange $V[G]$ as an $\Add(\kappa,1)$-generic extension of a model $M$.
Fix some $\alpha\in[\gamma,\lambda)$ with $s\in V[G_\alpha]$ by the $\lambda$-c.c.

Since $\PP^\alpha$ is equivalent to $\Add(\kappa,1)\times\PP_\lambda$,[See Corollary <ref>.]
we write

$$V[G]=V[G_\alpha\times K\times J],$$

where is $K\times J$ is an $\Add(\kappa,1)\times\PP_\lambda$-generic filter over $V[G_\alpha]$.
Then $V[G]$ is an $\Add(\kappa,1)$-generic extension of $M:=V[G_\alpha\times J]$.

Note that $s$ is a step-function in $M$,
since this concept is absolute between transitive $\ZFC$-models with the same ${}^{<\kappa}\kappa$.
Fix an $s$-function $h$ in $V[G]$
as in the previous lemma.

Then each $x\in\ran([h])$ is $\Add(\kappa,1)$-generic over $V[G_\gamma]$, since $V[G_\gamma]\subseteq M$.
It remains to see that $V[G]=M[K]$ is a
$\PP_\lambda$-generic extension of $V[G_\gamma][x]$.
Recall that $M[K]$ is an $\Add(\kappa,1)$-generic extension of $M[x]$.

$$M[x]=V[G_\alpha\times J][x]=V[G_\gamma][x][ G_{[\gamma,\alpha)}\times J]$$

is a $\PP_{[\alpha,\gamma)}\times\PP_\lambda$-generic extension of  $V[G_\gamma][x]$.
Since $\PP_{[\alpha,\gamma)}\times\PP_\lambda\times\Add(\kappa,1)$ is equivalent to $\PP_\lambda$,[See Corollary <ref>.]
this suffices.

In the uncountable case only, the proof yields the following stronger version of Theorem <ref>.
For a given $\ddim$-dihypergraph $I$ on ${}^\kappa\kappa$,
$\ODDI\kappa I$ denote the statement:

$\ODDI\kappa I$: Either $I$ has a $\kappa$-coloring, or
there exists
an injective continuous homomorphism
from $\dhH{\ddim}$ to $I$.
[See Definition <ref> below.]

We claim that
if $\kappa<\lambda$ are regular uncountable cardinals and $2\leq\ddim\leq\kappa$,
then the following hold in all $\PP_\lambda$-generic extensions $V[G]$:
\beginenumerate-(1)

$\ODDI\kappa\ddim(\defsetsk,\defsetsk)$ if $\lambda$ is inacessible.
[The definitions of
$\ODDI\kappa\ddim(\defsetsk,\defsetsk)$ and
are analogous to the ones in Definition <ref>.
Note that <ref> implies $\ODDI\kappa\ddim(\defsetsk)$ for all $\ddim$ with $2\leq\ddim<\kappa$ by Lemma <ref>.]

$\ODDI\kappa\ddim(\defsetsk)$ if $\lambda$ is Mahlo.
\endenumerate-(1)

To prove this, recall that after the Nice Quotient Lemma <ref>,

we defined a continuous homomorphism $f$ from $\dhH\ddim$ to $H\restr X$ by letting $f(x):=\nna^{[h](x)}$, where $\gamma$ and $\nna$ are as in Lemma <ref>.
In particular, note that $\nna$ is a name in $V[G_\gamma]$ for a new element of  ${}^\kappa\kappa$.

It remains to show that $f$ is injective. Let $x,y$ be distinct elements of ${}^\kappa\ddim$,
let $x':=[h](x)$ and $y':=[h](y)$.
Then $f(x)=\nna^{x'}\in V[G_\gamma][x'] - V[G_\gamma]$
$f(y)=\nna^{y'}\in V[G_\gamma][y'] - V[G_\gamma]$.
$x'$ and $y'$ are mutually generic over $V[G_\gamma]$,
we have
$V[G_\gamma][x']\cap V[G_\gamma][y']=V[G_\gamma]$.

$f(x)\neq f(y)$.

Theorem <ref>
and  Corollary <ref>
below provide an alternative proof using a weak variant of the $\Diamond$ principle.

Note that the Nice Quotient Lemmas <ref> and <ref> both fail in the countable case,
since $\ODDI\omega\omega(\defsets\omega,\defsets\omega)$ fails by Proposition <ref> below.
While Lemma <ref> does hold without the last condition on mutual genericity by Lemma <ref>, we do not know if this is the case for Lemma <ref>.

§.§.§ The Nice Quotient Lemma for Add(mu,1)

This subsection is dedicated to the proof of Lemma <ref>.

We make the same assumptions as in this lemma
throughout the subsection.
In particular,
$M$ is a transitive class model of $\ZFC$.
In $M$, $\mu$ is an uncountable cardinal with $\mu^{<\mu}=\mu$,
we need to assume that $\mu^{<\mu}=\mu$ holds in $M$ e.g. to have $|\QQ|=\mu$
$2\leq\ddimq\leq \mu$ and
is a step function for $\Add(\mu,1)$.
[See Definition <ref>]

Our proof uses ideas from <cit.>.
We first give a short description of the proof.
Working in $M$, we define a forcing $\QQ$ whose domain contains
I wrote “whose domain contains” because: the partial order on $\QQ$ is unusual; also, $\QQ$ consists of strict partial $s$-functions only
partial functions of size $\lle\mu$ approximating
an $s$-function.
[See Definition <ref>.]
We first prove that
$\QQ$ is equivalent to $\Add(\mu,1)$.
We then show that
if $h$ is the $s$-function added in the natural way by a given $\QQ$-generic filter,
then $h$ satisfies the requirements in Lemma <ref>.
A significant portion of the arguments in this subsection are aimed at proving the second requirement
that elements of $\ran([h])$ have well-behaved quotients.

We now present the detailed proof.
We argue either in $M$ or in a $\QQ$-generic extension of $M$ throughout this subsection.

Recall that an $s$-function for $\Add(\mu,1)$ is a strict order preserving
map $h:{}^{<\mu}\ddimq\to{}^{<\mu}\mu$
[Note that ${}^{<\mu}\mu$ is $\Add(\mu,1)$ with the ordering reversed.]
which is “built along” $s$ in
the sense of Definition <ref>.
In <ref> of the next definition, we consider partial functions

with the same property.
Our forcing $\QQ$ will consist of
such partial $s$-functions of size $\lle\mu$
with the additional technical requirement of strictness defined in <ref>
that is used to ensure $\QQ$ is separative.
$\supsetneq$ (instead of $\supseteq$) in

the definition of “partial $s$-function”
is also needed only to ensure separativity

\beginenumerate-(a)

A partial $s$-function is a
strict order preserving the requirement in the equation below does not guarantee that $p$ is order preserving at limit lengths
partial function

such that
$\dom(p)$ is a subtree of ${}^{<\mu}\ddimq$, and
if $u,u\conc\langle \alpha\rangle\in\dom(p)$, then

\[
\supsetneq
s\big(p(u), \alpha\big).
\]

A partial $s$-function $p$ is strict
if for all $t\in\dom(p)$ with $\lh(t)\in\Lim$,“strict order preserving” already implies this requirement for $\lh(t)\in\Succ$
we have

\[p(t)\supsetneq \displaystyle\bigcup_{u\subsetneq t} p(u).\]

\endenumerate-(a)

We now define the forcing $\QQ$ that will be used in our proof of the Nice Quotient
Lemma <ref> for $\Add(\mu,1)$.
It is embedded into a poset $\mpo$ that will be useful as well.

\beginenumerate-(a)

$\mpo$ consists of all strict partial $s$-functions.
pagebreak before this sentence

$\QQ$ consists of all $q\in\mpo$ with $\left|\dom(q)\right|<\mu$.

We equip both $\QQ$ and $\mpo$ with the partial order $\mleq$ defined by letting
$p\mleq q$
if the following conditions hold:
\beginenumerate-(i)

* $\dom (q)\subseteq\dom (p)$.

* $q(t)=p(t)$ for every non-terminal node $t$ of $\dom(q)$.

* $q(t)\subseteq p(t)$ for every terminal node $t$ of $\dom(q)$.
\endenumerate-(i)
\endenumerate-(a)

Thus, a condition $q$ is extended by moving its topmost values upwards and end-extending the tree $\dom(q)$.
Note that the definition of $\QQ$ is absolute to transitive models
$N\supseteq M$ of $\ZFC$ with $({}^{<\mu}\mu)^M=({}^{<\mu}\mu)^N$.
Specifically, since $\QQ$ is $\lle\mu$-closed by Lemma <ref> below, the definition of $\QQ$ is absolute between $M$ and $\QQ$-generic extensions of $M$.

$\mpo^N$ may contain some
total $s$-functions
that are not in $M$.

We now discuss some properties of $\mpo$ which will be useful in this subsection.
The next lemma
gives an equivalent reformulation
of the definition of strict partial $s$-functions.

Both the next claim and lemma will be needed right away

Let $q:{}^{<\mu}\ddimq\partialto{}^{<\mu}\mu$ be a
the assumption that $q$ is strict order preserving is not needed
$\supsetneq$ (instead of $\supseteq$) in (<ref>) is needed for $\QQ$ to be separative; see the last paragraph of the proof of Lemma <ref>.

This is why $\supsetneq$ was needed in the def of “(partial) $s$-function” and also why the partial $s$-functions in $\QQ$ need to be strict
partial function whose domain
is a subtree of ${}^{<\mu}\ddimq$.

$q$ is a strict partial $s$-function (i.e., $q\in\mpo$)
if and only if
for all $t\in\dom(q)$,

\begin{equation}
\label{Q equiv eq}
q(t)\supsetneq \bigcup_{\alpha< \lh(t)} s\big(q(t\restr\alpha), t(\alpha) \big).
\end{equation}

First suppose that
$q$ is a strict partial $s$-function

and $t\in\dom(q)$.
If $\lh(t)\in\Lim$, then

\[
q(t)\supsetneq \bigcup_{\alpha< \lh(t)} q(t\restr\alpha)=
\bigcup_{\alpha< \lh(t)} q(t\restr(\alpha+1))\supseteq
\bigcup_{\alpha< \lh(t)} s\big(q(t\restr\alpha), t(\alpha)\big).
\]

The inclusion on the left
holds since $q$ is strict, while the
inclusion on the right holds
since $q$ is a partial $s$-function.

Now assume
$\lh(t)\in\Succ$ and $t=u\conc\langle\beta\rangle$.
Since $s$ is a step function
$q$ is a partial $s$-function,
we have

\begin{equation*}
\begin{gathered}
q(u)\subseteq s(q(u),\beta)\subsetneq q(t),
\\[4pt]
s\big(q(t\restr\alpha), t(\alpha)\big)\subsetneq
\subseteq q(u)
\smallskip
\end{gathered}
\end{equation*}

for all $\alpha<\lh(u)$.
$\bigcup_{\alpha< \lh(t)} s\big(q(t\restr\alpha), t(\alpha)\big)=
s(q(u),\beta)\subsetneq q(t)$.

Conversely, (<ref>) implies immediately that $q$ is a strict partial $s$-function.
In particular, $q$ is strict order preserving since $s$ is a step-function.

Recall that $K\subseteq\mpo$ is a downward directed subset of the poset $\mpo$
if for every $p,q\in K$, there exists $r\in K$ with $r\mleq p,q$.

In particular, any $\QQ$-generic filter $K$ over $M$ is a downward directed subset
of $\mpo^{M[K]}$.
We next define $\mbigcap K\in \mpo$ for any downward directed
subset $K$ of $\mpo$. This will turn out to equal the infimum in relevant cases.

Suppose $K$ is a downward directed subset of $\mpo$.
pagebreak inserted
Let $\mbigcap K$ denote the partial function from ${}^{<\mu}\ddimq$ to ${}^{\leq\mu}\mu$ defined as follows:

\begin{equation*}
\begin{gathered}
\dom(\mbigcap K\big):=\bigcup\{\dom(q):q\in K\},
\\[4pt]
%& \
(\mbigcap K) (t):=\bigcup\{q(t):{q\in K,\,t\in\dom(q)}\}.
\end{gathered}
\end{equation*}

Note that the definition of $\mbigcap K$
is absolute between transitive models $N\subseteq N'$ of $\ZFC$ with $K\in N$. This might not be needed, but it's interesting to note that the infimum of a subset $K\in M$ of $\QQ$ will be the same in $\mpo^M$ and $\mpo^{M[K]}$
The next lemma shows that $\mbigcap K$ is indeed the infimum of $K$ in $\mpo$ in all relevant cases.

Let $K$ be a downward directed subset of $\mpo$.
\beginenumerate-(1)

If $q\in K$ and $t$ is a non-terminal node of $\dom(q)$, then
$(\mbigcap K)(t)=q(t)$.

Suppose that for any $t\in \dom(\mbigcap K)$, there is some $p\in K$ such that $t$ is a non-terminal node in $\dom(p)$.
Then $\ran(\mbigcap K) \subseteq {}^{<\mu}\mu$.
The assumption that $|K|<\mu$ also implies that $\ran(\mbigcap K) \subseteq {}^{<\mu}\mu$. However, this will not be needed. <ref> will be enough.

If $\ran(\mbigcap K) \subseteq {}^{<\mu}\mu$, then $\mbigcap K\in\mpo$, and
$\mbigcap K$
is the infimum of $K$ in $\mpo$.

*  If $K$ is a downward directed subset of $\QQ$ of size $\lle\mu$, then $\mbigcap K\in\QQ$ and $\mbigcap K$ is the infimum of $K$ in $\QQ$.
\endenumerate-(1)

For <ref>,

note that $q(t)\subseteq (\mbigcap K)(t)$.
Conversely, if $p\in K$ and $t\in\dom(p)$, then there exists $r\in K$ such that $r\mleq p,q$ and therefore
$p(t)\subseteq r(t)=q(t)$.
Hence $(\mbigcap K)(t)\subseteq q(t).$

For <ref>, let $t\in \dom(\mbigcap K)$ and
take $p\in K$ such that $t$ is a non-terminal node in $\dom(p)$.
Then $(\mbigcap K) (t)=p(t)\in {}^{<\mu}\mu$ by <ref>.

For <ref>,
we first claim that $\mbigcap K\in\mpo$.
Since $\dom(\mbigcap K)$ is a subtree of ${}^{<\mu}\ddimq$,
it suffices to show (<ref>) for $\mbigcap K$
by the previous lemma.

Let $t\in\dom(\mbigcap K)$ and take $p\in K$ with $t\in\dom(p)$.
Then $t\restr\alpha$ is a non-terminal node of $\dom(p)$ for all $\alpha<\lh(t)$,
so $(\mbigcap K)(t\restr\alpha)=p(t\restr\alpha)$ by <ref>.
Since (<ref>) holds for $p$,

\[
\bigcup_{\alpha< \lh(t)} s\big((\mbigcap K)(t\restr\alpha), t(\alpha) \big)=
\bigcup_{\alpha< \lh(t)} s\big(p(t\restr\alpha), t(\alpha) \big)\subsetneq
(\mbigcap K)(t).
\]

We now claim that $\mbigcap K$ is the infimum of $K$ in $\mpo$.
We first show that $\mbigcap K\mleq p$ for any $p\in K$.
By definition, $\dom(p)\subseteq \dom(\mbigcap K)$
and $p(t)\subseteq (\mbigcap K)(t)$ for any $t\in\dom(p)$.
Moreover, if $t$ is a non-terminal in $\dom(p)$, then $(\mbigcap K)(t)=p(t)$ by  <ref>.
Hence $\mbigcap K\mleq p$.

Now, suppose $h\in\mpo$ and $h\mleq p$ for all $p\in K$. It suffices to show
$h\mleq \mbigcap K$.
By definition, $\dom(\mbigcap K)\subseteq\dom(h)$ and $(\mbigcap K)(t)\subseteq h(t)$ for all $t\in\dom(\mbigcap K)$.
Moreover, if $t$ is a non-terminal node of $\dom(\mbigcap K)$, then there exists $p\in K$ such that $t$ is a non-terminal node of $\dom(p)$.
$(\mbigcap K)(t)=p(t)=h(t)$ by
<ref> and $h\mleq p$.
Therefore $h\leq\mbigcap K$.

For <ref>, note that $|\dom(\mbigcap K)|<\mu$ and $\ran(\mbigcap K)\subseteq {}^{<\mu}\mu$.
Thus the claim follows from <ref> since $\QQ\subseteq\mpo$.

There is no general relationship between $\mbigcap K$ and the infimum of $K$ in the Boolean completion $\BB(\QQ)$ of $\QQ$.
However, they are equal if $K$ is a downward directed subset of $\QQ$ of size ${<}\mu$
using Lemma <ref> <ref>
and the fact that $\QQ$ is a dense subset of $\BB(\QQ)$.
[$\QQ$ can be assumed to be a dense subset of $\BB(\QQ)$,
since we will prove in Lemma <ref> below that $\QQ$ is separative.

Our first goal is to show that the forcing $\QQ$ is equivalent to $\Add(\kappa,1)$.
This is done in Lemma <ref>.
Before proving this,
we characterize
when two given elements of $\QQ$ are compatible, in the next lemma.
This will be used to prove that $\QQ$ is
[See the proof Lemma <ref>.]
and separative.
[See Lemma <ref>.]

Two elements $p$ and $q$ of $\QQ$ are compatible if and only if
the following hold for all $t\in\dom(p)\cap\dom(q)$:
\beginenumerate-(1)

$p(t)\compat q(t)$.
[I.e., $p(t)\subseteq q(t)$ or $q(t)\subseteq p(t)$.]

If $t$ is a non-terminal node of $\dom(p)$, then
$q(t)\subseteq p(t)$.

If $t$ is a non-terminal node of $\dom(q)$, then
$p(t)\subseteq q(t)$.
\endenumerate-(1)
In particular, if $p$ and $q$ are compatible, then
whenever $t$ is non-terminal in both $\dom(p)$ and $\dom(q)$.

First, assume $p$ and $q$ are compatible
and $t\in\dom(p)\cap\dom(q)$.
Let $r\in \QQ$ be a common extension of $p$ and $q$.
$r(t)\supseteq p(t)$ and $r(t)\supseteq q(t)$, and hence $p(t)\compat q(t)$.
$t$ is a non-terminal node of $\dom(p)$,
then $q(t)\subseteq r(t)=p(t)$.
if $t$ is a non-terminal node of $\dom(q)$,
then $p(t)\subseteq r(t)=q(t)$.

Now assume <ref>–<ref> hold for all $t\in\dom(p)\cap\dom(q)$.
Let $r$ be the unique function with $\dom(r)=\dom(p)\cup\dom(q)$ and $r(t)=p(t)\cup q(t)$ for all $t\in\dom(r)$.
Then  $r(t)=\max\{p(t),q(t)\}$
for all $t\in\dom(p)\cap\dom(q)$
by <ref>.
$r(t)=p(t)$ for all
non-terminal nodes $t$ of $\dom(p)$
by <ref>,
$r(t)=q(t)$  for all
non-terminal nodes $t$ of $\dom(q)$
by <ref>.
Using these observations, it is easy to check that $r\in\QQ$ and
that $r\leq p, q$.

$\QQ$ is equivalent to $\Add(\mu,1)$.

It suffices to show that $\QQ$ is  a non-atomic $\lle\mu$-closed forcing of size $\mu$
by Lemma <ref>.
The assumption $\mu^{<\mu}=\mu$ implies that $|\QQ|=\mu$.
To show that $\QQ$ is $\lle\mu$-closed, suppose that $\langle p_\alpha:\alpha<\xi\rangle$ is a descending sequence of elements of $\QQ$ of length $\xi<\mu$. Then
$p=\mbigcap\{p_\alpha:\alpha<\xi\}$ is an element of $\QQ$ such that $p\leq p_\alpha$ for all $\alpha<\xi$ by Lemma <ref> <ref>.

To show that $\QQ$ is non-atomic, let $p\in \QQ$. Take a minimal element $t$ of ${{}^{<\mu}\ddimq}-\dom(p)$, let
$u=\bigcup_{\alpha< \lh(t)} s\big(q(t\restr\alpha), t(\alpha)\big),$
and choose $u_0,u_1\in{}^{<\mu}\mu$ with $u_0\perp u_1$ and $u_0,u_1\supsetneq u$.
Finally, let $q_i:=p\cup\{(t,u_i)\}$ for $i=0,1$.
Then $q_0,q_1\in\QQ$ by Lemma <ref>.
It is clear that
$q_0,q_1\leq p$, and
$q_0\perp q_1$ holds by
the previous lemma.

Throughout the rest of this subsection, we use the following notation.

Suppose $K$ is a $\QQ$-generic filter over $M$.
In $M[K]$, let

\begin{equation*}
h_K=\mbigcap K.
\end{equation*}

In $M$, let $\dot K$ denote the canonical name for a $\QQ$-generic filter,
and let $\dot h$ be a $\QQ$-name for $\mbigcap \dot K$
in the sense that

\begin{equation*}
\one_\QQ\forces\:\dot h=\mbigcap \dot K.
\end{equation*}

$h_K$ is well defined
since $K$ is a filter on $\QQ$ and hence an upward directed subset of

In the rest of this subsection, we will show that
$h_K$ is an $s$-function which
satisfies the conclusion of the Nice Quotient Lemma <ref> for $\Add(\mu,1)$.

We first prove
that $h_K$ is indeed an $s$-function.
We will need the following lemma.

If $t\in{}^{<\mu}\ddimq$, then
is a dense subset of $\QQ$.

We prove the lemma by induction on $\lh(t)$.
$D_{t\restr\alpha}$ is dense in $\QQ$
for all $\alpha<\lh(t)$.
To show that $D_t$ is dense in $\QQ$, let $p\in\QQ$.
Using the fact that $\QQ$ is $\lle\mu$-closed, construct a decreasing sequence
$\langle q_\alpha:\alpha<\lh(t)\rangle$ of conditions below $p$
such that
$q_\alpha\in D_{t\restr\alpha}$, and let
Then $q\in\QQ$
and $q\leq  p$
by Lemma <ref> <ref>, and
$t\restr\alpha\in\dom(q)$ for all $\alpha<\lh(t)$.

We define a condition
$r\in D_t$ below $q$
as follows.
If $t\in\dom(q)$, then let $r:=q$.
If $t\notin\dom(q)$, fix $u\in{}^{<\mu}\mu$ with
u\supsetneq \bigcup_{\alpha< \lh(t)} s\big(q(t\restr\alpha), t(\alpha) \big)
and let $r:=q\cup\{(t,u)\}$.
Then $r\in \QQ$ by Lemma <ref>,
it is clear that $t\in\dom(r)$ and $r\leq  q$.
$r$ is an element of $D_t$ below $p$.

Suppose that $K$ is a $\QQ$-generic filter over $M$. Then in $M[K]$:
\beginenumerate-(1)

$h_K\in\mpo$ and $h_K$ is the infimum of $K$ in $\bar{\QQ}$.

$h_K$ is an $s$-function.
\endenumerate-(1)

<ref> follows from the previous lemma.
For <ref>, it suffices to show $\ran(h_K)\subseteq {}^{<\mu}\mu$ by Lemma <ref> <ref>.
To see this, note that $\dom(h_K)={}^{<\mu}\ddimq=\bigcup_{p\in K}\dom(p)$ by <ref> and the definition of $\mbigcap$.
Thus,  $h_K=\mbigcap K$ by Lemma <ref> <ref>, as required.
<ref> follows immediately from <ref> and <ref>.

The following lemma shows that any $\QQ$-generic filter
$K$ is definable from $h_K$.

For all $\QQ$-generic filters $K$ over $M$
This lemma is used twice later.
and all $q\in\QQ$,

\[
q\in K
\ \Longleftrightarrow \
%\quad\text{if and only if}\quad
h_K\mleq q.
\]

If $q\in K$, then $h_K\mleq q$
by the previous corollary.

Conversely, suppose $h_K\mleq q$.
Since $K$ is $\QQ$-generic, it suffices to show that all elements $p$ of $K$ are compatible with $q$ in $\QQ$.
Given $p\in K$,
let $r:=h_K\restr(\dom(p)\cup \dom(q))$.
Then $r\in\QQ$ since
$h_K\in\mpo^{M[K]}$ and $|\dom(p)\cup\dom(q)|<\mu$.
Moreover, $r\leq  p,\,q$
since $h_K\mleq p,\,q$.

In the remainder of the subsection, we will prove that $h_K$ satisfies the conclusion of the Nice Quotient Lemma <ref> for $\Add(\mu,1)$,
i.e., that
for all $x\in \ran([h_K])$,
\beginenumerate-(1)

$x$ is $\Add(\mu,1)$-generic over $M$,

$M[K]$ is an $\Add(\mu,1)$-generic extension of $M[x]$.
\endenumerate-(1)

In order to show <ref>, we will later calculate the quotient forcing for a given element of $\ran([h_K])$.
When doing so, it will be more convenient to work with certain dense subforcings $\QQ^{*}$ of $\QQ$.
These will be
the intersections of ${<}\mu$ many subforcings
of the form $\QQ^*(\vartheta,\sigma)$ given in the next definition.
Regarding notation,
$\height(A)$, $\lh(A)$ were defined in Subsection <ref>,
and $\vartheta_{(p)},\,\vartheta_{[p]}$ in Definition <ref>.

Let $\vartheta$ and $\sigma$ be $\QQ$-names such that
$\one_\QQ\forces\vartheta\in{}^{\mu}\ddimq\wedge[\dot h](\vartheta)=\sigma$.
Then $\QQ^*(\vartheta,\sigma)$ denotes the subforcing of $\QQ$ which consists of all conditions $p\in \QQ$ such that the following hold:
Note that
we can equivalently
replace $\vartheta_{(p)}$ by $\vartheta_{[p]}$
and $\sigma_{(p)}$ by $\sigma_{[p]}$
everywhere in this definition,

by Lemma <ref>.
\beginenumerate-(a)

$\height(\dom p)\in \Lim$ and
$\height(\dom p)\subseteq \dom(\vartheta_{(p)})$.

$\lh(\ran p)\in \Lim$ and $\lh(\ran p)\subseteq \dom(\sigma_{(p)})$.

for all $\gamma<\height(\dom(p))$.
\endenumerate-(a)

Note that if $p\in \QQ^*(\vartheta,\sigma)$, then
we get the equality $\lh(\ran p)= \dom(\sigma_{(p)})$
in <ref>,
by Lemma <ref> below.

We next show that $\QQ^*(\vartheta,\sigma)$ is a dense subforcing of $\QQ$ and
more generally, that
the intersection $\QQ^*$ of ${<}\mu$ many subforcings of this form is also dense in $\QQ$.
We will need the following lemma.

Suppose $\langle p_\alpha:\alpha<\xi\rangle$ is a strictly decreasing sequence of elements of $\QQ$ of length $\xi<\mu$.

\[
\]

$\height(\dom p)=\bigcup_{\alpha<\xi}\height(\dom p_\alpha)$
$\lh(\ran p)=\bigcup_{\alpha<\xi}\lh(\ran p_\alpha)$.

The first part of the conclusion holds since
$\dom(p)=\bigcup_{\alpha<\xi}\dom(p_\alpha)$ by definition.
$\lh(p(t))=\bigcup\{\lh(p_\alpha(t)):\:\alpha<\xi,\, t\in\dom(p_\alpha)\}$ for all $t\in\dom(p)$ by definition.
$\lh(\ran p)=\bigcup\{\lh(p(t)):t\in\dom(p)\}=
\bigcup\{\lh(p_\alpha(t)):\:\alpha<\xi,\,t\in\dom(p_\alpha)\}
=\bigcup_{\alpha<\xi}\lh(\ran p_\alpha)$.

Suppose that $\xi<\mu$.
For all $i<\xi$,
let $\vartheta_i$ and $\sigma_i$
denote $\QQ$-names such that
$\one_\QQ\forces \vartheta_i\in{}^{\mu}\ddimq\wedge\sigma_i=[\dot h](\vartheta_i)$,
and let

\begin{equation*}
%\label{Q* eq}
\QQ^*=\bigcap_{i<\xi}\QQ^*(\vartheta_i,\sigma_i).
\end{equation*}

is a dense subforcing of $\QQ$.

This proof works only when $\mu>\omega$
Let $q\in \QQ$. Construct by recursion a strictly decreasing sequence
$\langle p_n:n<\omega\rangle$ of conditions in $\QQ$ as follows.
Let $p_0:=q$.
If $n<\omega$ and $p_n$ has been defined, use Lemma <ref>
and the fact that
$\QQ$ is $\lle\mu$-closed to choose $p_{n+1}$ such that
\beginenumerate-(i)

$p_{n+1}$ decides
$\vartheta_i\restr\height(\dom p_n)$ and $\sigma_i\restr\lh(\ran p_n)$
for all $i<\xi$,

$(\vartheta_i)_{[p_n]}\in\dom(p_{n+1})$ for all $i<\xi$,

$\height(\dom p_{n+1})>\height(\dom p_n)$, and

$\lh(\ran p_{n+1})>\lh(\ran p_n)$.
\endenumerate-(i)
Let $p:=\mbigcap\{p_n:n<\omega\}$.

Then $p\in\QQ$ and $p\leq  q$ by Lemma <ref> <ref>.
To prove that $p\in\QQ^*$,
let $i<\xi$.
We show that $p\in \QQ^*(\vartheta_i,\sigma_i)$.
The fact that
$\height(\dom p)\in \Lim$ and $\lh(\ran p)\in \Lim$
follows from <ref>, <ref> and Lemma <ref>.
By <ref>,
$\height(\dom p_n)\subseteq \dom{(\vartheta_i)_{(p_{n+1})}}$
$\lh(\ran p_n)\subseteq\dom{(\sigma_i)_{(p_{n+1})}}$
for all $n<\omega$.
From this and Lemma <ref>, we obtain:
we have to use $\lh(\ran p)$ instead of $\height(\ran p)$ in the def of $\QQ^{*}$ for this step to work

\begin{equation*}
\begin{gathered}
\height(\dom p)=\bigcup_{n<\omega}\height(\dom p_n)
\subseteq
\bigcup_{n<\omega}\dom{(\vartheta_i)_{(p_{n+1})}}
\subseteq \dom{(\vartheta_i)_{(p)}},
\\
\lh(\ran p)=\bigcup_{n<\omega}\lh(\ran p_n)
\subseteq
\bigcup_{n<\omega}\dom{(\sigma_i)_{(p_{n+1})}}
\subseteq \dom{(\sigma_i)_{(p)}}.
\end{gathered}
\end{equation*}

Definition <ref> <ref> and <ref>
hold for
$\vartheta_i,\,\sigma_i$ and $p$.
To show that
<ref> also holds, let
$\gamma<\height(\dom p)$ and take $n<\omega$ with
$\gamma\leq\height(\dom p_{n})$.
Then by <ref>,
or equivalently, $\gamma\leq\lh{\left((\vartheta_i)_{[p_{n+1}]}\right)}$.

\[

by <ref>.
Therefore $(\vartheta_i)_{(p)}\restr\gamma=t\in\dom(p)$, as required.

The next lemma
is a stronger form of the following observation:
Let $p\in\QQ^*(\vartheta,\sigma)$, where $\vartheta$ and $\sigma$ are $\QQ$-names
such that $\one_\QQ\forces\vartheta\in{}^{\mu}\ddimq\wedge[\dot h](\vartheta)=\sigma$.
Let $t$ be the least initial segment of $\vartheta_{[p]}$ with $t\notin\dom(p)$.
[$t$ exists since
if $p\in\QQ^*(\vartheta,\sigma)$
by Definition <ref> <ref>
and Lemma <ref>.
For instance, if $\vartheta=\check x$ for some $x$ in $({}^\mu\mu)^M$, then $\vartheta_{[p]}=x$.]
$t=\vartheta_{(p)}\restr\height(\dom p)$.
Furthermore, let
v:=\bigcup\big\{p(t\restr\alpha):{\alpha<\height(\dom p)}\big\}.
For any choice of $u$ with $v\subsetneq u$,
remains a strict partial $s$-function (i.e., $r\in\QQ$)
and moreover,
$\sigma_{(r)}\supseteq u$.
Note that
$\sigma_{(p)}=v$ by Lemma <ref> below.

We will need the next lemma
for ${<}\mu$ many pairs of names $(\vartheta_i,\sigma_i)$.

Using the same assumptions as in Lemma <ref>,
suppose that $p\in\QQ^*$.
Let $t_i :=(\vartheta_i)_{(p)}\restr\height(\dom p)$ for all $i<\xi$.
\beginenumerate-(1)

$t_i$ is the least initial segment of $(\vartheta_i)_{[p]}$ with $t_i\notin\dom(p)$ for all $i<\xi$.

If $t_i\neq t_j$,
$u_i\in{}^{<\mu}\mu$ and
$\bigcup\big\{p(t_i\restr\alpha):{\alpha<\height(\dom p)}\big\}\subsetneq u_i$
for all $i<j<\xi$

$$r:= p\cup\{(t_i, u_i):i<\xi\},$$

then $r\in\QQ$,
$r\leq  p$ and
$r\forces (t_i\subseteq \vartheta_i\wedge u_i\subseteq \sigma_i)$
for all $i<\xi$.
[In fact, $p\forces t_i\subseteq \vartheta_i$ since $t_i\subseteq(\vartheta_i)_{(p)}$.]
\endenumerate-(1)

For <ref>,
note that
and in particular, since
$\height(\dom p)\subseteq \dom\big({(\vartheta_i)}_{(p)}\big)$
by definition.

The latter also implies $t_i\subseteq{(\vartheta_i)}_{[p]}$ by Lemma <ref>.
It thus suffices that $t_i\restr\gamma\in\dom(p)$ for all $\gamma<\height(\dom p)$, but this follows from $p\in\QQ^*(\vartheta_i,\sigma_i)$ by definition.

For <ref>, we first show $r\in\QQ$.
Since $\xi<\mu$ and
$t_i\neq t_j$ for all $i<j<\xi$,
$r$ is a partial function from ${}^{<\mu}\ddimq$ to ${}^{<\mu}\mu$
with $|\dom(r)|<\mu$.
$\dom(r)$ is a subtree of ${}^{<\mu}\ddimq$
by <ref>.
Moreover, $r$ is a strict partial $s$-function by Lemma <ref> and
our assumption about the values $u_i=r(t_i)$.
$r\in\QQ$, as required.

It is clear that $r\leq  p$.
If $i<\xi$, then
$r$ forces $t_i\subseteq\vartheta_i$
since $t_i\subseteq(\vartheta_i)_{(p)}\subseteq(\vartheta_i)_{(r)}$.
$r\forces\dot h(t_i)\subseteq [\dot h](\vartheta_i)=\sigma_i$.
Since $r\forces \dot h\mleq r$ by Corollary <ref> <ref>,
$r$ also forces that $u_i\subseteq\dot h(t_i)$. Therefore
$r\forces u_i\subseteq\sigma_i$.

The next lemma states that for $p\in \QQ^*$, $\sigma_{(p)}$ can be calculated from $p$ and $\vartheta$.

If $p\in\QQ^*:=\QQ^*(\vartheta,\sigma)$
$\vartheta$ and $\sigma$ are $\QQ$-names such that
$\one_\QQ\forces \vartheta\in{}^{\mu}\ddimq\wedge\sigma=[\dot h](\vartheta)$,

the second statement is used in Lemma <ref>

\[
\sigma_{(p)}=\bigcup\big\{p(\vartheta_{(p)}\restr\alpha): \alpha<\height(\dom p)\big\}.
\]

In particular, $\dom(\sigma_{(p)})=\lh(\ran p)$.

Let $p\in\QQ^*$,

and let
$v:= \bigcup\big\{p(\vartheta_{(p)}\restr\alpha): \alpha<\height(\dom p)\big\}.$

We first show $v=\sigma_{(p)}$.
$p\forces\dot h\mleq p$
by Lemma <ref> and
$p\forces\vartheta_{(p)}\subseteq \vartheta$,

\[p\forces p(\vartheta_{(p)}\restr\alpha)\subseteq\dot h(\vartheta_{(p)}\restr\alpha)\subseteq [\dot h](\vartheta)=\sigma.\]

for all $\alpha<\height(\dom p)$.
$p\forces v\subseteq\sigma$, so $v\subseteq\sigma_{(p)}$.
Suppose $v\subsetneq \sigma_{(p)}$.
Take $u\in{}^{<\mu}\mu$ with $v\subsetneq u$
and  $u\perp \sigma_{(p)}$.
$t:=(\vartheta_{(p)})\restr\height(\dom p)$
and $r:=p\cup\{(t,u)\}.$
Then $r\in\QQ$ and
$r\forces u\subseteq \sigma$
by Lemma <ref>.
However, $r\leq  p$ and therefore $r\forces\sigma_{(p)}\subseteq\sigma$, a contradiction.

To show the second part of the conclusion,
observe that
$\sigma_{(p)}=v$  implies

\[
\dom(\sigma_{(p)})=\lh(v)
=\bigcup\{\lh(p(\vartheta_{(p)}\restr\alpha):\:\alpha<\height(\dom p)\}
\subseteq\lh(\ran p).
\]

$\lh(\ran p)\subseteq\dom(\sigma_{(p)})$
follows from our assumption that $p\in\QQ^*$.

The next lemma will be used to prove the ${<}\mu$-closure of the quotient forcings for elements of $\ran([h_K])$.

Suppose that
$\vartheta$ and $\sigma$ are $\QQ$-names such that
$\one_\QQ\forces \vartheta\in{}^{\mu}\ddimq\wedge\sigma=[\dot h](\vartheta)$.
$\langle p_\alpha:\alpha<\xi\rangle$ is a strictly decreasing chain of elements of $\QQ^*$ of length $\xi<\mu$ and

\[
\]

then $p\in\QQ^*$ and $\sigma_{(p)}=\bigcup_{\alpha<\xi}\sigma_{(p_\alpha)}$.

We first show that $p\in\QQ^*$.
Since $|\dom(p)|<\mu$,
we have $p\in\QQ$ by Lemma <ref>.
The fact that $\height(\dom p)$ and $\lh(\ran p)$ are limit ordinals follows
from Lemma <ref> in the case that $\xi\in\Lim$,
from the fact that $p_\alpha\in\QQ^*$ in the case that $\xi=\alpha+1$.
we have to use $\lh(\ran p)$ instead of $\height(\ran p)$ in the def of $\QQ^{*}$ for this step to work

\begin{equation*}
\begin{gathered}
\height(\dom p)=\bigcup_{\alpha<\xi}\height(\dom p_\alpha)
\subseteq
\bigcup_{\alpha<\xi}\dom(\vartheta_{(p_\alpha)})
\subseteq \dom(\vartheta_{(p)}),
\\
\lh(\ran p)=\bigcup_{\alpha<\xi}\lh(\ran p_\alpha)
\subseteq
\bigcup_{\alpha<\xi}\dom(\sigma_{(p_\alpha)})
\subseteq \dom(\sigma_{(p)})
\end{gathered}
\end{equation*}

Lemma <ref>
and since
$p_\alpha\in\QQ^*$ and $p\leq  p_\alpha$ for all $\alpha<\xi$.
Thus, Definition <ref> <ref> and <ref> hold.
To show
<ref>, let $\gamma<\height(\dom p)$.
Then $\gamma<\height(\dom p_\alpha)$ for some $\alpha<\xi$.
we have
$\vartheta_{(p)}\restr\gamma=\vartheta_{(p_\alpha)}\restr\gamma\in \dom(p_\alpha)\subseteq\dom(p).$

To show the second part of the conclusion, let $v:=\bigcup_{\alpha<\xi}\sigma_{(p_\alpha)}$.
Then $v\subseteq\sigma_{(p)}$.

\[
\dom(\sigma_{(p)})=\lh(\ran p)=\bigcup_{\alpha<\xi}\lh(\ran p_\alpha)=
\bigcup_{\alpha<\xi}\dom(\sigma_{(p_\alpha)})=\dom v
\]

by the previous lemma

and Lemma <ref>.
Hence $\sigma_{(p)}=v$.

The next lemma proves the first part of the Nice Quotient Lemma <ref> for $\Add(\mu,1)$.

Suppose $K$ is $\QQ$-generic over $M$ and
$\langle x_i:i<\xi\rangle$ is a sequence of distinct
elements of $({}^{\mu}\ddimq)^{M[K]}$ of length $\xi<\mu$.

is $\Add(\mu,\xi)$-generic over $M$.[$[h_K](x_i)$ is identified with the filter of its initial segments in $\Add(\mu,1)$.]

Let $i<\xi$.
By basic facts about forcing,
there exist $\QQ$-names $\vartheta_i$ and $\sigma_i$  such that
$\one_\QQ\forces\vartheta_i\in{}^\mu\ddimq\wedge [\dot{h}](\vartheta_i)=\sigma_i$.
In more detail, let $\vartheta'_i$ be a $\QQ$-name such that $(\vartheta'_i)^K=x_i$ and let

\begin{align*}
\vartheta_i&:=\big\{\big((\alpha,\beta),q\big):\:
,\ \,
\big\},
\\
\sigma_i&:=\big\{\big((\alpha,\beta),q\big):\:
,\ \,
q\forces(\alpha,\beta)\in[\dot h](\vartheta_i)
\big\}.
\end{align*}

Since $x_i\neq x_j$ for all $i<j<\xi$, there exists an ordinal $\alpha<\mu$ and a condition $q\in K$ such that
$q\forces\vartheta_i\restr\alpha\neq \vartheta_j\restr\alpha$
for all $i<j<\xi$.
We may also assume $\height(\dom q)\geq\alpha$, since
$\{p\in\QQ:\height(\dom p)\geq\alpha\}$ is dense in $\QQ$ by Lemma <ref>.

If $D$ is a dense subset of $\Add(\mu,\xi)$, then

D':=\big\{r\in\QQ\::\: {\exists u\in D}\ \ {\forall i<\xi} \ \ {r\forces u(i)\subseteq\sigma_i}\big\}
%D':=\big\{r\in\QQ\::\: {\exists \langle u_i:i<\xi\rangle\in D}\ \ {\forall i<\xi} \ \ {r\forces u_i\subseteq\sigma_i}\big\}

is dense below $q$.

Suppose $q'\in\QQ$ and $q'\leq  q$.
Take $p\leq  q'$ with $p\in\bigcap_{i<\xi}\QQ^*(\vartheta_i,\sigma_i)$ by Lemma <ref>.
For all $i<\xi$, choose $t_i\in{}^{<\mu}\ddimq$ and $u_i\in{}^{<\mu}\mu$ as in Lemma <ref>. That is, let
$t_i:=(\vartheta_i)_{(p)}\restr\height(\dom p)$ and
choose $u_i$ so that
$u_i\supsetneq \bigcup\big\{p(t_i\restr\alpha)\::\:{\alpha<\height(\dom p)}\big\}.$
In addition, we may assume $\langle u_i : i<\xi\rangle\in D$.
Since $p\leq  q$, we have $t_i\neq t_j$ for all $i<j<\xi$.
Thus $r:=p\cup\{(t_i,u_i):i<\xi\}$ is a condition in $\QQ$ below $p$ which forces $u_i\subseteq \sigma_i$ for all $i<\xi$ by Lemma <ref>.
Hence $r\leq  q'$ and $r\in D'$ as required.

To see that $G':=\prod_{i<\xi}[h_K](x_i)$ is $\Add(\mu,\xi)$-generic,
let $D$ be a dense subset of $\Add(\mu,\xi)$.
Then $D'\cap K\neq\emptyset$
by the previous claim.
Hence there exists $u\in D$
with $u(i)\subseteq (\sigma_i)^K=[h_K](x_i)$
for all $i<\xi$, i.e., with
$u\in G'$.
This completes the proof of Lemma <ref>.

The previous lemma
shows that the Nice Quotient Lemma for $\Add(\mu,1)$ holds with the following stronger $\xi$-dimensional variant
of <ref>:
If $\langle x_i:i<\xi\rangle$ is a sequence of distinct
elements of $({}^{\mu}\ddimq)^{M[K]}$ of length $\xi<\mu$,
is $\Add(\mu,\xi)$-generic over $M$.

In the rest of this subsection, we will prove the second part of the Nice Quotient Lemma for $\Add(\mu,1)$.
We need to show that
the quotient forcing
in a $\QQ$-generic extension $M[K]$ of $M$
for an element $\sigma^K$
of $\ran([h_K])$
is equivalent to $\Add(\mu,1)$.
The precise statement is given in Lemma <ref> below.

For this purpose, it will be more convenient to work
with the Boolean completion of $\QQ$.
We therefore use the next lemma.

$\QQ$ is separative.

Suppose $p,q\in\QQ$ and $p\not\leq  q$.
We aim to find $r\in\QQ$ with $r\leq p$ and $r\perp  q$.
We may assume $p\compat q$.

First, suppose $q(t)\not\subseteq p(t)$
for some $t\in\dom(p)\cap\dom(q)$.
Then $p(t)\subsetneq q(t)$ and $t$ is a terminal node of $\dom(p)$ by Lemma <ref> <ref> and <ref>.
Take $v\in{}^{<\mu}\mu$ with $p(t)\subsetneq v$ and $v\perp  q(t)$.
Let $r$ be the element of $\QQ$
obtained from $p$ by extending the value of $t$ to $v$.
That is,
let $\dom(r):=\dom(p)$, and for all $u\in\dom(r)$, let

\[
\begin{cases}
v &\text{ if $u=t$,}
\\
p(u) &\text{ if $u\neq t$}.
\end{cases}
\]

$r< p$, and $r\perp  q$ by Lemma <ref> <ref>.

Now, suppose $q(t)\subseteq p(t)$ for all $t\in\dom(p)\cap\dom(q)$.
Then $q(t)=p(t)$
for all $t\in \dom(p)\cap \dom(q)$ such that $t$ is non-terminal in $\dom(q)$
by Lemma <ref> <ref>.
Thus, since $p\not\leq  q$, we must have $\dom(q)\not\subseteq\dom(p)$.
Take $t\in\dom(q)$ of minimal length with $t\notin\dom(p)$, and let

\[
u:= \bigcup_{\alpha< \lh(t)} s\big(p(t\restr\alpha), t(\alpha) \big)=
\bigcup_{\alpha< \lh(t)} s\big(q(t\restr\alpha), t(\alpha) \big).
\]

The second equality holds since
for all $\alpha<\lh(t)$,
$t\restr\alpha$ is a non-terminal node of $\dom(q)$ with $t\restr\alpha\in\dom(p)$,
and hence $p(t\restr\alpha)=q(t\restr\alpha)$.
Then $u\subsetneq q(t)$ holds by Lemma <ref> since $q\in \QQ$.

Take $v\in{}^{<\mu}\mu$ with $u\subsetneq v$ and $v\perp q(t)$,
and let $r:=p\cup\{(t,v)\}$.
Then $r\in\QQ$ by Lemma <ref>, and clearly $r< p$.
We further have  $r\perp  q$ by Lemma <ref> <ref>.

that $\BB(\QQ)$ denotes the Boolean completion of $\QQ$. We may assume that
$\QQ$ is a dense subset of $\BB(\QQ)$, since $\QQ$ is separative.
Write $\BB:=\BB(\QQ)$.
Let $\bigwedgepo\BB$ and $\bigveepo\BB$ denote the infimum and supremum in $\BB$.
[Recall that $\bigwedgepo\BB K$ agrees with $\mbigcap K$ for downward directed subsets $K$ of $\QQ$ of size $\lle\mu$ by Remark <ref>.]
$\boolval \varphi:=\boolvalpo \varphi \BB$ for all formulas $\varphi$ in the forcing language.
Assume $\vartheta$ and $\sigma$ denote $\BB$-names such that
$\boolval{\vartheta\in{}^\mu\ddimq \wedge [\dot{h}](\vartheta)=\sigma} =\one_\BB$, and let $\QQ^*:=\QQ^*(\vartheta,\sigma)$.
Note that $\QQ^*$ is a dense subforcing of $\BB$ by Lemma <ref>.

For all $u,v\in{}^{<\mu}\mu$:
\beginenumerate-(1)

If $u\subseteq v$, then $\boolval{u\subseteq\sigma}\geq\boolval{v\subseteq\sigma}$.

If $\boolval{u\subseteq\sigma}>\boolval{v\subseteq\sigma}$, then $u\subsetneq v$.
\endenumerate-(1)

<ref> is immediate.
<ref>, suppose $\boolval{u\subseteq\sigma}>\boolval{v\subseteq\sigma}$.
Then $u\compat v$ since otherwise $\boolval{u\subseteq\sigma}\perp\boolval{v\subseteq\sigma}$.
If $v\subseteq u$, then $\boolval{v\subseteq\sigma}\leq\boolval{u\subseteq\sigma}$ by <ref>.
Therefore $u\subsetneq v$.

Recall from Definition <ref>
$\BB(\sigma)=\BB^\QQ(\sigma)$ denotes the complete Boolean subalgebra of $\BB$ that is completely generated by $\{\boolval{(\alpha,\beta)\in\sigma}:\:\alpha,\beta<\mu\}$.

Note that $\sigma_{(p)}\in{}^{<\mu}\mu$ for all $p\in\QQ^*$
by Lemma <ref>.
Define the map $\mmap:\QQ^*\to\BB(\sigma)$ by letting

\mmap(p):=\boolval{\sigma_{(p)}\subseteq\sigma}%

for all $p\in \QQ^*$.
[$\mmap(p)\in\BB(\sigma)$ since $\mmap(p)=\bigwedgepo\BB\big\{\boolval{(\alpha,\beta)\in\sigma}:(\alpha,\beta)\in\sigma_{(p)}\big\}$
and $\sigma_{(p)}\in{}^{<\mu}\mu$.]
Let $\RR^*:=\ran(\mmap)=\{\mmap(q):q\in\QQ^*\}$.

For all $p,q\in\QQ^*$:
\beginenumerate-(1)

*  $q\leq \mmap(q)$.

*  If $p\leq q$, then $\mmap(p)\leq\mmap(q)$.
\endenumerate-(1)

<ref> holds
since $q\forces \sigma_{(q)}\subseteq \sigma$.
For <ref>, suppose $p\leq q$. Then $\sigma_{(p)}\supseteq \sigma_{(q)}$ and hence $\mmap(p)\leq\mmap(q)$ by
the previous lemma.

\beginenumerate-(1)

If $q\in\QQ^*$, then $\mmap(q)$ is a reduction[See Definition <ref>.] of $q$ to $\RR^*$ with respect to the inclusion $\id_{\RR^*}: \RR^*\to \BB$.
pagebreak inserted

$\RR^*$ is a complete subforcing of $\BB$.
\endenumerate-(1)

For <ref>, suppose $r\in\RR^*$ and $r\leq \mmap(q)$.
We aim to show $r\compat q$.
We may assume $r< \mmap(q)$ by the previous lemma.
Take $p\in\QQ^*$ with $r=\mmap(p)$.

\[
\boolval{\sigma_{(p)}\subseteq\sigma}=\mmap(p)<\mmap(q)=\boolval{\sigma_{(q)}\subseteq\sigma}.
\]

Therefore $\sigma_{(p)}\supsetneq \sigma_{(q)}$ by Lemma <ref>.

Let $t:=\vartheta_{(q)}\restr\height(\dom q)$ and take $u\in{}^{<\mu}\mu$ with  $u\supseteq\sigma_{(p)}$.

\[u\supsetneq \sigma_{(q)}=\bigcup\big\{q(\vartheta_{(q)}\restr\alpha): \alpha<\height(\dom q)\big\}
\]

by Lemma <ref>.
Thus $q$, $t$ and $u$ satisfy the requirements of Lemma <ref>, and  hence
$\bar q:=q\cup\{(t,u)\}\in\QQ$, $\bar q\leq q$ and
$\bar q\forces u\subseteq \sigma$.
Since $u\supseteq \sigma_{(p)}$, we have

\[
\bar q\leq\boolval{u\subseteq\sigma}\leq
\boolval{\sigma_{(p)}\subseteq\sigma}=\mmap(p)=r.
\]

This shows that $q$ and $r$ are compatible in $\QQ$ and hence in $\BB$.

For <ref>, it suffices to show that every $b\in\BB$ has a reduction to $\RR^*$.
Given $b\in \BB$, take $q\in\QQ^*$ with $q\leq b$.
Since $\mmap(q)$ is a reduction of $q$ to $\RR^{*}$,
$\mmap(q)$ is also a reduction of $b$ to $\RR^{*}$.

Let $\BB(\RR^*)$ be the Boolean completion of $\RR^{*}$.
Since $\RR^*$ is a complete subforcing of $\BB$
and therefore separative,
we may assume
that $\RR^{*}$ is dense in $\BB(\RR^{*})$
and that $\BB(\RR^*)$ is a complete Boolean subalgebra of $\BB$.
[See Fact <ref> <ref>.]
It is easy to see
that $\BB(\RR^*)$ is then equal to the complete Boolean subalgebra of $\BB$ that is completely generated by $\RR^*$.

since $\RR^*\subseteq\BB(\sigma)$ by definition.

For $\BB(\sigma)\subseteq\BB(\RR^*)$,
it suffices to show
$\boolval{(\alpha,\beta)\in\sigma}\in\BB(\RR^*)$ for all $\alpha,\beta\in\mu$.
Suppose $\alpha,\beta\in\mu$ and let

Then $a\in\BB(\RR^{*})$.
We claim that $\boolval{(\alpha,\beta)\in\sigma}=a$.

Since $\QQ^*$ is dense in $\BB$ and $q\leq \pi^*(q)$ for all $q\in\QQ^*$,

\begin{align*}
\boolval{(\alpha,\beta)\in\sigma}
&=\bigveepo\BB\{q\in\QQ^*:\: q\leq\boolval{(\alpha,\beta)\in\sigma}\} \\
&=\bigveepo\BB\{q\in\QQ^*:\: (\alpha,\beta)\in\sigma_{(q)}\}  \\
& \leq  \bigveepo\BB\{\mmap(q):\: q\in\QQ^*,\, (\alpha,\beta)\in\sigma_{(q)}\}=a.
\end{align*}

On the other hand, if $q\in\QQ^*$ and $(\alpha,\beta)\in\sigma_{(q)}$, then
$a\leq \boolval{(\alpha,\beta)\in\sigma}$.

Let $\pi:\BB\to \BB(\sigma)$ be the retraction associated to the inclusion map $\id_{\BB(\sigma)}:\BB(\sigma)\to\BB$. That is, let
\pi(b):=\bigwedgepo{\BB}\{a\in\BB(\sigma) : a\geq b\}
for all $b\in\BB$.
[See Definition <ref>.]

for all $p\in\QQ^*$.
Equivalently: for all $p\in\QQ^*$, $\boolval{p\in\BB/\BB(\sigma)} = \boolval{\sigma_{(b)}\subseteq\sigma.}$

Let $p\in\QQ^*$.
Then $\pi(p)\leq\mmap(p)$
since $\mmap(p)\in\BB(\RR^*)=\BB(\sigma)$
and $p\leq\mmap(p)$.
Moreover, $\mmap(p)\leq\pi(p)$, since
$\mmap(p)$ is a reduction of $p$ to $\RR^{*}$
I think we use that $\RR^*$ is dense in $\BB(\RR^*)$ here
and hence also to $\BB(\RR^*)=\BB(\sigma)$
$\pi(p)$ is the largest reduction of $p$ to $\BB(\sigma)$ by Lemma <ref>.

The next lemma proves the second part of the Nice Quotient Lemma for $\Add(\mu,1)$.

${\BB(\sigma)}$ forces that the quotient forcing
$\BB/\BB(\sigma)$ is equivalent to $\Add(\mu,1)$.

Suppose $J$ is a $\BB(\sigma)$-generic filter over $M$.
It suffices to show that in $M[J]$,
the forcing $\BB/J$ is non-atomic and
has a dense subforcing of size ${\leq}\mu$ that is $\lle\mu$-closed,
by Lemma <ref>.
We use the standard facts about quotient forcings in Fact <ref> <ref>.

To see that $\BB/J$ is non-atomic in $M[J]$, it suffices to show that  $M[J][K]\neq M[J]$ for all $\BB/J$-generic filters $K$ over $M[J]$.
Let $K$ be $\BB/J$-generic over $M[J]$.
Then $K$ is also $\BB$-generic over $M$, $M[J][K]=M[K]$ and  $J=K\cap\BB(\sigma)$.
Then $M\big[\sigma^K\big]=M[J]$ by
Lemma <ref>.

Take $y,z\in ({}^\mu\ddimq)^{M[K]}$ with
%=[\dot h^K](y)
[h_{K\cap \QQ}](y)$ and $y\neq z$.
[Recall the notation $h_{K\cap \QQ}$ from page h_K def after Lemma <ref>.]
Then $[h_{K\cap \QQ}](z)$
is $\Add(\mu,1)$-generic over $M\big[\sigma^K\big]$
by Lemma <ref>.
Since $[h_{K\cap \QQ}](z)\in M[K]$,
we have $M[K]\neq M\big[\sigma^K\big]$ as required.

In $M[J]$,
$\PP^*:=\QQ^{*}\cap (\BB/J).$
Note that $\PP^*$ is a dense subforcing of $\BB/J$
since $\QQ^*$ is a dense subforcing of $\BB$,
and $|\PP^*|\leq|\QQ^*|\leq|\QQ|\leq\mu$.

To prove that
$\PP^*$ is $\lle\mu$-closed,
let $K$ be $\BB/J$-generic over $M[J]$.
Then $K$ is $\BB$-generic over $M$
and $J=K\cap\BB(\sigma)$.
It suffices
to show that

since the right hand side is $\lle\mu$-closed
by Lemma <ref>.
Let $q\in\QQ^*$.
\Longleftrightarrow
\pi(q)\in J
\Longleftrightarrow
\pi(q)\in K$
by Lemma <ref> <ref>.
Since $\pi(q)=\mmap(q)=\boolval{\sigma_{(q)}\subseteq\sigma}$
by Lemma <ref>,
$\pi(q)\in K
\Longleftrightarrow
\sigma_{(q)}\subseteq\sigma^K$, as required.

If $K$ is a $\BB$-generic filter over $M$,
and $\sigma$ is a $\BB$-name for $x$ with $\boolval{\sigma\in\ran([\dot h])}=\one_\BB$
[Recall the notation $\dot h$ from page dot h def.
Note that all $x\in\ran([h_{K\cap\QQ}])$ are of this form, as in the proof of Lemma <ref>.
then $M[K]$ is an $\Add(\kappa,1)$-generic extension of
by the previous lemma.
This completes the proof of the Nice Quotient Lemma <ref> for $\Add(\mu,1)$.

§.§ A global version

Recall that a class forcing extension is called tame if it is a model of $\ZFC$.

Assume there exists a proper class of inaccessible cardinals.
Then there is a tame class forcing extension of $V$
in which
holds for all regular infinite cardinals $\kappa$ and all $\ddim$ with $2\leq\ddim\leq\kappa$.
[Note that this implies $\ODD\kappa\ddim(\defsetsk)$ for all $\ddim$ with $2\leq\ddim<\kappa$ by Lemma <ref>.]

Let $\bar{\kappa}:=\langle \kappa_{\alpha}: \alpha\in \Ord\rangle$ denote the order preserving enumeration of the closure $C$ of the class of inaccessible cardinals and $\omega$.
We define the following Easton support iteration $\bar{\PP}=\langle \PP_{\alpha},\dot{\PP}_{\alpha}: \alpha\in \Ord\rangle$ with bounded support at regular limits and full support at singular limits.
Let $\dot{\nu}_\alpha$ be a $\PP_\alpha$-name for the least regular cardinal $\nu\geq \kappa_\alpha$
that is not collapsed by $\PP_\alpha$, and let $\dot{\PP}_{\alpha}$ be a $\PP_{\alpha}$-name for $\Col(\dot{\nu}_\alpha,{{<}\kappa_{\alpha+1}})$.
(We may assume that the names $\dot{\PP}_\alpha$ are chosen in a canonical way so that the iteration is definable, for instance by working with nice names for elements of the forcing.)
Let $\PP$ denote the class forcing defined by $\bar{\PP}$.
Moreover, let $\dot{\PP}^{(\alpha)}$ denote a $\PP_\alpha$-name for the tail forcing of the iteration at stage $\alpha$.
It follows from the definition of the iteration that $|\PP_\alpha|<\kappa_{\alpha+1}$ and $\one_{\PP_\alpha} \forces \dot{\PP}^{(\alpha)}$ is ${<}\kappa_\alpha$-closed, for all $\alpha\in\Ord$.

Now suppose that $G$ is $\PP$-generic over $V$.
We write $G_\alpha:=G\cap \PP_\alpha$ and $\PP^{(\alpha)}:=(\dot{\PP}^{(\alpha)})^{G_\alpha}$ for $\alpha\in\Ord$. Moreover, let $\nu_\alpha:=\dot{\nu}^{G_\alpha}$ for $\alpha\geq 1$ and write $\bar{\nu}:=\langle \nu_\alpha: \alpha\geq 1\rangle$.

\beginenumerate-(a)

If $\kappa_\alpha$ is inaccessible in $V$, then $\kappa_\alpha$ remains a regular cardinal in $V[G]$, $\nu_\alpha=\kappa_\alpha$ and ${(\kappa_\alpha^+)}^{V[G]}=\kappa_{\alpha+1}$.

* If $\kappa_\alpha$ is a singular limit in $V$, then

\endenumerate-(a)

If $\kappa_\alpha$ is inaccessible in $V$, it follows from the $\Delta$-system lemma that $\PP_\alpha$ has the $\kappa_\alpha$-c.c.
Dorottya's note to self: See the proof of Theorem 10.17 in Kanamori's book. This also follows from <cit.>*Theorems 16.4 and 16.30.
The claims in <ref> easily follow from this and the fact that $\PP^{(\alpha+1)}$ is ${<}\kappa_{\alpha+1}$-closed.
If $\kappa_\alpha$ is a singular limit in $V$, then
$\nu_\alpha$ is the least regular cardinal strictly above $\kappa_\alpha$ in $V[G_\alpha]$ by the definition of $\dot{\nu}_\alpha$.
Moreover, $\nu_\alpha$ is not collapsed in $V[G]$, since $\PP^{(\alpha+1)}$ is ${<}\kappa_{\alpha+1}$-closed.

By the previous claim, $\bar{\nu}$ enumerates the class of regular infinite cardinals in $V[G]$.
Thus, suppose that $\alpha\geq 1$, $\kappa=\nu_\alpha$ and $2\leq \ddim\leq\kappa$.
In $V[G]$,
let $X\in\defsetsk$ be a subset of $\kk$, and
let $H\in\defsetsk$ be a box-open $\ddim$-dihypergraph on $\kk$.

$\ODD\kappa{H\restr X}$ holds in $V[G]$.

Since $\one_{\PP_{\beta}}$
forces that
$\dot{\PP}_{\beta}$ is homogeneous for all $\beta\in\Ord$, the tail forcing $\PP^{(\beta)}$ is homogeneous for all $\beta\in\Ord$.

We first claim that $X,H\in V[G_{\alpha+1}]$.
Note that $\PP^{(\alpha+1)}$ does not add new elements of ${}^{\kappa}\Ord$ since $\PP^{(\alpha+1)}$ is ${<}\kappa_{\alpha+1}$-closed.
Suppose that $X=X_{\varphi,a}^{V[G]}$ where $\varphi$ is a formula and $a\in {}^\kappa\Ord$.
Since  $\PP^{(\alpha+1)}$ is homogeneous, $X= X_{\psi,a}^{V[G_{\alpha+1}]}$ where $\psi(x,a)$ denotes the formula $\one_{\PP^{(\alpha+1)}}\forces^{V[G_{\alpha+1}]}\varphi(x,a)$.
Hence $X\in V[G_{\alpha+1}]$, and similarly $H\in V[G_{\alpha+1}]$.

Since $\kappa=\nu_\alpha$, $\dot{\PP}_\alpha$ is a name for $\Col(\kappa,\lle\kappa_{\alpha+1})$ and hence $\ODD\kappa{H\restr X}$ holds in $V[G_{\alpha+1}]$
by Theorem <ref> <ref>.
Since $\PP^{(\alpha+1)}$ is $\lle\kappa_{\alpha+1}$-closed, a $\kappa$-coloring of $H\restr X$ in $V[G_{\alpha+1}]$ remains a $\kappa$-coloring in $V[G]$, and a continuous homomorphism from $\dhH\ddim$ to $H\restr X$ in $V[G_{\alpha+1}]$ remains one in $V[G]$.
Hence $\ODD\kappa{H\restr X}$ holds in $V[G]$.

holds in $V[G]$ by the previous claim.

It is open whether a global version of Theorem <ref> <ref> holds in analogy with the previous theorem.
The problem is that in an iterated forcing extension as above, new box-open dihypergraphs are added by the tail forcings.

In the model of the previous theorem, we actually have $\ODDI\kappa\ddim(\defsets\kappa,\defsets\kappa)$
for all regular uncountable $\kappa$ and $d\leq\kappa$
by Remark <ref>.
[For $\ddim<\kappa$, we have $\ODDI\kappa\ddim(\defsets\kappa)$.
In fact, the continuous homomorphism can even be chosen to be a homeomorphism onto a closed set in this case by
Definition <ref> and Corollary <ref> below.]

§ VARIANTS

We begin with some preliminary results
regarding conditions under which
the map $[\iota]$ is closed and the tree $T(\iota)$ is sufficiently closed for a given strict order preserving map $\iota$ in Subsection <ref>.

We show in Subsection <ref>
that the open graph dichotomy
is equivalent to $\ODD\kappa 2(X)$.
In fact, we prove a more general version of this equivalence for an extension of $\OGD_\kappa(X)$ to $\ddim$-hypergraphs of higher dimensions $2\leq\ddim\leq\kappa$.

In the remainder of the section,
we show that the definition of $\ODD\kappa H$ is optimal in several ways.
$\dhH\ddim$ witnesses that for any $\ddim\geq 3$, the existence of a continuous homomorphism cannot be replaced by the existence of a large complete subhypergraph.
Stronger counterexamples are discussed
in Subsection <ref>.
One may further wonder whether
the continuous homomorphism provided by $\ODD\kappa H$ can be chosen
to have additional properties such as injectivity
for relatively box-open[I.e., $H$ is box-open on its domain $\domh H$.]
$\ddim$-dihypergraphs $H$ on ${}^\kappa\kappa$.
We show that this cannot be done in general for $\ODD\omega H$.
However, in the uncountable case
a weak version $\Diamondi\kappa$ of $\Diamond_\kappa$ allows us to choose injective continuous homomorphisms.
For $\ddim<\kappa$,
the continuous homomorphism in $\ODD\kappa H$ can even
be chosen to be a homeomorphism onto a closed subset
without assuming $\Diamondi\kappa$.
For $\ddim=\kappa$ one cannot in general find a homomorphism that is
a homeomorphism onto its image.
These results are proved in Subsection <ref>.
Subsection <ref> studies the principle $\Diamondi\kappa$ in preparation for
Subsection <ref>.

§.§ Closure properties of maps

Let $Y$ and $Z$ be topological spaces.
Recall that a map $f:Y\to Z$ is called closed if $f(X)$ is a closed subset of $Z$ for any closed subset $X$ of $Y$.
It is easy to see that $f$ is closed if and only if $\closure{f(X)}=f(\closure{X})$ for all subsets $X$ of $Y$.

For any map
$f: {}^{\kappa}\kappa\to {}^{\kappa}\kappa$,
$X\subseteq{}^\kappa\kappa$ and $t\in{}^{<\kappa}\kappa$,
let $\Lim_t^{f}(X)$ denote the union of all
limit points of sets of the form
$\{f(x_\alpha) : \alpha<\kappa\}$
$x_\alpha\in N_{t\conc\langle\alpha\rangle}\cap X$
for all $\alpha<\kappa$,
and let

For strict order preserving maps $\iota: {}^{<\kappa}\kappa\to {}^{<\kappa}\kappa$,
we also write
$\Lim_t^{\iota}(X):=\Lim_t^{[\iota]}(X)$ and

Suppose that
$\iota:{}^{<\kappa}\ddim\to{}^{<\kappa}\kappa$ is $\perp$- and strict order preserving.
\beginenumerate-(1)

If $2\leq\ddim<\kappa$, then

$[\iota]$ is a closed map.

If $\ddim=\kappa$,

$$\closure{[\iota](X)}=[\iota](\closure{X})\cup\bigcup_{t\in T(X)} \Lim^\iota_t(X).$$

for all subsets $X$ of ${}^\kappa\kappa$.
In particular,
$[\iota]$ is a closed map if
$\Lim^\iota_t=\emptyset$ for all $t\in{}^{<\kappa}\kappa$.
\endenumerate-(1)

We prove <ref> and <ref> simultaneously.
Fix $2\leq\ddim\leq\kappa$ and a subset $X$ of ${}^\kappa\ddim$.
Note that it suffices to show  $[\iota](\closure{X})=\closure{[\iota](X)}$ for <ref>.
It is clear that $[\iota](\closure{X})\subseteq\closure{[\iota](X)}$ and that
$\Lim^\iota_t(X)\subseteq\closure{[\iota](X)}$ for all $t\in{}^{<\kappa}\kappa$ in the case $\ddim=\kappa$.

Now suppose that $y\in\closure{[\iota](X)}$.
It suffices to show $y \in [\iota](\closure{X})\cup\bigcup_{t\in T(X)} \Lim^\iota_t(X)$.
$y$ is the limit of some convergent sequence
$\langle [\iota](x_\alpha):\alpha<\kappa\rangle$ with $x_\alpha\in X$ for each $\alpha<\kappa$.
If $\bar x:=\langle x_\alpha:\alpha<\kappa\rangle$ converges to some $x$, then $x\in \closure X$ and $y\in[\iota](\closure X)$.
Therefore we now assume that $\bar{x}$ does not converge.
We will see that $\ddim=\kappa$ and $y\in\Lim^\iota_t(X)$ for some $t\in{}^{<\kappa}\kappa$.

Let $\gamma\leq \kappa$ be least such that
$\langle x_\alpha\restr\gamma : \alpha<\kappa\rangle$ is not eventually constant.
Then clearly $\gamma>0$.
We first claim that $\gamma$ is a successor ordinal.
Fix some $\delta_\beta<\kappa$ such that $\langle x_\alpha\restr\beta : \delta_\beta\leq\alpha<\kappa\rangle$ is constant for each $\beta<\gamma$ and let $\delta:=\sup_{\beta<\gamma} \delta_\beta<\kappa$.
If $\gamma$ were a limit ordinal, then $\langle x_\alpha\restr\gamma : \delta\leq\alpha<\kappa\rangle$ would be constant.

Let $\gamma=\beta+1$.
We may assume that
$\langle x_\alpha\restr\beta:\alpha<\kappa\rangle$ has a constant value $t\in{}^{<\kappa}\kappa$ by replacing $\bar x$ by a tail.
Then $x_\alpha\restr\gamma=t\conc\langle x_\alpha(\beta)\rangle$ for all $\alpha<\kappa$.

Next, we claim that
$\langle x_\alpha(\beta):\alpha<\kappa\rangle$
is unbounded in $\kappa$; in particular $\ddim=\kappa$.
Towards a contradiction, suppose that there is a strict upper bound $\delta<\kappa$ with
$x_\alpha(\beta)<\delta$ for all $\alpha<\kappa$.
$\langle x_\alpha(\beta):\alpha<\kappa\rangle$
is not eventually constant,
there exist $i<j<\kappa$ and disjoint unbounded subsets $I,\,J$ of $\kappa$ with
$x_\alpha(\beta)=i$ for all $\alpha\in I$
$x_\alpha(\beta)=j$ for all $\alpha\in J$.
$[\iota](x_\alpha)\supseteq \iota(t\conc\langle i\rangle)$
for all $\alpha\in I$ and
$[\iota](x_\alpha)\supseteq \iota(t\conc\langle j\rangle)$
for all $\alpha\in J$.
Since $\iota$ is $\perp$-preserving,
$\iota(t\conc\langle i\rangle)\perp\iota(t\conc\langle j\rangle)$.
But then $\langle[\iota](x_\alpha):\alpha<\kappa\rangle$ cannot be convergent.

To show that  $y\in\Lim^\iota_t(X)$,
we may assume that $\langle x_\alpha(\beta):\alpha<\kappa\rangle$ is injective by replacing $\bar x$ with a subsequence.
Let $\bar z= \langle z_\alpha : \alpha<\kappa \rangle$ be a sequence with $z_\alpha\in N_{t\conc\langle \alpha\rangle}\cap X$ and
$z_\alpha=x_\delta$ if $\alpha= x_\delta(\beta)$.
Since $\iota$ is $\perp$-preserving, $[\iota]$ is injective.
Thus $y$ is a limit point of the set $\{[\iota](z_\alpha):\alpha<\kappa\}$ and hence $y\in\Lim^\iota_t(X)$.

The next corollary follows from the previous lemma and Lemma <ref>.

$\iota:{}^{<\kappa}\ddim\to{}^{<\kappa}\kappa$ is $\perp$- and strict order preserving.
If $\ddim<\kappa$ or if $\ddim=\kappa$ and $\Lim^\iota_t=\emptyset$ for all $t\in{}^{<\kappa}\kappa$, then
$[\iota]$ is a homeomorphism from ${}^\kappa\ddim$ onto a closed subset of ${}^\kappa\kappa$.

The next lemma shows that the assumption that $\iota$ is $\perp$-preserving can be omitted from the previous lemma if $\kappa$ is weakly compact.

Suppose that $\kappa=\omega$ or $\kappa$ is weakly compact.
Let $f:{}^{\kappa}\ddim\to{}^{\kappa}\kappa$ be a continuous map.
\beginenumerate-(1)

If $2\leq\ddim<\kappa$, then
$f$ is a closed map.

If $\ddim=\kappa$,
$$\closure{f(X)}=f(\closure{X})\cup\bigcup_{t\in T(X)} \Lim^f_t(X).$$

for all subsets $X$ of ${}^\kappa\kappa$.
In particular,
$f$ is a closed map if
$\Lim^f_t=\emptyset$ for all $t\in{}^{<\kappa}\kappa$.
\endenumerate-(1)

<ref> holds since closed subsets of ${}^
\kappa\ddim$ are $\kappa$-compact,
$\kappa$-compactness is preserved under continuous images and
$\kappa$-compact subsets of ${}^\kappa\kappa$ are closed.

For <ref>,
note that clearly $f(\closure{X})\subseteq\closure{f(X)}$ and that
$\Lim^f_t(X)\subseteq\closure{f(X)}$ for all $t\in{}^{<\kappa}\kappa$.

It suffices to show $y \in f(\closure{X})\cup\bigcup_{t\in T(X)} \Lim^f_t(X)$ for any $y\in\closure{f(X)}$.
Fix a sequence
$\langle f(x_\alpha):\alpha<\kappa\rangle$ converging to $y$ with $x_\alpha\in X$ for all $\alpha<\kappa$.
We may assume $f(x_\alpha)\neq y$ for all $\alpha<\kappa$.

If $\bar x:=\langle x_\alpha:\alpha<\kappa\rangle$ is bounded, then it is contained in a $\kappa$-compact subset $C$ of $\closure X$. Then $f(C)$ is also $\kappa$-compact and thus closed, so $y\in f(C)\subseteq f(\closure X)$.
Now suppose that $\bar x$ is unbounded.
Let $\beta<\kappa$ be least such that
$\langle x_\alpha(\beta):\alpha<\kappa\rangle$ is unbounded in $\kappa$.
We may assume that $\langle x_\alpha(\beta):\alpha<\kappa\rangle$ is injective by replacing $\bar x$ with a subsequence.

There are only $|\kappa{}^\beta|$ many possibilities for $x_\alpha\restr\beta$, and $|\kappa{}^\beta|<\kappa$ since $\kappa$ is inaccessible.
We may thus assume that $\langle x_\alpha\restr\beta:\alpha<\kappa\rangle$
has a constant value $t\in{}^{<\kappa}\kappa$ by replacing $\bar x$
with a further subsequence.
Then $x_\alpha\restr{(\beta+1)}=t\conc\langle x_\alpha(\beta)\rangle$ for all $\alpha<\kappa$.

Let $\bar z= \langle z_\alpha : \alpha<\kappa \rangle$ be a sequence with $z_\alpha\in N_{t\conc\langle \alpha\rangle}\cap X$ and
$z_\alpha=x_\gamma$ if $\alpha= x_\gamma(\beta)$.
Since $y\neq f(x_\gamma)$ for all $\gamma<\kappa$,
$y$ is a limit point of the set $\{f(z_\alpha):\alpha<\kappa\}$, and thus $y\in\Lim^f_t(X)$.

The next remark shows that the following assumptions cannot be omitted: $\iota$ is $\perp$-preserving in Lemma <ref> and $\kappa$ is weakly compact in Lemma <ref>.

Suppose that $\kappa$ is not weakly compact.
\beginenumerate-(1)

There exists a strict order preserving map $\iota: {}^{<\kappa}2\to {}^{<\kappa} 2$ such that $\ran([\iota])$ is not closed, and thus
$[\iota]$ is not a closed map.
To see this, we first claim that there exists a barrier $A$ in ${}^{<\kappa}2$ of size $\kappa^{<\kappa}=\kappa$,[A similar argument also works without the assumption that $\kappa^{<\kappa}=\kappa$.]
i.e. a maximal antichain in ${}^{<\kappa}2$ of size $\kappa^{<\kappa}=\kappa$ with the following equivalent properties: (a) every $x\in {}^{\kappa}2$
has an initial segment in $A$ and (b) $T(A)$ has no $\kappa$-branches.
If $\kappa$ is inaccessible, let $A$ be the boundary of a $\kappa$-Aronszajn subtree of ${}^{<\kappa}2$.
If $\kappa=\mu^+$, let $A:={}^\mu 2$.

Fix an injective enumeration $\langle t_\alpha : \alpha<\kappa\rangle$ of $A$.
Further fix a strictly increasing sequence $\langle \gamma_\alpha : \alpha<\kappa\rangle$ with $\gamma_\alpha \geq \lh(t_\alpha)$ and let  $\iota(t_\alpha):= \langle 0\rangle^{\gamma_\alpha}\conc\langle1\rangle$ for all $\alpha<\kappa$.
We now extend $\iota$
to a strict order preserving map
from ${}^{<\kappa}2$ to ${}^{<\kappa}2$.
If $s\subsetneq t$ for some $t\in A$, let $\iota(s):=\langle 0\rangle^{\lh(s)}$.
We further
extend $\iota$ above $A$ arbitrarily
to a strict order preserving map.
Then $\langle 0\rangle^\kappa$ witnesses that $\ran([\iota])$ is not closed.

Note that we can also guarantee that $[\iota]$ is injective
by extending $\iota$ in a $\perp$-preserving way above $A$.
[I.e., so that for all $u\in A$ and all
incompatible $v,w\supsetneq u$, we have $\iota(v)\perp\iota(w)$).
E.g. let $\iota(u \conc v):=\iota(u)\conc v$ for all
$u\in A$ and $v\in{}^{<\kappa}\kappa$.]

There exists a strict order preserving map $\theta:{}^{<\kappa}\kappa\to{}^{<\kappa}\kappa$ such that $\Lim^{\theta}_t=\emptyset$ for all $t\in{}^{<\kappa}\kappa$ and $\ran([\theta])$ is not closed.
To see this, define $\theta(s\conc t):=\iota(s)\conc t$ for all $s\in{}^{<\kappa}2$ and $t\in{}^{<\kappa}\kappa$ with $t(0)\geq 2$,
where $\iota$ is as in <ref>.
Note that if $[\iota]$ is injective, then so is $[\theta]$.
\endenumerate-(1)

If $\mu$ is a regular cardinal, let $\cof_\mu$ denote the class of limit ordinals
with $\cof(\alpha)=\mu$,
for any ordinal $\gamma$,
let $\cof^\gamma_\mu:=\cof_\mu\cap \gamma$.
We define $\cof_{{\leq}\mu}$, $\cof_{<\mu}$,
$\cof^\gamma_{{>}\mu}$, $\cof^\gamma_{>\mu}$ etc. similarly
for arbitrary ordinals $\mu$.
Given a class $C$ of limit ordinals and a subtree $T$ of ${}^{<\kappa}\kappa$, we say that
$T$ is $C$-closed if every strictly increasing sequence in $T$ with length in $A$ has an upper bound in $T$.

The next lemma shows that the relevant closure properties of $T(\iota)$ are determined by the following sets.

For $\iota : {}^{<\kappa}\ddim \to {}^{<\kappa}\kappa$ and $u\in {}^{<\kappa}\ddim$, we write $T^{\iota,u}$ for the downwards closure of
$\{\iota(u\conc \langle \alpha \rangle): \alpha<\kappa \}$ in ${}^{<\kappa}\ddim$.

If $2 \leq \ddim\leq \kappa$, $\iota: {}^{<\kappa}\ddim \to {}^{<\kappa}\kappa$ is $\perp$- and strict order preserving and $\mu<\kappa$ is a regular infinite cardinal, then the following statements are equivalent:
\beginenumerate-(1)

$T(\iota)$ is $\cof^\kappa_\mu$-closed.

For all $u\in {}^{<\kappa}\ddim$, $T^{\iota,u}$ is $\cof^\kappa_\mu$-closed.
\endenumerate-(1)

<ref> $\Rightarrow$ <ref>:
Suppose that $w$ is a limit of nodes in $T^{\iota,u}$ with $w \in T(\iota) \,\setminus\, T^{\iota,u}$.
There exists some $\eta<\kappa$ with $\iota(u\conc\langle \eta\rangle)\subsetneq w$, since otherwise $w$ properly extends $\iota(u)$ and is incompatible with $\iota(t)$ for all $t\supsetneq u$, but this would entail $w\notin T(\iota)$.
Then $\iota(u\conc\langle \xi\rangle)\perp w$ for all $\xi\neq \eta$, so $w$ cannot be a limit of nodes in $T^{\iota,u}$.

<ref> $\Rightarrow$ <ref>:
Suppose that $w$ is a limit of nodes in $T(\iota)$ with $\lh(w) \in \cof^\kappa_\mu$.
We shall show that $w\in T(\iota)$.

For all $\gamma<\lh(w)$, pick some $s_\gamma$ such that $w{\restr}\gamma \subseteq \iota(s_\gamma)$ and $\lh(s_\gamma)>\lh(u)$.
$A := \{ t\in {}^{<\kappa}\ddim: \iota(t)\subseteq w\}$. Note that the elements of $A$ are pairwise comparable, since $\iota$ is $\perp$-preserving.
Hence $u:= \bigcup A$ and $v:= \bigcup \{\iota(t) : t\in A\}$ are elements of ${}^{<\kappa}\ddim$.
We may assume that $v\subsetneq w$, since otherwise $w=v\subseteq\iota(u)$.

We claim that $u\subsetneq s_\gamma$ for all $\gamma$ with $\lh(v) <\gamma< \lh(w)$.
To see this, note that for any $t\in A$,
we have $\iota(t)\subseteq v\subseteq w{\restr} \gamma \subseteq \iota(s_\gamma)$, and hence $t\subseteq s_\gamma$ since $\iota$ is $\perp$-preserving.
As this holds for all $t\in A$, we have $u\subseteq s_\gamma$.
As $\lh(s_\gamma)>\lh(u)$, we have $u\subsetneq s_\gamma$.

For each $\gamma$ as above, we can thus pick some $\eta_\gamma$ with $u\conc\langle\eta_\gamma\rangle \subseteq s_\gamma$.
It suffices to show $w{\restr}\gamma \subseteq \iota(u\conc\langle\eta_\gamma\rangle)$, since this implies $w\in T^{\iota,u}\subseteq T(\iota)$.
To see this, note that $w{\restr}\gamma \compat \iota(u\conc\langle \eta_\gamma\rangle)$, since both are initial segments of $\iota(s_\gamma)$.
But $\iota(u\conc\langle \eta_\gamma\rangle)\subseteq w{\restr}\gamma\subseteq w$ would entail $u\conc\langle \eta_\gamma\rangle \subseteq u$ by the definition of $u$.

The next corollary is immediate:

Suppose that $\iota: {}^{<\kappa}\ddim \to {}^{<\kappa}\kappa$ is $\perp$- and strict order preserving.
\beginenumerate-(1)

* If $2\leq \ddim<\kappa$, then $T(\iota)$ is $\cof^\kappa_{>\ddim}$-closed.

* If the nodes $\iota(t\conc\langle\alpha\rangle)$
split at the same node
for each $t\in {}^{<\kappa}\ddim$,
[I.e., the $\iota(t\conc\langle\alpha\rangle)$'s extend pairwise different immediate successors of the some $s_t\in{}^{<\kappa}\kappa$, for each $t\in{}^{<\kappa}\ddim$.
Equivalently, $\iota$ is an order homomorphism for the dihypergraph $\dhhd$ defined right after Definition <ref>.]
then $T(\iota)$ is ${<}\kappa$-closed.
\endenumerate-(1)

The next remark shows
the restriction to $\cof^\kappa_{>\ddim}$
is necessary in the previous corollary.

Suppose $\ddim\leq\kappa$ and
Then there exists a $\perp$- and strict order preserving function $\iota: {}^{<\kappa}\ddim \to {}^{<\kappa}\kappa$ such that  $T(\iota)$ is not $\{\eta\}$-closed.
To see this, choose a strictly increasing sequence $\langle \gamma_\alpha:\alpha<\cof(\eta)\rangle$ cofinal in $\eta$,
and let
$\iota=\iota_{\ddim,\eta}$ be any $\perp$- and strict order preserving map
from ${}^{<\kappa}\ddim$ to ${}^{<\kappa}\kappa$
such that

\[
\iota\left(\langle \alpha\rangle\right)=
\begin{cases}
\langle 0\rangle^{\gamma_\alpha}\conc\langle 1\rangle
&\text{if $\alpha<\cof(\eta)$,}
\\
\langle\alpha\rangle
&\text{if $\alpha\in[\cof(\eta),\ddim)$.}
\end{cases}
\]

Then the node
${\langle 0\rangle}^{\eta}$
witnesses that
$T(\iota)$ is not $\{\eta\}$-closed.
For $\ddim=\kappa$, it is also easy to make sure that
$\Lim^\iota_t=\emptyset$ for all $t\in{}^{<\kappa}\kappa$.
[Let, for example, $\iota(\langle \alpha \rangle \conc t)=\iota(\langle \alpha\rangle)\conc t$ for all $\alpha<\kappa$ and $t\in{}^{<\kappa}\kappa$.
Note that $\Lim^\iota_\emptyset=\emptyset$ is guaranteed by the choice of the $\iota(\langle \alpha\rangle)$'s.]

Suppose that $\ddim=\kappa$.
Then there exists a $\perp$- and strict order preserving function $\psi: {}^{<\kappa}\ddim \to {}^{<\kappa}\kappa$ such that $T(\psi)$ is not $\{\eta\}$-closed for all limit ordinals $\eta<\kappa$, and at the same time
$\Lim^\psi_t=\emptyset$ for all $t\in{}^{<\kappa}\kappa$.
To see this, let $\langle \eta_\alpha:\alpha<\kappa\rangle$ be an enumeration of
all the limit ordinals below $\kappa$.
For all $\alpha<\kappa$, let $\iota_\alpha$ be the map
defined in the previous paragraph.
$\psi(\langle\alpha\rangle\conc t):=
\langle \alpha\rangle\conc\iota_\alpha(t)$
for all $\alpha<\kappa$ and

Then $T(\psi)$ is not $\{\eta\}$-closed for any limit ordinal $\eta<\kappa$,
and we can also ensure that $\Lim^\psi_t=\emptyset$ for all $t\in{}^{<\kappa}\kappa$.

Suppose $2\leq\ddim<\omega$.
\beginenumerate-(1)

If $\iota:{}^{<\kappa}\ddim\to{}^{<\kappa}\kappa$
is $\perp$- and strict order preserving, then
$\ran([\iota])$ is a $\kappa$-perfect set.

If $X\subseteq{}^\kappa\kappa$ has a $\kappa$-perfect subset, then
there exists a $\perp$- and strict order preserving map
$\iota:{}^{<\kappa}\ddim\to{}^{<\kappa}\kappa$ with $\ran([\iota])\subseteq X$.
\endenumerate-(1)

For <ref>, let $\iota:{}^{<\kappa}\ddim\to{}^{<\kappa}\kappa$ be
$\perp$- and strict order preserving.
By Lemma <ref>,
$\ran([\iota])$ is a closed set, so
It thus suffices to show that $T(\iota)$ is a $\kappa$-perfect tree.
$T(\iota)$ is $\lle\kappa$-closed
by Corollary <ref>.
$T(\iota)$ is also cofinally splitting since $\iota$ preserves $\perp$.
For <ref>, let
$T$ be a $\kappa$-perfect tree $T$ with $[T]\subseteq X$.
It is easy to construct a $\perp$- and strict order preserving map
$\iota:{}^{<\kappa}\ddim\to T$ by recursion.
Then $\ran([\iota])\subseteq[T]\subseteq X$.

§.§ The open graph dichotomy

Throughout this subsection, we assume that $X$ is a subset of ${}^\kappa\kappa$.
We say that a graph $G$ on $X$ has a $\kappa$-perfect complete subgraph if there exists a $\kappa$-perfect subset $Y\subseteq X$ such that the complete graph $\gK Y$
is a subset of $G$.
Recall that the open graph dichotomy $\OGD_\kappa(X)$ is the following statement:

If $G$ is an open graph on $X$, then
either $G$ admits a $\kappa$-coloring or
$G$ has a $\kappa$-perfect
complete subgraph.

In this subsection, we show that
the open graph dichotomy
is equivalent to $\ODD\kappa 2(X)$.
Thus, it also follows from
for any $\ddim>2$
by Lemmas <ref> and <ref>.]
In fact, we prove a more general version of
this equivalence for an extension of $\OGD_\kappa(X)$ to $\ddim$-hypergraphs for higher dimensions $2\leq\ddim\leq\kappa$.

Recall that a $\ddim$-hypergraph is a $\ddim$-dihypergraph that is closed under permutation of hyperedges.
For any $\ddim$-dihypergraph $I$,[In this subsection only, $H$ always denotes a hypergraph and $I$ denotes a dihypergraph.]
let $\hypg I$ denote the smallest hypergraph containing $I$, i.e.,

$$\hypg I:=\bigcup
\{
\bar{x}^\pi:\:
%\langle x_{\pi(i)}:i\in\ddim\rangle:\:
{\bar{x}\in I,\ \pi\in\Sym(\ddim)}\}.

\left\{\prod_{\alpha\in \ddim}N_{t\conc\langle \pi(\alpha)\rangle}:\:
and $\hhH2$ equals the complete graph $\gK{{}^\kappa 2}$.

$\OHD\kappa\ddim(X)$ states that the following holds for all box-open $\ddim$-hypergraphs $H$ on $X$:

$\OHD\kappa H$:
Either $H$ admits a $\kappa$-coloring, or
there exists a continuous homomorphism
from $\hhH{\ddim}$ to $H$.

We first show that $\OHD\kappa 2(X)$ is equivalent to the open graph dichotomy $\OGD_\kappa(X)$.
Since $\hhH 2=\gK{{}^\kappa 2}$,

it suffices to show that
a graph $G$ on $X$ has a $\kappa$-perfect complete subgraph
if and only if there is a continuous homomorphism from $\gK{{}^\kappa 2}$ to $G$.
The next lemma is
a more general version of this for dihypergraphs.

We say a $\ddim$-dihypergraph
has a $\kappa$-perfect complete subhypergraph if there exists a $\kappa$-perfect subset $Y\subseteq X$ such that $\dhK \ddim Y\subseteq I$.

Suppose that $2\leq\ddim\leq\kappa$ and
$I$ is a $\ddim$-dihypergraph on $X$.
\beginenumerate-(1)

If $f$ is a continuous homomorphism from $\dhH\ddim$ to $I$,
then $\ran(f)$ has a $\kappa$-perfect subset.

$I$ has a $\kappa$-perfect complete subhypergraph if and only if
there exists a continuous homomorphism from $\dhK\ddim{{}^\kappa\! \ddim}$ to $I$.
\endenumerate-(1)

For <ref>, note first that
$f$ is also a homomorphism from $\dhH\ddim$ to
$\dhK\ddim {\ran(f)}$.
$\dhK\ddim {\ran(f)}$
is relatively box-open,
there exists
a continuous order homomorphism
$\dhK\ddim {\ran(f)}$
by Lemma <ref>.

For all $t\in{}^{<\kappa}\ddim$,
there exists $u,v\supsetneq t$ with $\iota(u)\perp \iota(v)$
We can not guarantee that there exists $\alpha<\beta<\ddim$ with $\iota(t\conc\langle\alpha\rangle)\perp\iota(t\conc\langle\beta\rangle)$.
The negation of this statement (all the $\iota(t\conc\langle\alpha\rangle)$ being comparable)
does NOT imply that some sequence in $\prod_{\alpha<\ddim}(N_{\iota(t\conc\langle\alpha\rangle)}\cap \ran(f))$ is constant, since the union $w$ of all the  $\iota(t\conc\langle\alpha\rangle)$'s might not be in $\ran(f)\cup T(\ran(f))$.
since otherwise,
every sequence in $\prod_{\alpha<\ddim}(N_{\iota(t\conc\langle\alpha\rangle)}\cap \ran([\iota]))$
would be constant,
so $\iota$ would not be an order homomorphism
for $\dhK\ddim{\ran(f)}$.
Using this observation and the continuity of $\iota$,
it is easy to construct a continuous $\perp$- and strict order preserving map
$\theta:{}^{<\kappa}2\to \ran(\iota)$ by recursion.
Then $\ran([\theta])$ is a $\kappa$-perfect subset of
by Lemma <ref>.

For <ref>,
suppose first that $f$ is a homomorphism from
to $I$.
$\dhK\ddim{\ran(f)}\subseteq I$.
Since $\ran(f)$ has a $\kappa$-perfect subset
by <ref>,
$I$ has a $\kappa$-perfect complete subhypergraph.
let $Y$ be a $\kappa$-perfect subset of $X$ such that $\dhK\ddim{Y}\subseteq I$.
Take any map $f: {}^\kappa\ddim\to Y$ that is a homeomorphism onto its image.
Since $f$ is injective, it is a homomorphism from
$\dhK\ddim{{}^\kappa\! \ddim}$ to $\dhK\ddim Y\subseteq I$.

The special case of <ref> for the complete dihypergraph $I:=\dhK\ddim X$ shows that $X$ has a $\kappa$-perfect subset if and only if there exists a continuous injection from ${}^\kappa \ddim$ onto a closed subset of $X$.
[This can also be proved directly as in <cit.>*Lemma 2.9 using Lemma <ref>.]
It also implies the next corollary since $\dhK\ddim X$ has a $\kappa$-coloring if and only if $|X|\leq\kappa$.

$\PSP_\kappa(X)\Longleftrightarrow \ODD\kappa{\dhK\ddim X}$ for all $\ddim$ with $2\leq\ddim\leq\kappa$.

Let $H$ be a $\ddim$-hypergraph on $X$.
any homomorphism from $\dhH{\ddim}$ to $H$ is also a homomorphism from $\hhH\ddim$ to $H$,
hence $\OHD\kappa H$ is equivalent to $\ODD\kappa H$.
$\OHD\kappa\ddim(X)$ is equivalent to $\ODD\kappa\ddim(X,\mathcal H_\ddim)$,
where $\mathcal H_\ddim$ denotes the class of all $\ddim$-hypergraphs.
The next theorem
shows that $\OHD\kappa\ddim(X)$ is in fact equivalent to

Suppose $2\leq\ddim\leq\kappa$ and $I$ is a $\ddim$-dihypergraph on $X$.
\beginenumerate-(1)

*  If $I$ is box-open on $X$, then
$\hypg I$ is box-open on $X$.

*  $I\equivf\hypg I$.[See Definition <ref>.]

*  If $I$ is box-open on $X$, then
$\ODD\kappa I\Longleftrightarrow\OHD\kappa{\hypg I}$.
[Equivalently, there exists a continuous homomorphism from $\dhH\ddim$ to $I$ if and only if there exists a continuous homomorphism from $\hhH\ddim$ to $\hypg I$.]

can the assumption that $I$ is box-open be omitted? It's used in the proof below
\endenumerate-(1)
In particular, $\OHD\kappa\ddim(X)$ is equivalent to $\ODD\kappa\ddim(X)$.

<ref> is clear.
<ref> holds since a subset of ${}^\kappa\kappa$ is $\hypg I$-independent if and only if it is $I$-independent.
For <ref>,
note that
$\ODD\kappa I\Longleftrightarrow\ODD\kappa{\hypg I}$ by
<ref>, <ref> and Corollary <ref>.
The latter is equivalent to
$\OHD\kappa{\hypg I}$ by the paragraph preceding the theorem.
The last claim follows from <ref> and <ref>.

Suppose $2\leq\ddim\leq\kappa$ and $I$ is a $\ddim$-dihypergraph on $X$.
\beginenumerate-(1)

*  If $I$ is box-open on $X$, then
$\hypg I$ is box-open on $X$.

*  A function $c:X\to\kappa$ is
a $\kappa$-coloring of $I$ if and only if it is a $\kappa$-coloring of $\hypg I$.

*  Suppose $I$ is box-open on $X$.
can the assumption that $I$ is box-open be omitted? It's used in the proof below
Then there is a continuous homomorphism from $\dhH\ddim$ to $I$ if and only if there is a continuous homomorphism from $\hhH\ddim$ to $\hypg I$.
\endenumerate-(1)
In particular, $\OHD\kappa\ddim(X)$ is equivalent to $\ODD\kappa\ddim(X)$.

<ref> follows from the fact that a subset $Y$ of $X$ is $\hypg I$-independent if and only if it is $I$-independent.

For <ref>, note first that any homomorphism from $\dhH\ddim$ to $I$ is also a homomorphism from $\hhH\ddim$ to $\hypg I$.
Conversely, suppose $f$ is a continuous homomorphism from $\hhH\ddim$ to $\hypg I$.
Then $f$ is also a homomorphism from $\dhH\ddim$ to $\hypg I$.
Since $\hypg I$ is box-open on $X$,
there exists an order homomorphism $\iota:{}^{<\kappa}\ddim\to{}^{<\kappa}\kappa$ for $(X,\hypg I)$
by Lemma <ref>.
It suffices to construct a continuous strict order preserving map $e:{}^{<\kappa}\ddim\to{}^{<\kappa}\ddim$ such that $\iota\comp e$ is an order homomorphism for $(X,I)$,
since then $[\iota\comp e]$ will be a continuous homomorphism from $\dhH\ddim$ to $I$.

We construct $e(t)$ by recursion on $\lh(t)$.
Let $e(\emptyset):=\emptyset$.
Let $t\in{}^{<\kappa}\ddim\,\setminus\{\emptyset\}$, and suppose that $e(u)$ has been defined for all $u\subsetneq t$.
If $\lh(t)\in\Lim$, then let $e(t):=\bigcup_{u\subsetneq t} e(u)$.
Suppose that $\lh(t)\in\Succ$, and let $u$ be the immediate predecessor of $t$.
We define
simultaneously for all $\alpha<\ddim$.
For all $\alpha<\ddim$, choose $x_\alpha\in N_{e(u)\conc\langle\alpha\rangle}$.
$\bar y:=\langle[\iota](x_\alpha):\alpha<\ddim\rangle$ is a hyperedge of
$\hypg I$ since $\iota$ is an order homomorphism for $(X,\hypg I).$
Take a permutation $\pi$ of $\ddim$ such that
$\bar y^{\pi}\in I$.
Since $I$ is box-open on $X$, there exists a sequence
$\langle s_\alpha:\alpha<\kappa\rangle$
such that
$e(u)\conc\langle\pi(\alpha)\rangle\subseteq s_\alpha\subseteq x_{\pi(\alpha)}$
$\prod_{\alpha<\kappa} N_{\iota(s_\alpha)}\cap X \subseteq I$.
Let $e(u\conc\langle \alpha\rangle):=s_\alpha$ for each $\alpha<\ddim$.
Then $\iota\comp e$ is an order homomorphism for $(X,I)$ by the construction and
since $\ran([\iota\comp e])\subseteq\ran([\iota])\subseteq X$.

We have $\ODD\kappa\ddim(X) \Rightarrow\OHD\kappa\ddim(X)$,
$\OHD\kappa{H}\Leftrightarrow\ODD\kappa H$ holds
for any $\ddim$-hypergraph $H$.
For the converse direction,
it suffices to show
$\OHD\kappa{\hypg I}\Rightarrow\ODD\kappa I$
for any box-open $\ddim$-dihypergraph $I$ on $X$ by <ref>.
But this is immediate from <ref> and <ref>.

Suppose $I$ is an open digraph
on $X$,
can the assumption that $I$ is open be omitted? It's used in the proof below
and let $G:=\hypg I$ be the smallest graph containing $I$.
Then the following statements are equivalent:
\beginenumerate-(1)

There is a continuous homomorphism from $\dhH 2$ to $I$.

There is a continuous homomorphism from $\gK{{}^\kappa\! 2}$ to $G$.

$G$ has a $\kappa$-perfect complete subgraph.
\endenumerate-(1)
In particular,
the open graph dichotomy $\OGD_\kappa(X)$ is equivalent to $\ODD\kappa 2(X)$.

<ref> $\Leftrightarrow$ <ref> follows from Theorem <ref> <ref>
since $\hhH 2=\gK{{}^\kappa 2}$.
$H$ needs to be open to apply Theorem <ref> <ref>
<ref> $\Leftrightarrow$ <ref> follows from Lemma <ref> <ref>
since $\dhK 2 {{}^\kappa 2}=\gK {{}^\kappa 2}$.

§.§ Counterexamples

A complete subhypergraph of a $\ddim$-dihypergraph $H$ on $X$ is a subhypergraph of the form $\dhK \ddim Y\subseteq H$ for some subset $Y\subseteq X$.

In the definition of $\ODD\kappa H$, one cannot replace
the existence of a continuous homomorphism with the existence of a large complete subhypergraph if $|d|\geq 3$,
since $\dhH\ddim$ does not admit a $\kappa$-coloring by Lemma <ref>
and it does not have a complete subhypergraph of size $2$.
Instead of complete subhypergraphs,
it is thus more interesting to consider subhypergraphs of the form
$\dhII\ddim Y$
consisting of all injective $\ddim$-sequences in some subset $Y$ of $X$.
However, $\dhII\ddim Y\not\subseteq\dhH\ddim$ for any $Y$ of size $|\ddim|$.
Similarly, the hypergraph $\hhH\ddim$ does not admit a $\kappa$-coloring
and if $d$ is infinite, then
$\dhII\ddim Y\not\subseteq\dhH\ddim$ for any $Y$ of size $|\ddim|$.

If $d$ is finite, then
$\dhII\ddim Y\not\subseteq\dhH\ddim$ for any $Y$ of size $|\ddim|+1$.

We now describe some more interesting counterexamples with stronger properties.
Moreover, we will see
that one cannot replace open by closed dihypergraphs in $\ODD\kappa\ddim(X)$.
The claims are shown by generalizing known examples to uncountable cardinals.

For any ordinal $\ddim\geq 3$, there exists
a product-open
[I.e., $H$ is an open subset of ${}^\ddim X$ in the product topology.]
$H\in\defsetsk$ on ${}^\kappa 2$
with the properties:
Let $[X]^3=\{(x,y,z)\in {}^3 X:x,y,z$ are pairwise distinct$\}$. For $\ddim=3$, $H\cap [X]^3$ will be a clopen subset of $[X]^3$, but $H$ will not be a closed subset of $\dhK 3 X$. Even if it were, it would not guarantee that $H'$ (as in Lemma 3.3) is a closed subset of $\dhK\ddim X$ for $\ddim>3$.
\beginenumerate-(a)

$H$ does not admit a $\kappa$-coloring.

$\dhII\ddim Y\not\subseteq H$ for any $Y$ of size at least $\max\{5,|\ddim|\}$.
\endenumerate-(a)
For $\ddim=3$, $H$ can be chosen to be a hypergraph.

It suffices to provide an example for $\ddim=3$, since one can then apply a projection
(see Lemma <ref>).
We follow  <cit.>*Section 10 and <cit.>*Example 3.4.
Let $X$ denote the set of those $x\in {}^\kappa 2$ with $x(\alpha)=0$ for $\alpha=0$ and all limits $\alpha<\kappa$.
Since $X$ is homeomorphic to ${}^\kappa 2$,
it suffices to define an open dihypergraph on $X$ with the required properties.
Recall that for $x\neq y$ in ${}^\kappa2$, $\Delta(x,y)$ denotes the least $\alpha<\kappa$ with $x(\alpha)\neq y(\alpha)$.
[Note that $\Delta(x,y)$ equals the length of the node $x\wedge y$ where $x$ and $y$ split.]

For any
$\bar x\in\dhII 3 X$,
$\bar x_0,\, \bar x_1,\, \bar x_2$
list $\ran(\bar x)$ such that
$\Delta(\bar x_0,\bar x_1)<\Delta(\bar x_1,\bar x_2)$.
As can be seen in Figure <ref>,
$\bar{x}_0$ is unique while the role of $\bar{x}_1$ and $\bar{x}_2$ can be switched.
$\Delta_0^{\bar x}:=\Delta(\bar{x}_0,\bar{x}_1)=\Delta(\bar{x}_0,\bar{x}_2)$
$\Delta_1^{\bar x}:=\Delta(\bar{x}_1,\bar{x}_2)$.
Then $\Delta_0^{\bar x},\,\Delta_1^{\bar x}$ are successor ordinals with $\Delta_0^{\bar x}<\Delta_1^{\bar x}$.
For any
$\bar x\in\dhII 3 X$, let

\bar x\in H\ :\Longleftrightarrow\
\bar{x}_2{\left(\Delta_0^{\bar x}-1\right)}
\neq
\bar{x}_2{\left(\Delta_1^{\bar x}-1\right)}.

It is easy to see that $H\in\defsets\kappa$ is an open $3$-hypergraph.

(0.5*,0.75*) edge[-] (0*,1.5*)
(0*,1.5*) edge[-, thick, RedOrangeArrow] (-0.5*,2.25*)
(0*,1.5*) edge[-] (0.5*,2.25*);

(0*,1.5*) edge[-, dashed, LightGrey] (3*,1.5*)
(0.5*,2.25*) edge[-, dashed, LightGrey] (3*,2.25*);

(0.5*,2.25*) edge[thick, -, TealArrow] (1.2*,4*);

(-0.5*,2.25*) edge[thick, -, RedOrangeArrow] (-1.65*,4*)
(0.5*,2.25*) edge[thick, -, DarkGreenArrow] (0*,4*);

(-1.6*,4*) node[above]
(0.1*,4*) node[above]
(1.35*,4*) node[above]

(3*,1.5*) node[right]
$\Delta_0^{\bar x}$
(3*,2.25*) node[right]
$\Delta_1^{\bar x}$;

(0*,0*) node[above]
(4*,4*) node[above] ;

A triple in $H$

For <ref>, it suffices to show that any $H$-independent subset $Y$ of $X$ is nowhere dense in $X$, since $X$ is homeomorphic to ${}^\kappa 2$ and thus not the union of $\kappa$ many nowhere dense subsets.
To this end, take any $s\in T(X)$.
There exist  $s_0,s_1,s_2\in T(X)$
with $s\subseteq s_i$ for all $i\leq 2$ and
In the next 2 figures: The gap between the figure and the caption was decreased by LOCALLY setting the value of “abovecaptionskip”, and so was the gap between the text and the figure (“intextsep”). Remove before submitting?
$N_{s_0}\times N_{s_1}\times N_{s_2}\subseteq H$
by the definition of $H$.
Since $Y$ is $H$-independent, there is some $i\leq 2$ with $N_{s_i}\cap Y=\emptyset$.

To show <ref>,
Suppose that $Y\subseteq X$ and $|Y|=5$.
By considering the splitting nodes of pairs in $Y$ from bottom to top, it is easy to check that there exist $x_0,\, x_1,\, x_2,\, x_3\in Y$ with
$\Delta_0<\Delta_1<\Delta_2$, where
$\Delta_i:=\Delta(x_i, x_{i+1})$
for $i\leq 2$.

(0*,0) edge[-] (-0.75*,0.75*);

(-0.75*,0.75*) edge[-, thick, RedOrangeArrow] (-4*,4*)
(-0.75*,0.75*) edge[-] (0*,1.5*)
(0*,1.5*) edge[-, thick, YellowArrow] (-0.5*,2.25*)
(0*,1.5*) edge[-] (0.5*,2.25*);

(-0.6*,0.75*) edge[-, dashed, LightGrey] (3*,0.75*)
(0*,1.5*) edge[-, dashed, LightGrey] (3*,1.5*)
(0.5*,2.25*) edge[-, dashed, LightGrey] (3*,2.25*);

(0.5*,2.25*) edge[thick, -, TealArrow] (1.2*,4*);

(-0.5*,2.25*) edge[thick, -, YellowArrow] (-1.65*,4*)
(0.5*,2.25*) edge[thick, -, DarkGreenArrow] (0*,4*);

(-4*,4*) node[above]
(-1.6*,4*) node[above]
(0.1*,4*) node[above]
(1.35*,4*) node[above]

(3*,0.75*) node[right] $\Delta_0$

(3*,1.5*) node[right] $\Delta_1$

(3*,2.25*) node[right] $\Delta_2$;

(0*,0*) node[above]
(4*,4*) node[above] ;

A splitting pattern witnessing $\dhII 3 Y\not\subseteq H$

Let $i,j$ be such that $i<j\leq 2$ and
Since $(x_i,x_j,x_3)\not\in H$, we have $\dhII 3 Y\not\subseteq H$.

We have not studied whether there exists a $\ddim$-hypergraph with the previous properties for any $\ddim\geq 4$.
In the previous example,
the set of all $x\in {}^\kappa2$ with $x(\alpha)=0$ for all even $\alpha<\kappa$ is a $\kappa$-perfect $H$-independent set for all $\ddim\geq 3$.
The next example shows that
one can strengthen the non-existence of a $\kappa$-coloring to the non-existence of a large independent set.

For any ordinal $\ddim\geq 3$, there exists
a product-open
$\ddim$-dihypergraph $H\in\defsetsk$ on ${}^\kappa \kappa$
with the properties:
$H_3$ will be a clopen subset of $\dhK\ddim{}$, but $H_\ddim$ may not even be a closed subset of $\dhK\ddim{}$ (in the product or in the box topology)
\beginenumerate-(a)

$\dhII\ddim Y\cap H\neq\emptyset$ for any $Y$ of size $\kappa^+$.

$\dhII\ddim Y\not\subseteq H$ for any $Y$ of size $\kappa^+$.
\endenumerate-(a)
For $\ddim=3$, $H$ can be chosen to be a hypergraph.
Note that <ref> implies that
there is no $H$-independent set of size $\kappa^+$ and hence
$H$ does not admit a $\kappa$-coloring.

It suffices to provide an example for $\ddim=3$, since one can then apply a projection
(see Lemma <ref>).
We follow <cit.>*Example 1.3.[This is also a special case of an argument of Blass <cit.>*p. 273 with a stronger conclusion.]

For any
$\bar x\in\dhII 3{{}^\kappa\kappa}$,
let $\bar x^{0}\lexleq \bar x^{1}\lexleq \bar x^{2}$ list $\ran(\bar{x})$ in lexicographic order
and define

$$ \bar x\in H\ :\Longleftrightarrow\
\Delta{\left(\bar x^{0},\bar x^{1}\right)}
\leq
\Delta{\left(\bar x^{1},\bar x^{2}\right)}.$$

Note that $H\in \defsetsk$ is an open hypergraph.

We first show <ref>.
Suppose that $\dhII 3 Y\subseteq H$.
We claim that the lexicographic order on $Y$ is a well-order with order type at most $\kappa+1$.
All splitting nodes of $T(Y)$ are compatible and thus lie on a branch $z$.
Since $\dhII 3 Y\subseteq H$, we have $y\lexleq z$ for all $y\in Y$.
Therefore for any splitting node $s\subsetneq z$ of $T(Y)$,
the order type of the set of all $y\in Y\setminus \{z\}$ with $y\wedge z=s$
is an ordinal ${<}\kappa$.
Since $z$ may be in $Y$, the order type of $Y$ is at most $\kappa+1$.
Hence $|Y|\leq\kappa$.

For <ref>,
suppose that $\dhII\ddim Y\cap H=\emptyset$.
Then $\Delta{\left(\bar x^{0},\bar x^{1}\right)}>
\Delta{\left(\bar x^{1},\bar x^{2}\right)}$
for all
$\bar x\in\dhII 3Y$.
We claim that the reverse lexicographic order on $Y$ is a well-order with order type at most $\kappa+1$.
As before, all splitting nodes of $T(Y)$ lie on a branch $z$.
If $s\subsetneq z$ is a splitting node, then exactly one element of $Y$ splits off at $s$.
Therefore the order type is at most $\kappa+1$, and hence $|Y|\leq\kappa$.

We have not studied whether there exists a $\ddim$-hypergraph with the properties of the previous example for any $\ddim\geq 4$.
He <cit.>, Di Prisco and Todorčević  <cit.>
showed for
dimensions $3\leq n<\omega$
that open $n$-hypergraphs with additional structural properties
either admit a countable coloring or contain a perfect complete subhypergraph.

The next example shows that
$\ODD\kappa H$ may fail for some closed hypergraph $H$
in any dimension $\ddim\geq 2$.

For any ordinal $\ddim\geq 2$,
there exists a product-closed
[I.e., $H$ is a relatively closed subset
of $\dhK\ddim {{}^\kappa\kappa}$, where ${}^\ddim {({}^\kappa\kappa)}$ has the product topology.
Equivalently, $H\cup \dhC\ddim {{}^\kappa\kappa}$ is a closed subset of ${}^\ddim {({}^\kappa\kappa)}$ with the product topology.]
$\ddim$-hypergraph $H\in\defsetsk$
on ${}^\kappa\kappa$ with the properties:
\beginenumerate-(a)

$H$ does not admit a $\kappa$-coloring.

\beginenumerate-(i)

$H$ does not have a complete subhypergraph of size $\kappa^+$.

There is no continuous homomorphism from $\dhH\ddim$ to $H$.
\endenumerate-(i)
\endenumerate-(a)

We first provide an example for $\ddim=2$.
An example for $\kappa=\omega$ can be found in <cit.>, so suppose $\kappa$ is uncountable.
We follow <cit.>*Exercise 29.9 and <cit.>*Example 3.3.
The example is based on the notion of a strongly $\kappa$-dense linear order.
This is a linear order with no $({<}\kappa,{<}\kappa)$-gaps.
A $({<}\kappa,{<}\kappa)$-gap in a linear order $(\mathbb{L},\leq)$ is a pair $(A,B)$ of subsets of $\mathbb{L}$ with the properties:
\beginenumerate-(1)

* $a<b$ for all $a\in A$ and $b\in B$.

* There is no $x\in \mathbb{L}$ with $a<x<b$ for all $a\in A$ and $b\in B$.

* $|A|, |B|<\kappa$.
Strongly $\kappa$ dense linear orders have no endpoints: consider the $A=\emptyset$ case, and the $B=\emptyset$ case
\endenumerate-(1)
It is easy to see that a strongly $\kappa$-dense linear order $\mathbb{L}$ of size $\kappa$ exists for any $\kappa$ with $\kappa^{<\kappa}=\kappa$.
It is also easy to show using a back-and-forth argument that any two strongly $\kappa$-dense linear orders of size $\kappa$ are isomorphic.]
Both existence and uniqueness up to isomorphism also follow from the fact that
a linear order is strongly $\kappa$-dense if and only if it is a $\kappa$-saturated model of the theory $T$ of dense linear orders with no endpoints.
For instance, let $\QQ_\kappa$ denote the set of all $x\in {}^\kappa\{-1,0,1 \}$ that take values in $\{-1,1\}$ up to some $\alpha<\kappa$ and are constant with value $0$ from $\alpha$.
$\QQ_\kappa$ with the lexicographical order is strongly $\kappa$-dense.

Let $(\mathbb{L},<)$ be a strongly dense linear order of size $\kappa$, and let

\[
X:=\{y\in \pwrset(\mathbb{L})\,:\, {\leq}{\restriction}y
\text{ is a well-order}\}.
\]

Since $|\mathbb{L}|=\kappa$, $\pwrset (\mathbb{L}) $ can be identified with ${}^\kappa 2$.
Since $\kappa$ is uncountable, $X$ is a closed subset.[For $\kappa=\omega$, $X$ would be a $\coanalytic$ subset.]
Therefore it suffices to define a closed graph $H$ on $X$ with the required properties.
For $x,y\in X$, let

\[ (x,y) \in H  \ :\Longleftrightarrow\   x\subsetneq y  \text{ or }  y\subsetneq x. \]

It is clear that $H$ is a closed graph on $X$ which has no complete subgraph of size $\kappa^+$.
Thus there is no continuous homomorphism from $\dhH 2$ to $H$ by Corollary <ref>.

We now show that $H$ does not admit a $\kappa$-coloring.
Towards a contradiction, suppose that $X=\bigcup_{\alpha<\kappa} X_\alpha$ where $X_\alpha$ is $H$-independent for each $\alpha<\kappa$.
Construct a strictly $\subseteq$-increasing chain
$\langle x_\alpha:\alpha<\kappa\rangle$ in $X$,
a strictly
$\leq$-increasing chain
$\langle q_\alpha:\alpha<\kappa\rangle$ in $\mathbb{L}$
and a strictly
$\leq$-decreasing chain
$\langle r_\alpha :\alpha<\kappa\rangle$ in $\mathbb{L}$
such that for all $\alpha<\kappa$, we have
$q_\alpha<r_\alpha$ and $p<q_\alpha$ for all $p\in x_\alpha$.
Whenever possible,
choose $x_\alpha$, $q_\alpha$, $r_\alpha$ such that
$x_\alpha\in X_\alpha$.
Finally, let  $x:=\bigcup_{\alpha<\kappa}x_\alpha$ and suppose that $x\in X_\beta$.
Then $x_\beta$ must also be in $X_\beta$ by the construction.
Since $(x_\beta,x)\in H$, $X_\beta$ is not $H$-independent.

Given the example for $\ddim=2$,
the $\ddim>2$ case follows from the next claim.

Suppose $\ddim>2$ and
Does the claim work for digraphs instead of graphs? (We used that $G$ is a graph in the proof.)
$G$ is a graph on ${}^\kappa\kappa$.
Let $H_G$ be the $\ddim$-hypergraph consisting of those sequences $\bar x$ such that $\ran(\bar x)=\{x,y\}$ for some $(x,y)\in G$.
\beginenumerate-(1)

\beginenumerate-(i)

* If $G$ is closed on ${}^\kappa\kappa$, then $H_G$ is product-closed on ${}^\kappa\kappa$.

* If $G\in\defsetsk$, then $H_G\in\defsetsk$.
\endenumerate-(i)

A function $c:{}^\kappa\kappa\to\kappa$ is a $\kappa$-coloring of $H_G$
if and only if
it is a $\kappa$-coloring of $G$.

\beginenumerate-(i)

$H_G$ does not have a complete subhypergraph of size $3$.

If there is a continuous homomorphism from $\dhH\ddim$ to $H_G$,
then there is a continuous homomorphism from $\dhH 2$ to $G$.
This does NOT follow from Lemma <ref> since $G$ may not be box-open
\endenumerate-(i)
\endenumerate-(1)

is clear.
<ref> holds since a subset of ${}^\kappa\kappa$ is $H_G$-independent if and only if it is $G$-independent.

In <ref>,
<ref> is immediate.
<ref>, suppose that $f$ is a continuous homomorphism from $\dhH\ddim$ to $H_G$.
We construct a continuous $\perp$- and strict order preserving map
$\iota:{}^{<\kappa} 2\to{}^{<\kappa}\ddim$ such that

f(N_{\iota(t\conc\langle 0\rangle)})
\times
f(N_{\iota(t\conc\langle 1\rangle)})
\subseteq G

holds for all $t\in{}^{<\kappa} \ddim$.
We construct $\iota(t)$ by recursion on $\lh(t)$.
Let $\iota(\emptyset):=\emptyset$.
If $\lh(t)\in\Lim$,
and $\iota(s)$ has been defined for all $s\subsetneq t$, then
let $\iota(t):=\bigcup_{s\subsetneq t}\iota(s)$.
We now assume $\iota(t)$ has been defined
and construct
$\iota(t\conc\langle 0\rangle)$
and $\iota(t\conc\langle 1\rangle)$.

For each $i<\ddim$, take $x_i$ extending $\iota(t)\conc\langle i\rangle$.
$\bar x:=\langle x_i:i<\ddim\rangle$ is a hyperedge of $\dhH\ddim$
$f$ is a homomorphism from $\dhH\ddim$ to $\dhK\ddim{{}^\kappa\kappa}$,
there exist $j<k<\ddim$
with $f(x_{j})\neq f(x_{k})$.
Using the continuity of $f$,
take $\alpha<\kappa$
such that
$\alpha>\lh(\iota(t))$ and
$f(N_{x_j\restr\alpha})\cap f(N_{x_k\restr\alpha})=\emptyset$.

Since $f$ is a homomorphism from $\dhH\ddim$ to $H_G$
and $x_i\restr\alpha$ extends $\iota(t)\conc\langle i\rangle$ for each $i<\ddim$,
we have
$\prod_{i<\ddim}f(N_{x_i\restr\alpha})\subseteq H_G$.
$f(N_{x_j\restr\alpha})$ and $f(N_{x_k\restr\alpha})$ are disjoint,
we must have
We use that $G$ is symmetric in this step.

$$f(N_{x_j\restr\alpha})\times f(N_{x_k\restr\alpha})\subseteq G.$$

$\iota(t\conc\langle 0\rangle)=x_j\restr\alpha$
$\iota(t\conc\langle 0\rangle)=x_k\restr\alpha$.

Now, suppose that $\iota$ has been constructed and let
Then $g$ is continuous.
To see that $g$ is a homomorphism from $\dhH 2$ to $G$,
suppose $(x,y)\in\dhH 2$ and let $t\in{}^\kappa 2$
be the node where $x$ and $y$ split.
$(g(x),g(y))\in g(N_{t\conc\langle 0\rangle})\times
g(N_{t\conc\langle 1\rangle})
f(N_{\iota(t\conc\langle 0\rangle)})\times
f(N_{\iota(t\conc\langle 1\rangle)})\subseteq G$,
as required.

There is a simpler way to get a dihypergraph with the same properties as in the previous claim instead of a hypergraph.
Given a graph $G$ on ${}^\kappa\kappa$ and $\ddim>2$, let
$I_G$ denote the $\ddim$-dihypergraph that consists of all $\langle x\rangle\conc \langle y : 1\leq i<\ddim\rangle$ with $(x,y)\in G$.
It is again clear that
and <ref><ref>
hold for $I_G$.
holds as well, since
for any homomorphism $f$ from $\dhH\ddim$ to $I_G$, $f\restr {}^\kappa2$ is a homomorphism from $\dhH2$ to $G$.

Note that the closed set $X$ in the above example is not homeomorphic to ${}^\kappa\kappa$.
In fact, one can show
as in <cit.>
that $X$ is not a continuous image of ${}^\kappa\kappa$.
To see this, one only has to replace the relation $R_x$ coded by a subset $x$ of $\kappa$ by the relation ${<_\mathbb{L}}\restr x$ throughout the proof of <cit.>.

§.§ Diamondi_kappa

In this subsection, we define the combinatorial principle $\Diamondi\kappa$
that is used for some of our results, for instance Theorem <ref>,
and discuss its relationship with some other versions of $\Diamond_\kappa$.
It is a weak version of $\Diamond_\kappa$ that holds at all inaccessible cardinals.

Let $\kappa$ be an infinite cardinal and let $\gamma\geq 2$.
A $\Diamondi{\kappa,\gamma}(\cof)$-sequence (respectively $\Diamondi{\kappa,\gamma}(\stat)$-sequence)  is a sequence
$\langle A_\alpha:\alpha<\kappa\rangle$
of sets $A_\alpha\subseteq{}^\alpha\gamma$ such that
\beginenumerate-(a)

$|A_\alpha|<\kappa$ for all $\alpha<\kappa$, and

for all $y\in{}^{\kappa}\gamma$, the set
$\{\alpha<\kappa:\:y\restr\alpha\in A_\alpha\}$
is cofinal (stationary) in $\kappa$.
\endenumerate-(a)
$\Diamondi{\kappa,\gamma}(\cof)$ (respectively, $\Diamondi{\kappa,\gamma}(\stat)$)
denote the statement that there exists a
$\Diamondi{\kappa,\gamma}(\cof)$-sequence (respectively, $\Diamondi{\kappa,\gamma}(\cof)$-sequence).

Let $\Diamondi\kappa$ denote the statement $\Diamondi{\kappa,\kappa}(\cof)$.
a $\Diamondi\kappa$-sequence is a $\Diamondi{\kappa,\kappa}(\cof)$-sequence.

We shall see in the next lemma

that all these principles are equivalent as long as
$\kappa$ is an uncountable regular cardinal
$2\leq \gamma\leq\kappa$.
Note that $\Diamondi{\kappa,2}(\stat)$ is exactly the
principle $\Dell\kappa$  introduced in <cit.>.

Suppose that $\kappa$ is an uncountable regular cardinal.
Then $\Diamondi{\kappa,\beta}(\varphi)$ and $\Diamondi{\kappa,\gamma}(\psi)$ are equivalent
for all $\beta,\gamma$ with
$2\leq \beta,\gamma\leq \kappa$
and all $\varphi, \psi\in \{\cof, \stat\}$.

It is clear that
holds for all ordinals $\gamma\geq 2$.

If $2\leq\beta<\gamma\leq\kappa$, then

Suppose that
$\langle A_\alpha:\alpha<\kappa\rangle$
is a $\Diamondi{\kappa,\gamma}(\stat)$-sequence.
Then $\langle A_\alpha\cap{}^\alpha\beta:\alpha<\kappa\rangle$
is a $\Diamondi{\kappa,\beta}(\stat)$-sequence.

This is similar to
the argument
sketched in
Suppose $\langle A_\alpha:\alpha<\kappa\rangle$ is a

Let $f$ be any bijection between $\kappa\times\kappa$ and $\kappa$.
For all
$\alpha<\kappa$, let
$\mathcal A_\alpha:=\big\{s^{-1}(\{1\}):
s\in A_\alpha
\big\}$,
[Thus, $\mathcal A_\alpha$ is a subset of $\pwrset(\alpha)$, and
$A_\alpha$ consists of the characteristic functions of elements of $\mathcal A_\alpha$.]
and let

\[\mathcal B_\alpha=
\left\{f^{-1}{\left(S \right)}: S\in \mathcal A_\alpha\right\}

Using the fact that $C=\{\alpha<\kappa:f(\alpha\times\alpha)=\alpha\}$ is a club subset of $\kappa$,
it is easy to see that
for all $Y\subseteq\kappa\times\kappa$,
the set of ordinals $\alpha<\kappa$ with
$Y\cap(\alpha\times\alpha)\in \mathcal B_\alpha$ is stationary in $\kappa$.

For all $\alpha<\kappa$,
let $B_\alpha:=\mathcal B_\alpha\cap {}^\alpha\alpha$.
$\langle B_\alpha:\alpha<\kappa\rangle$ is a $\Diamondi\kappa(\stat)$-sequence.
Indeed, if $x\in{}^\kappa\kappa$, then
for a club set of ordinals $\alpha<\kappa$
we have
or equivalently,
$x\restr\alpha=x\cap(\alpha\times\alpha)\in{}^\alpha\alpha$.[This step uses that $\kappa$ is regular.]
Thus, the set
$\{\alpha<\kappa: x\restr\alpha\in B_\alpha\}$ is a stationary subset of $\kappa$.

It now suffices show that
First, note that
always holds
if $\kappa$ is inaccessible, since
$\langle {}^\alpha 2:\alpha<\kappa\rangle$ is a

In the case that $\kappa$ is not inaccessible, the implication
follows from the next claim.
It is a
of the Proposition in <cit.> whose proof works verbatim in this setting.

Suppose that $\lambda<\kappa$ is an infinite cardinal such that $2^\lambda\geq\kappa$.
Suppose that there exists a sequence
$\langle A_\alpha:\alpha<\kappa\rangle$
of sets $A_\alpha\subseteq{}^\alpha 2$ such that
\beginenumerate-(a)

$|A_\alpha|<\kappa$ for all $\alpha<\kappa$, and

for all $y\in{}^{\kappa}2$, there exists $\alpha\geq\lambda$
such that $y\restr\alpha\in A_\alpha$.
\endenumerate-(a)
Then $\Diamondi{\kappa,2}(\stat)$ holds.

Lemma <ref> follows immediately from the previous claims.

\beginenumerate-(1)

If $\Diamond_\kappa$ holds or $\kappa$ is inaccessible, then $\Diamondi\kappa$ holds.

$\Diamondi\kappa$ implies that $\kappa^{<\kappa}=\kappa$.
\endenumerate-(1)

$\langle a_\alpha\in{}^\alpha 2:\alpha<\kappa\rangle$
is a $\Diamond_\kappa$-sequence, then
$\big\langle \{a_\alpha\}:\alpha<\kappa\big\rangle$
is a $\Diamondi{\kappa,2}(\stat)$-sequence.

If $\kappa$ is inaccessible, then
$\langle {}^\alpha 2:\alpha<\kappa\rangle$
is a $\Diamondi{\kappa,2}(\stat)$-sequence.
Under both of these assumptions, $\kappa$ is an uncountable regular cardinal
and therefore
$\Diamondi{\kappa,2}(\stat)$ implies
by Lemma <ref>.

To prove <ref>, suppose that
$\langle A_\alpha:\alpha<\kappa\rangle$ is a $\Diamondi\kappa$-sequence.
It is sufficient to show that
%\bigcup_{\alpha<\kappa}(A_\alpha\cup T(A_\alpha))
\bigcup_{\alpha<\kappa}T(A_\alpha)
since the latter set
has size $\kappa$.
Suppose $s\in{}^{<\kappa}\kappa$.
Let $x$ be an arbitrary element of ${}^\kappa\kappa$ extending $s$,
and let
$\alpha\geq\lh(s)$ be such that $x\restr\alpha\in A_\alpha$.
Then $s\subseteq x\restr\alpha$ and hence

Suppose that either $\kappa=\omega$ or that $\kappa$ is a singular strong limit cardinal.
$\Diamondi{\kappa,2}(\stat)$ holds
since $\langle {}^\alpha 2:\alpha<\kappa\rangle$
is a $\Diamondi{\kappa,2}(\stat)$-sequence,
but $\Diamondi\kappa=\Diamondi{\kappa,\kappa}(\cof)$ is false.

To show the latter statement, first note that
that $\Diamondi\kappa$ implies $\kappa$ is regular by
Claim <ref> <ref>.

Seeking a contradiction, suppose that $\Diamondi\omega$ holds and let
$\langle A_n:n<\omega\rangle$ be a $\Diamondi\omega$-sequence.
Define $x\in{}^\omega\omega$
by letting
$x(n):=\sup\{\ran(a):a\in A_{n+1}\}+1$
for each $n<\omega$.

Take $m<\omega$ with
$x\restr m\in A_m$.

\[x(m-1)>\sup\{\ran(a):a\in A_{m}\}\geq\sup\left(\ran(x\restr m)\right)\geq x(m-1),\]

a contradiction.

\beginenumerate-(1)

* For $\kappa=\lambda^+\geq\aleph_1$,
inserted pagebreak
Definition <ref> <ref>
is equivalent to
$|A_\alpha|\leq\lambda$ for all $\alpha<\kappa$.
Therefore $\Diamondi\kappa$ is equivalent to $\Diamond_\kappa$
by a result of Kunen's
[See also <cit.>*Theorem III.7.8 and Exercise III.7.16.]

* For $\kappa=\lambda^+\geq\aleph_2$,
$\kappa^{<\kappa}=\kappa$ is equivalent to $\Diamond_\kappa$
and therefore also to $\Diamondi\kappa$.

* For inaccessible cardinals $\kappa$,
$\Diamond_\kappa$ and $\Diamondi\kappa$ are not in general equivalent,
since the failure of $\Diamond_\kappa$ at the least inaccessible cardinal $\kappa$ is consistent <cit.>.
\endenumerate-(1)

To the best knowledge of the authors, it is not known whether
any of the assumptions
and $\Diamond_\kappa$ are equivalent
when $\kappa$ is weakly inaccessible but not inaccessible.

The next lemma describes a
$\delta$-dimensional equivalent version
of $\Diamondi\kappa$ for ordinals $2\leq\delta<\kappa$.

Recall that
$\dhC\delta X$ denotes the set of constant sequences $\langle c\rangle ^\ddim \in{}^\ddim X$

$\dhK\delta X={}^\delta X-\dhC\delta X$.

Suppose $\kappa$ is a regular uncountable cardinal and $2\leq\delta<\kappa$.
Then the following statements are equivalent:
\beginenumerate-(1)

There exists a sequence
$\langle B_\alpha:\alpha<\kappa\rangle$ of sets
$B_\alpha\subseteq\dhK\delta{{}^\alpha\kappa}$ such that
\beginenumerate-(a)

$|B_\alpha|<\kappa$ for all $\alpha<\kappa$, and

for all $\bar x\in\dhK\delta{{}^\kappa\kappa}$,
there exists $\alpha<\kappa$
$\bar x\cwrestr\alpha\in B_\alpha$.
[Recall that
$\bar x\cwrestr\alpha:=\langle x_i\restr\alpha:i<\delta\rangle$
for any sequence
$\bar x=\langle x_i:i\in\delta \rangle\in{}^\delta{({}^\kappa\kappa)}$ and $\alpha<\kappa$.]
\endenumerate-(a)
\endenumerate-(1)

<ref> $\Rightarrow$ <ref>:
Suppose that $\langle B_\alpha:\alpha<\kappa\rangle$ is a sequence as in <ref>.
For all $\alpha<\kappa$, let
A_\alpha:=\{s_0: \langle s_\beta:\beta<\delta\rangle\in B_\alpha\}.
We claim that $\langle A_\alpha:\alpha<\kappa\rangle$ is a $\Diamondi\kappa$-sequence.
To see this, suppose $x\in{}^\kappa\kappa$ and $\gamma<\kappa$.
It suffices to find $\alpha<\kappa$ such that $\alpha>\gamma$
and $x\restr\alpha\in A_\alpha$.
Take $\bar x=\langle x_\beta:\beta<\delta\rangle$ in
with $x_0=x$ and
$\bar x\cwrestr\gamma\in\dhC\delta{{}^\gamma\kappa}$.
By <ref><ref>, there exists $\alpha<\kappa$ such that
$\bar x\cwrestr\alpha\in B_\alpha$.
Then $x\restr\alpha=x_0\restr\alpha\in A_\alpha$,
and $\alpha>\gamma$ since
$\bar x\cwrestr\gamma\in\dhC\delta{{}^\gamma\kappa}$
$\bar x\cwrestr\alpha\in B_\alpha\subseteq\dhK\delta{{}^\alpha\kappa}$.

<ref> $\Rightarrow$ <ref>:
Let $\langle A_\alpha:\alpha<\kappa\rangle$ be a $\Diamondi\kappa$-sequence.
For every
$\gamma\leq\kappa$, let
\to
be the bijection
which maps each
to the unique element
\langle t_\beta:\beta<\delta\rangle$ of
for all $\alpha<\gamma$ and $\beta<\delta$.
For each $\gamma<\kappa$, let

\[
\bigcup_{\beta<\delta}
\left\{
s\in A_{\delta\cdot\gamma+\beta}
\right\}.
\]

We claim that $\langle B_\gamma:\gamma<\kappa\rangle$ satisfies <ref>.
Since $\kappa$ is regular,
$|B_\gamma|<\kappa$ for all $\gamma<\kappa$.
To see <ref><ref>,
take $\bar x\in\dhK\delta{{}^\kappa\kappa}$ and let $z:=g_\kappa^{-1}(\bar x)$.
Since $z\in{}^\kappa\kappa$ and $\langle A_\alpha:\alpha<\kappa\rangle$ is a $\Diamondi\kappa$-sequence,
there exist ordinals
and $\beta<\delta$ such that
$z\restr(\delta\cdot\gamma+\beta)\in A_{\delta\cdot\gamma+\beta}$.
\bar x\cwrestr\gamma=

§.§ Strong variants

We consider the following extensions of $\ODD\kappa H$. These are ordered by increasing strength.

Suppose that $2\leq\ddim\leq\kappa$ and
$H$ is a $\ddim$-dihypergraph on  ${}^\kappa\kappa$.

\begin{equation*}
\parbox{\dimexpr\linewidth-60pt}{%
\hypertarget{ODDI def}{}%
\strut
{$\ODDI\kappa H$:}
Either $H$ admits a $\kappa$-coloring, or
there exists
an \emph{injective} continuous homomorphism
from~$\dhH{\ddim}$ to~$H$.
\strut
\end{equation*}

\begin{equation*}
\parbox{\dimexpr\linewidth-60pt}{%
\hypertarget{ODDH def}{}%
\strut
{$\ODDH\kappa H$:}
Either $H$ admits a $\kappa$-coloring, or
there exists
a homomorphism
from~$\dhH{\ddim}$ to~$H$
which is
\emph{a homeomorphism} from ${}^\kappa\ddim$ onto its image.
\strut
\end{equation*}

\begin{equation*}
\parbox{\dimexpr\linewidth-60pt}{%
\hypertarget{ODDC def}{}%
\strut
{$\ODDC\kappa H$:}
Either $H$ admits a $\kappa$-coloring, or
there exists
a homomorphism
from~$\dhH{\ddim}$ to~$H$
which is
\emph{a homeomorphism} from ${}^\kappa\ddim$ onto
a \emph{closed} subset of~${}^\kappa\kappa$.
\strut
\end{equation*}

Suppose that $2\leq\ddim\leq\kappa$ and
$H$ is a $\ddim$-dihypergraph on  ${}^\kappa\kappa$.

\begin{align*}
\hypertarget{ODDI def}{}%
\strut
{$\ODDI\kappa H$:}
Either $H$ admits a $\kappa$-coloring, or
there exists
an \emph{injective} continuous homomorphism
from~$\dhH{\ddim}$ to~$H$.
\strut
\\
\hypertarget{ODDH def}{}%
\strut
{$\ODDH\kappa H$:}
Either $H$ admits a $\kappa$-coloring, or
there exists
a homomorphism
from~$\dhH{\ddim}$ to~$H$
which is
\emph{a homeomorphism} from ${}^\kappa\ddim$ onto its image.
\strut
\\
\hypertarget{ODDC def}{}%
\strut
{$\ODDC\kappa H$:}
Either $H$ admits a $\kappa$-coloring, or
there exists
a homomorphism
from~$\dhH{\ddim}$ to~$H$
which is
\emph{a homeomorphism} from ${}^\kappa\ddim$ onto
a \emph{closed} subset of~${}^\kappa\kappa$.
\strut
\end{align*}

Note that if $H$ and $I$ are box-open $\ddim$- and $c$- dihypergraphs with $H\equivif I$ on a subset $X$ of ${}^\kappa\kappa$, respectively,
then $\ODDI\kappa H\Longleftrightarrow\ODDI\kappa I$ and
$\ODDH\kappa H\Longleftrightarrow\ODDH\kappa I$
by Remark <ref>.
[If $c=d$ and $I\subseteq H$, then the assumption that $H$ is box-open may be omitted, since in this case any homomorphism from $\dhH\ddim$ to $I$ is a homomorphism to $H$ as well.]

Recall that a dihypergraph $H$ is defined to be relatively box-open if it is box-open on its domain $\domh H$.[See Definition <ref>.]
The next lemma characterizes the existence of a homomorphism as in $\ODDH\kappa H$ via order homomorphisms.

that $2\leq\ddim\leq\kappa$,
and $H$ is a $\ddim$-dihypergraph on  ${}^\kappa\kappa$.
\beginenumerate-(1)

If $\iota$
is a $\perp$-preserving order homomorphism for $(X,H)$,
then $[\iota]$ is a homomorphism from $\dhH\ddim$ to $H\restr X$ which is a homeomorphism from ${}^\kappa\ddim$ onto its image.

If $H\restr X$ is box-open on $X$ and there exists a homomorphism $f$ from $\dhH\ddim$ to $H\restr X$ which is a homeomorphism from ${}^\kappa\ddim$ onto its image,
then there exists a continuous $\perp$-preserving
order homomorphism $\iota$ for $(X,H)$
and a continuous strict $\wedge$-homomorphism
[See Definition <ref>. Note that $e:{}^{<\kappa}\ddim\to{}^{<\kappa}\kappa$  is a strict $\wedge$-homomorphism if and only if it is strict order preserving and $e(t\conc\langle \alpha\rangle) \wedge e(t\conc\langle \beta\rangle) =e(t)$ for all $t\in{}^{<\kappa}\ddim$ and $\alpha<\beta<\ddim$. It is easy to see that such a map $e$ is $\perp$-preserving, and hence $[e]$ is a homeomorphism onto its image by Lemma <ref> <ref>.]
with $[\iota]=f\comp [e]$.

This lemma is only used in this subsection. <ref> used only in the proof of Prop <ref>.
\endenumerate-(1)

For <ref>, note that
$[\iota]$ is a homeomorphism from ${}^{\kappa}\ddim$ onto its image
by Lemma <ref> <ref>.
It is a homomorphism from $\dhH\ddim$ to $H\restr X$
by Lemma <ref> <ref>.

The argument for <ref> is similar to the proof of Lemma <ref> <ref>.
We construct continuous strict order preserving maps
with the following properties for all $t\in{}^{<\kappa}\ddim$ and
\beginenumerate-(i)

$e(t)\conc\langle \alpha\rangle\subseteq e(t\conc\langle \alpha\rangle)$,

$f(N_{e(t)})\subseteq N_{\iota(t)}$.

$\iota(t\conc\langle \alpha\rangle)\perp\iota(t\conc\langle \beta\rangle)$.

$\prod_{\gamma<\ddim}(N_{\iota(t\conc\langle \gamma \rangle)}\cap X)\subseteq H$.
\endenumerate-(i)
Construct $\iota(t)$ and $e(t)$ by recursion on $\lh(t)$.
Let $\iota(\emptyset):=e(\emptyset)=\emptyset$.
Let $t\in{}^{<\kappa}\ddim\,\setminus\{\emptyset\}$, and suppose that $\iota(u)$ and $e(u)$ have been defined for all $u\subsetneq t$.
If $\lh(t)\in\Lim$,
let $\iota(t):=\bigcup_{u\subsetneq t}\iota(u)$ and
let $e(t):=\bigcup_{u\subsetneq t}e(u)$.
$N_{e(t)}=\bigcap_{u\subsetneq t}N_{e(u)}$
and <ref> holds for all $u\subsetneq t$,
we have

\[
\bigcap_{u\subsetneq t}f(N_{e(u)})\subseteq
\bigcap_{u\subsetneq t}N_{\iota(u)}
\]

Suppose $\lh(t)\in\Succ$, and let $u$ be the immediate predecessor of $t$.
for each $\alpha<\ddim$ simultaneously as follows.