# On homogeneity of Cantor cubes

E. Shchepin  Steklov Mathematical Institute of Russian Academy of Sciences, 8
Gubkina St. Moscow, 119991, Russia<EMAIL_ADDRESS>and  V. Valov  Department
of Computer Science and Mathematics, Nipissing University, 100 College Drive,
P.O. Box 5002, North Bay, ON, P1B 8L7, Canada<EMAIL_ADDRESS>
###### Abstract.

We discuss the question of extending homeomorphism between closed subsets of
the Cantor cube $D^{\tau}$. It is established that any homeomorphism between
two closed negligible subset of $D^{\tau}$ can be extended to an
autohomeomorphism of $D^{\tau}$.

###### Key words and phrases:

Candor discontinuum, 0-dimensional spaces, homeomorphisms, nowhere
$\tau$-dense sets

###### 1991 Mathematics Subject Classification:

Primary 54C20, 54F45; Secondary 54B10, 54D30

The second author was partially supported by NSERC Grant 261914-19.

##  1\. Introduction

Knaster-Reichbach [4] established the following theorem (see also [9] for
other types of zero-dimensional separable metric spaces): Let $X$ and $Y$ be
compact, perfect zero-dimensional metric spaces, and let $P$ and $K$ be closed
nowhere dense subsets of $X$ and $Y$, respectively. If $f$ is a homeomorphism
between $P$ and $K$, then there exists a homeomorphism between $X$ and $Y$
extending $f$.

If we omit the metrizability condition in the Knaster-Reichbach’s theorem,
then the conclusion is not anymore true. In order to obtain a correct
generalization of the theorem, first of all, it is necessary to find the
correct analogue of the condition ”nowhere dense”. Moreover, the perfectness
condition can be formulated as the nowhere density of the points.

Such an analogue is the following concept of negligibility. A subset of a
topological space is called negligible if it does not contain a non-empty
intersection of a family of open sets such that the cardinality of the family
is less than the weight of the space. Note that for metric compacta the
condition of nowhere density is equivalent to the condition of negligibility.

Now we able to present a generalization of the Knaster-Reichbach’s theorem.

######  Theorem 1.1.

Let $X$ and $Y$ be compact, zero-dimensional absolute extensors for
0-dimensional spaces of the same weight with negligible points, and let $P$
and $K$ be closed negligible subsets of $X$ and $Y$, respectively. If $f$ is a
homeomorphism between $P$ and $K$, then there exists a homeomorphism between
$X$ and $Y$ extending $f$.

This theorem for metric compacta turns into Knaster-Reichbach’s theorem,
because every metric compact is an absolute extensor in dimension 0. In
general, it is very difficult to avoid the condition of being an absolute
extensor in dimension 0, because before you can continue homeomorphisms, you
need to be able to continue at least continuous maps.

Since the condition of the negligibility of a point in a compactum is
equivalent to having a character at that point equal to the weight of the
compactum, then Theorem 1 from [10] allows us to assert that the compacta $X$
and $Y$ in Theorem 1.1 are homeomorphic to the Cantor cube $D^{\tau}$, where
$\tau$ is the weight of $X$ and $Y$. Therefore, the above theorem can be
obtained from the following special case of its own.

######  Theorem 1.2.

Let $f$ be a homeomorphism between closed negligible subsets $P$ and $K$ of
$D^{\tau}$, then $f$ can be extended to a homeomorphism on $D^{\tau}$.

Since every subset of the weight less than the weight of the entire space is
negligible, the following consequence follows from the formulated theorem.

######  Corollary 1.3.

If $P,K$ are closed subsets of $D^{\tau}$ both of weight $<\tau$, then every
homeomorphism between $P$ and $K$ can be extended to homeomorphism on
$D^{\tau}$.

Actually, we prove a more general version of Theorem 1.2, see Theorem 4.1
below.

##  2\. Some preliminary results

Anywhere below, by a homeomorphism we always mean a surjective homeomorphism.
We need a more precise notion of negligibility. For a space $X$, a subset
$P\subset X$ and an infinite cardinal $\lambda$ we denote by $P^{(\lambda)}$
the $\lambda$-interior of $P$ in $X$, i.e. the set all $x\in P$ such that
there exists a $G_{\lambda}$-subset $K$ of $X$ with $x\in K\subset P$. If
$\lambda$ is finite, then $P^{(\lambda)}$ is the ordinary interior of $P$ and
it is denoted by $P^{(0)}$. If there exists $\tau\geq\aleph_{0}$ with
$P^{(\lambda)}$ is empty for all $\lambda<\tau$, we say that $P$ is
$\tau$-negligible in $X$.

Let $X=\prod_{\alpha\in A}X_{\alpha}$ be a product of spaces and $B\subset A$.
If $P\subset X$, then $P_{B}$ denotes $\pi_{B}(P)$, where
$\pi_{B}:X\to\prod_{\alpha\in B}X_{\alpha}$ is the projection.

######  Proposition 2.1.

Let $X=\prod_{\alpha\in A}X_{\alpha}$ be a product of compact metric spaces,
$P$ and $K$ be closed subsets of $X$ and $f:P\to K$ be a homeomorphism. Then
for any countable set $C\subset A$ there is a countable set $B\subset A$ and a
homeomorphism $f_{B}:P_{B}\to K_{B}$ such that $C\subset B$ and $\pi_{B}\circ
f=f_{B}\circ\pi_{B}$.

###### Proof.

Obviously this is true for a countable set $A$, so we suppose $A$ is
uncountable. Let $g:K\to P$ be the inverse of $f$. Using that any continuous
function on $X$ depends on countably many coordinates [8], we construct by
induction a sequence of countable sets $B(n)\subset A$ and maps
$f_{n}:P_{B(n+1)}\to K_{B(n)}$ and $g_{n}:K_{B(n+1)}\to P_{B(n)}$ such that:

  * •

$B(1)=C$ and $B(n)\subset B(n+1)$;

  * •

$\displaystyle\pi_{B(n)}\circ f=f_{n}\circ\pi_{B(n+1)}$;

  * •

$\displaystyle\pi_{B(n)}\circ g=g_{n}\circ\pi_{B(n+1)}$;

Then $B=\bigcup_{n=1}^{\infty}B(n)$ is countable and there exist maps
$f_{B}:P_{B}\to K_{B}$ and $g_{B}:K_{B}\to P_{B}$ with
$\displaystyle\pi_{B}\circ f=f_{B}\circ\pi_{B}$ and $\displaystyle\pi_{B}\circ
g=g_{B}\circ\pi_{B}$. Then $f_{B}$ is a homeomorphism between $P_{B}$ and
$K_{B}$. ∎

In the situation of Proposition 2.1, a subset $B\subset A$ is called
$f$-admissible if there exists a homeomorphism $f_{B}:P_{B}\to K_{B}$ with
$\displaystyle\pi_{B}\circ f=f_{B}\circ\pi_{B}$. It easily seen that arbitrary
union of $f$-admissible sets is also $f$-admissible.

In [6] $\tau$-negligible sets were considered under the name
$\widetilde{G}_{\tau}$-sets. By [6, Lemma 6], if $X$ is a product of metric
compacta, then $F$ is $\tau$-negligible in $X$ if and only if the
$\pi$-character $\pi\chi(F,X)$ of $F$ in $X$ is $\geq\tau$. Recall that
$\pi\chi(F,X)$ is the smallest cardinality $\lambda$ such that there is an
open family $\mathcal{U}$ in $X$ of cardinality $\lambda$ with the following
property: every neighborhood of $F$ in $X$ contains an element of
$\mathcal{U}$.

Next lemma is a modification of [6, Theorem 2].

######  Lemma 2.2.

Let $X=\prod_{\alpha\in A}X_{\alpha}$ be a product of compact metric spaces
and $P$ be a closed set in $X$. Suppose $C\subset A$ is a set of cardinality
$<\tau$ such that $(\\{z\\}\times X_{A\backslash C})\cap P$ is
$\tau$-negligible in $\\{z\\}\times X_{A\backslash C}$ for every $z\in P_{C}$.
Then $P_{A\backslash C}\neq X_{A\backslash C}$.

###### Proof.

Indeed, otherwise we may assume that the projection $\pi_{A\backslash C}$
restricted on $P$ is an irreducible map onto $X_{A\backslash C}$, and denote
this map by $f$. Consequently,

$\pi\chi(f((\\{z\\}\times X_{A\backslash C})\cap P),X_{A\backslash
C})\leq\pi\chi((\\{z\\}\times X_{A\backslash C})\cap P,P)$

for every $z\in P_{C}$. On the other hand, $\pi\chi((\\{z\\}\times
X_{A\backslash C})\cap P,P)\leq\pi\chi(z,P_{C})<\tau$. So,
$\pi\chi(f((\\{z\\}\times X_{A\backslash C})\cap P),X_{A\backslash C})<\tau$.
That, according to [6, Lemma 6], contradicts the fact that $f((\\{z\\}\times
X_{A\backslash C})\cap P)$ is $\tau$-negligible in $X_{A\backslash C}$. ∎

######  Corollary 2.3.

Let $X=\prod_{\alpha\in A}X_{\alpha}$ be a product of compact metric spaces
and $P$ be a closed $\tau$-negligible set in $X$. If $C\subset A$ is a set of
cardinality $<\tau$, then there is a set $B\subset A$ containing $C$ such that
$B\backslash C$ is countable and $\pi_{B\backslash C}(P)$ is nowhere dense in
$X_{B\backslash C}$. If, in addition, $f:P\to K$ is a homeomorphism with
$K\subset X$ and $C$ is $f$-admissible, then we can assume that $B$ is also
$f$-admissible.

###### Proof.

Since $P$ is a $\tau$-negligible set in $X$, so are the sets
$P(z)=(\\{z\\}\times X_{A\backslash C})\cap P$ for all $z\in P_{C}$. This
implies that each $\pi_{A\backslash C}(P(z))$ is $\tau$-negligible in
$X_{A\backslash C}$. Indeed, otherwise $\pi_{A\backslash C}(P(z^{*}))$ would
contain a closed $G_{\lambda}$-set in $X_{A\backslash C}$ for some $z^{*}\in
P_{C}$ and $\lambda<\tau$. Because $\\{z^{*}\\}\times X_{A\backslash C}$ is
$G_{\mu}$-set in $X$, where $\mu$ is the cardinality of $C$, $P(z^{*})$
contains a $G_{\lambda^{\prime}}$-subset of $X$ with
$\lambda^{\prime}=\max\\{\lambda,\mu\\}<\tau$, a contradiction. Therefore, we
can apply Lemma 2.2 countably many times to construct by induction a disjoint
sequence $\\{C_{n}\\}$ of finite subsets of $A\backslash C$ such that

  * •

$C_{1}\subset A\backslash C$;

  * •

$C_{n+1}\subset A\backslash\bigcup_{k\leq n}C\cup C_{k}$;

  * •

$\pi_{C_{n}}(P)\neq X_{C_{n}}$ for all $n$.

Then $B=\bigcup_{n\geq 1}C\cup C_{n}$ is the required set. If $f:P\to K$ is a
homeomorphism and $C$ is $f$-admissible, then for every $\alpha\in A$ fix a
countable $f$-admissible set $B(\alpha)$ containing $\alpha$, see Proposition
2.1. Next, using Lemma 2.2 we construct a disjoint sequence $\\{C_{n}\\}$ of
finite sets with

  * •

$C_{1}\subset A\backslash C$;

  * •

$C_{n+1}\subset A\backslash\bigcup_{k\leq n}C\cup C_{k}^{\prime}$, where
$C_{k}^{\prime}=\bigcup_{\alpha\in C_{k}}B(\alpha)$;

  * •

$\pi_{C_{n}}(P)\neq X_{C_{n}}$ for all $n$.

Then $B=\bigcup_{n\geq 1}C\cup C_{n}^{\prime}$ is $f$-admissible and satisfies
the required conditions. ∎

##  3\. Homeomorphisms on product spaces

For any space $X$ let $\mathcal{H}(X)$ denote the set of all homeomorphisms on
$X$ equipped with the compact-open topology. In this section we prove that
$\mathcal{H}(X)$, where $X$ is a product of compact metric spaces, is an
absolute extensor for compact 0-dimensional spaces. Our proof uses the
technique developed in [5].

######  Proposition 3.1.

Let $K\subset\mathcal{H}(X)$ be a Lindel$\ddot{o}$f subset, where
$X=\prod_{\alpha\in A}X_{\alpha}$ is a product of compact metric spaces with
$|A|=\tau$. Then $A$ can be covered by a family of sets
$\\{A(\alpha):\alpha\in\omega(\tau)\\}$ such that for every $\alpha$ we have:

  * •

$A(\alpha)=\bigcup_{\gamma<\alpha}B(\gamma)$ if $\alpha$ is a limit ordinal;

  * •

$A(\alpha)\subset A(\alpha+1)$ and $A(\alpha+1)\backslash A(\alpha)$ is
countable for all $\alpha$;

  * •

For every $f\in K$ there is $f_{\alpha}\in\mathcal{H}(X_{A(\alpha)})$ with
$\pi_{A(\alpha)}\circ f=f_{\alpha}\circ\pi_{A(\alpha)}$.

###### Proof.

In case $A$ is countable we take $A(\alpha)=A$ for all $\alpha$. Suppose $A$
is uncountable and let $B\subset A$ be a countable set. Take a sequence of
open covers $\\{\mathcal{U}_{n}\\}_{n\geq 1}$ of $X_{B}$ such that
$\rm{diam}(\mathcal{U}_{n})<1/n$ for all $n$. Since $X_{B}$ is metrizable and
$X$ is compact, the compact-open topology on the function space $C(X,X_{B})$
coincides with the limitation topology [11]. Recall that [1] $U\subset
C(X,X_{B})$ is open with respect to the limitation topology if for every $f\in
U$ there is $\mathcal{V}\in\rm{cov}(X_{B})$ such that $U$ contains the set
$B(f,\mathcal{V})=\\{g\in
C(X,X_{B}):g{~{}}\mbox{is}{~{}}\mathcal{V}-\mbox{close to}{~{}}f\\}.$ Here,
$\rm{cov}(X_{B})$ is the family of all open covers of $X_{B}$ and $g$ is
$\mathcal{V}$-close $f$ provided for any $x\in X$ there is $V\in\mathcal{V}$
containing both points $f(x)$ and $g(x)$. In particular, every
$B(f,\mathcal{V})$ contains a neighborhood $B_{*}(f,\mathcal{V})$ of $f$.
There exists a natural map $p_{B}:\mathcal{H}(X)\to C(X,X_{B})$,
$p_{B}(h)=\pi_{B}\circ h$, which is continuous when both $\mathcal{H}(X)$ and
$C(X,X_{B})$ carry the compact-open topology.

######  Claim 3.2.

There is a countable set $\Gamma(B)\subset A$ containing $B$ such that for
every $f\in K$ there exist $f_{\Gamma(B)},g_{\Gamma(B)}\in
C(X_{\Gamma(B)},X_{B})$ with $\pi_{B}\circ
f=f_{\Gamma(B)}\circ\pi_{\Gamma(B)}$ and $\pi_{B}\circ
f^{-1}=g_{\Gamma(B)}\circ\pi_{\Gamma(B)}$.

Since for each $n$ the family $\\{B_{*}(\pi_{B}\circ f,\mathcal{U}_{n}):f\in
K\\}$ is an open cover of $p_{B}(K)$, there is a sequence $\\{f_{ni}\\}_{i\geq
1}\subset\mathcal{H}(X)$ such that $\\{B_{*}(\pi_{B}\circ
f_{ni},\mathcal{U}_{n}):i\geq 1\\}$ covers $p_{B}(K)$. Similarly, there exists
sequences $\\{g_{ni}\\}_{i\geq 1}\subset\mathcal{H}(X)$, $n\geq 1$, such that
$\\{B_{*}(\pi_{B}\circ g_{ni},\mathcal{U}_{n}):i\geq 1\\}$ covers
$p_{B}(K^{-1})$, where $K^{-1}=\\{f^{-1}:f\in K\\}$. As in the proof of
Proposition 2.1, there is a countable set $\Gamma(B)$ containing $B$ and
corresponding to each $n$ sequences $\\{\varphi_{ni}\\}_{i\geq 1}\subset
C(X_{\Gamma(B)},X_{B})$ and $\\{\phi_{ni}\\}_{i\geq 1}\subset
C(X_{\Gamma(B)},X_{B})$ such that $\pi_{B}\circ
f_{ni}=\varphi_{ni}\circ\pi_{\Gamma(B)}$ and $\pi_{B}\circ
g_{ni}=\phi_{ni}\circ\pi_{\Gamma(B)}$ for all $n,i$. Then for every $f\in K$
there exist two sequences $\displaystyle\\{\varphi_{ni_{n}}\\}_{n\geq
1}\subset C(X_{\Gamma(B)},X_{B})$ and $\displaystyle\\{\phi_{ni_{n}}\\}_{n\geq
1}\subset C(X_{\Gamma(B)},X_{B})$ such that $\pi_{B}\circ f$ is
$\mathcal{U}_{n}$-close to $\displaystyle\varphi_{ni_{n}}\circ\pi_{\Gamma(B)}$
and $\pi_{B}\circ f^{-1}$ is $\mathcal{U}_{n}$-close to
$\displaystyle\phi_{ni_{n}}\circ\pi_{\Gamma(B)}$ for each $n$. The last
condition implies that the sequence
$\displaystyle\\{\varphi_{ni_{n}}\\}_{n\geq 1}$ uniformly converges to a map
$f_{\Gamma(B)}\in C(X_{\Gamma(B)},X_{B})$ and $\pi_{B}\circ
f=f_{\Gamma(B)}\circ\pi_{\Gamma(B)}$. The sequence
$\displaystyle\\{\phi_{ni_{n}}\\}_{n\geq 1}$ also converges uniformly to a map
$g_{\Gamma(B)}\in C(X_{\Gamma(B)},X_{B})$ and $\pi_{B}\circ
f^{-1}=g_{\Gamma(B)}\circ\pi_{\Gamma(B)}$.

######  Claim 3.3.

For every countable set $B\subset A$ there is a countable set
$\Lambda(B)\subset A$ containing $B$ such that for every $f\in K$ there exist
homeomorphisms $f_{\Lambda(B)},g_{\Lambda(B)}\in\mathcal{H}(X_{\Lambda(B)})$
with $\pi_{\Lambda(B)}\circ f=f_{\Lambda(B)}\circ\pi_{\Lambda(B)}$ and
$\pi_{\Lambda(B)}\circ f^{-1}=g_{\Lambda(B)}\circ\pi_{\Lambda(B)}$.

Indeed, using the notations from Claim 3.2, we construct an increasing
sequence $B(n)$ of countable subsets of $A$ such that $B(0)=B$ and
$B(n)=\Gamma(B(n-1))$ for every $n\geq 1$. Let $\Lambda(B)=\bigcup_{n\geq
1}B(n)$. Then for every $f\in K$ there exist sequences $\\{f_{n}\\}$ and
$\\{g_{n}\\}$ satisfying the following conditions:

  * •

$f_{n},g_{n}\in C(X_{B(n)},X_{B(n-1)})$;

  * •

$\pi_{B(n-1)}\circ f=f_{n}\circ\pi_{B(n)}$ and $\pi_{B(n-1)}\circ
f^{-1}=g_{n}\circ\pi_{B(n)}$.

The last two conditions imply that if $f\in K$, then for every $x,y\in X$ with
$\pi_{\Lambda(B)}(x)=\pi_{\Lambda(B)}(y)$ we have
$\pi_{\Lambda(B)}(f(x))=\pi_{\Lambda(B)}(f(y))$ and
$\pi_{\Lambda(B)}(f^{-1}(x))=\pi_{\Lambda(B)}(f^{-1}(y))$. Therefore, there
exist homeomorphisms
$f_{\Lambda(B)},g_{\Lambda(B)}\in\mathcal{H}(X_{\Lambda(B)})$ satisfying the
required conditions.

Now, for every $\alpha<\omega(\tau)$ take a countable set
$\Lambda(\alpha)\subset A$ satisfying the hypotheses of Claim 3.3 with
$B=\\{\alpha\\}$. Let $A(\alpha)=\bigcup_{\gamma<\alpha}\Lambda(\gamma)$ if
$\alpha$ is a limit ordinal, and $A(\alpha)=A(\alpha-1)\cup\Lambda(\alpha)$
otherwise. Since for every $\alpha\in A$ and $f\in K$ there exist
$f_{\Lambda(\alpha)},g_{\Lambda(\alpha)}\in\mathcal{H}(X_{\Lambda(\alpha)})$
with $\pi_{\Lambda(\alpha)}\circ
f=f_{\Lambda(\alpha)}\circ\pi_{\Lambda(\alpha)}$ and
$\pi_{\Lambda(\alpha)}\circ
f^{-1}=g_{\Lambda(\alpha)}\circ\pi_{\Lambda(\alpha)}$, we have
$\pi_{A(\alpha)}(f(x))=\pi_{A(\alpha)}(f(y))$ and
$\pi_{A(\alpha)}(f^{-1}(x))=\pi_{A(\alpha)}(f^{-1}(y))$ for any pair $x,y\in
X$ with $\pi_{A(\alpha)}(x)=\pi_{A(\alpha)}(y)$. This yields a homeomorphism
$f_{\alpha}\in\mathcal{H}(X_{A(\alpha)})$ such that $\pi_{A(\alpha)}\circ
f=f_{\alpha}\circ\pi_{A(\alpha)}$. ∎

Next theorem is an analogue of Mednikov’s result [5, Corollary 3] stating that
$\mathcal{H}([0,1]^{A})$ is an absolute extensor for compact spaces.

######  Theorem 3.4.

Let $X=\prod_{\alpha\in A}X_{\alpha}$ be a product of compact metric spaces.
Then $\mathcal{H}(X)$ is an absolute extensor for zero-dimensional compact
spaces.

###### Proof.

Suppose $Y$ is a 0-dimensional compact space and $g:P\to\mathcal{H}(X)$ be a
map, where $P$ is closed in $Y$. If $A$ is countable, then $\mathcal{H}(X)$ is
a complete separable metric space. So, we can consider $\mathcal{H}(X)$ as a
closed subset of $l_{2}$ and find a continuous extension $\widetilde{g}:Y\to
l_{2}$ of $g$. By the well-known factorization theorem, there is a metric
0-dimensional compactum $Y_{0}$ and a maps $g_{0}:Y_{0}\to l_{2}$ and $h:Y\to
Y_{0}$ with $\widetilde{g}=g_{0}\circ h$. Then $P_{0}=h(P)$ is a closed subset
of $Y_{0}$. Since $Y_{0}$ is $0$-dimensional, there is a retraction
$r:Y_{0}\to P_{0}$. Finally, the composition $(g_{0}|P_{0})\circ r\circ h$ is
a map from $Y$ to $\mathcal{H}(X)$ extending $g$.

Assume $A$ is an uncountable set of cardinality $\tau$. Then $A$ can be
covered by a family $\xi=\\{A(\alpha):\alpha\in\omega(\tau)\\}$ satisfying the
hypotheses of Proposition 3.1 with $K=g(P)$. Then for every
$\alpha\in\omega(\tau)$ and $f\in K$ we have

  * (*)

$\pi^{A(\alpha+1)}_{A(\alpha)}\circ
f_{\alpha+1}=f_{\alpha}\circ\pi^{A(\alpha+1)}_{A(\alpha)}$.

Denote by $\mathcal{H}_{\xi}(X)$ the subspace of $\mathcal{H}(X)$ consisting
of all $f$ with the following property: For every $\alpha$ there is
$f_{\alpha}\in\mathcal{H}(X_{A(\alpha)})$ such that $f_{\alpha}$ and
$f_{\alpha+1}$ satisfy $(*)$. Therefore, $K\subset\mathcal{H}_{\xi}(X)$. So,
it suffice to show that $\mathcal{H}_{\xi}(X)$ is an absolute extensor for
0-dimensional compacta.

Since $X_{A(\alpha+1)}=X_{A(\alpha)}\times X_{A(\alpha+1)\backslash
A(\alpha)}$, the map $f_{\alpha+1}$ is of the form
$f_{\alpha+1}(x,y)=(f_{\alpha}(x),g(x,y))$, where $x\in X_{A(\alpha)}$ and
$y\in X_{A(\alpha+1)\backslash A(\alpha)}$ and $g$ is a map from
$X_{A(\alpha+1)}$ into $X_{A(\alpha+1)\backslash A(\alpha)}$ such that for any
$x\in X_{A(\alpha)}$ the map $\varphi_{g}(x)$, $\varphi_{g}(x)(y)=g(x,y)$, is
a homeomorphism on $X_{A(\alpha)\backslash A(\alpha)}$. Therefore, by [3,
Theorem 3.4.9], the correspondence $\varphi_{g}\leftrightarrow g$ is a
homeomorphism between $C(X_{A(\alpha)},\mathcal{H}(X_{A(\alpha)\backslash
A(\alpha)}))$ and the subset of $C(X_{A(\alpha+1)},X_{A(\alpha+1)\backslash
A(\alpha)})$ consisting of all $g$ such that for each $x\in X_{A(\alpha)}$ the
map $\varphi_{g}(x)$ belongs to $\mathcal{H}(X_{A(\alpha)\backslash
A(\alpha)})$. Hence, the correspondence
$(f_{\alpha},\varphi_{g})\leftrightarrow f_{\alpha+1}$ provides a
homeomorphism between the spaces
$C(X_{A(\alpha)},\mathcal{H}(X_{A(\alpha)\backslash A(\alpha)}))$ and
$\mathcal{H}_{\alpha}(X_{A(\alpha+1)})$, where
$\mathcal{H}_{\alpha}(X_{A(\alpha+1)})$ consists of all homeomorphisms
$f_{\alpha+1}$ on $X_{A(\alpha+1)}$ satisfying equality $(*)$. This means that
there is one-to-one correspondence between $\mathcal{H}_{\xi}(X)$ and the
product
$\mathcal{H}(X_{A(0)})\times\Pi_{\alpha<\omega(\tau)}C(X_{A(\alpha)},\mathcal{H}(X_{A(\alpha+1)\backslash
A(\alpha)}))$. This correspondence is a homeomorphism when all function spaces
carry the compact-open topology.

It remains to show that each multiple in this product is an absolute extensor
for 0-dimensional compacta. This is true for $\mathcal{H}(X_{A(0)})$ because
$A(0)$ is countable. Let’s prove that each
$C(X_{A(\alpha)},\mathcal{H}(X_{A(\alpha+1)\backslash A(\alpha)}))$ is also an
absolute extensor for 0–dimensional compacta. To this end, take a pair
$L\subset Z$ of 0-dimensional compacta and a map $\theta$ from $L$ to
$C(X_{A(\alpha)},\mathcal{H}(X_{A(\alpha+1)\backslash A(\alpha)}))$. Since
$\mathcal{H}(X_{A(\alpha+1)\backslash A(\alpha)})$ is a separable complete
metric space and $X_{A(\alpha)}$ is a compactum of weight $\leq\tau$,
$C(X_{A(\alpha)},\mathcal{H}(X_{A(\alpha+1)\backslash A(\alpha)}))$ is a
complete metric space of weight $\leq\tau$. So,
$C(X_{A(\alpha)},\mathcal{H}(X_{A(\alpha+1)\backslash A(\alpha)}))$ can be
embedded in $l_{2}(\tau)$ and $\theta$ can be extended to a map $\eta:Z\to
l_{2}(\tau)$. Because the image $\eta(Z)$ is a metric compactum, we can repeat
the arguments from the first paragraph of our proof and obtain an extension
$\widetilde{\theta}:Z\to C(X_{A(\alpha)},\mathcal{H}(X_{A(\alpha+1)\backslash
A(\alpha)}))$ of $\theta$. ∎

Everywhere below by $\mathfrak{C}$ we denote the Cantor set.

######  Corollary 3.5.

Let $P$ and $K$ be proper closed subsets of $\mathfrak{C}^{A}$ and $f$ be a
homeomorphism between $P$ and $K$. Suppose there exist a proper subset
$B\subset A$ and a homeomorphism $f_{B}$ between $P_{B}$ and $K_{B}$ such that

  * •

$P=P_{B}\times\mathfrak{C}^{A\backslash B}$ and
$K=K_{B}\times\mathfrak{C}^{A\backslash B}$;

  * •

$f_{B}\circ\pi_{B}=\pi_{B}\circ f$;

  * •

$f_{B}$ can be extended to a homeomorphism
$\widetilde{f}_{B}\in\mathcal{H}(\mathfrak{C}^{B})$;

Then $f$ can be extended to a homeomorphism
$\widetilde{f}\in\mathcal{H}(\mathfrak{C}^{A})$.

###### Proof.

Since $f_{B}\circ\pi_{B}=\pi_{B}\circ f$, $f$ is of the form
$f(x,y)=(f_{B}(x),h(x,y))$ with $(x,y)\in P_{B}\times\mathfrak{C}^{A\backslash
B}$ such that for each $x\in P_{B}$ the map $\varphi_{x}$, defined by
$\varphi_{x}(y)=h(x,y)$, belongs to $\mathcal{H}(\mathfrak{C}^{A\backslash
B})$. So, we have a map $\varphi:P_{B}\to\mathcal{H}(\mathfrak{C}^{A\backslash
B})$, see [3, Theorem 3.4.9]. By Theorem 3.4, we can extend $\varphi$ to a map
$\widetilde{\varphi}:\mathfrak{C}^{B}\to\mathcal{H}(\mathfrak{C}^{A\backslash
B})$ and define $\widetilde{h}:\mathfrak{C}^{A}\to\mathfrak{C}^{A\backslash
B}$, $\widetilde{h}(x,y)=\widetilde{\varphi}(x)(y)$, where
$(x,y)\in\mathfrak{C}^{B}\times\mathfrak{C}^{A\backslash B}$. Finally,
$\widetilde{f}(x,y)=(\widetilde{f}_{B},\widetilde{h}(x,y))$ provides a
homeomorphism in $\mathcal{H}(\mathfrak{C}^{A})$ extending $f$. ∎

##  4\. Extending homeomorphisms

In this section we prove the following generalization of Theorem 1.2.

######  Theorem 4.1.

Let $f$ be a homeomorphism between closed $\lambda$-negligible subsets $P$ and
$K$ of $D^{\tau}$ with $\lambda\leq\tau$ such that $P=P^{(\lambda)}$ and
$K=K^{(\lambda)}$. Then $f$ can be extended to a homeomorphism on $D^{\tau}$.

######  Lemma 4.2.

Let $X,Y$ be 0-dimensional paracompact spaces and $\mathfrak{C}$ be the Cantor
set. Suppose $P^{\prime}\subset X\times\mathfrak{C}$, $K^{\prime}\subset
Y\times\mathfrak{C}$ are closed sets such that $\pi_{X}(P^{\prime})=X$,
$\pi_{Y}(K^{\prime})=Y$ and the sets
$\pi_{\mathfrak{C}}((\\{x\\}\times\mathfrak{C})\cap P^{\prime})$ and
$\pi_{\mathfrak{C}}((\\{y\\}\times\mathfrak{C})\cap K^{\prime})$ are nowhere
dense in $\mathfrak{C}$ for all $x\in X$ and $y\in Y$. Let $f:P^{\prime}\to
K^{\prime}$ and $g:X\to Y$ be homeomorphisms with $g\circ\pi_{X}=\pi_{Y}\circ
f$. Then $f$ can be extended to a homeomorphism
$\widetilde{f}:X\times\mathfrak{C}\to Y\times\mathfrak{C}$ such that
$g\circ\pi_{X}=\pi_{Y}\circ\widetilde{f}$.

###### Proof.

For any $x\in X$ let $\Phi(x)$ be the set of all
$h\in\mathcal{H}(\mathfrak{C})$ such that $f(x,c)=(g(x),h(c))$ for all
$c\in\pi_{X}^{-1}(x)\cap P^{\prime}$. Since for every $x\in X$ the restriction
$f|\pi_{X}^{-1}(x)\cap P^{\prime}$ is a homeomorphism between the nowhere
dense subsets $\pi_{X}^{-1}(x)\cap P^{\prime}$ and $\pi_{Y}^{-1}(g(x))\cap
K^{\prime}$ of $\mathfrak{C}$, by [4], there is
$h_{x}\in\mathcal{H}(\mathfrak{C})$ extending $f|\pi_{X}^{-1}(x)\cap
P^{\prime}$. Hence, $\Phi(x)\neq\varnothing$. Moreover, the sets $\Phi(x)$ are
closed in $\mathcal{H}(\mathfrak{C})$ equipped with the compact-open topology.
So, we have a set-valued map
$\Phi:X\rightsquigarrow\mathcal{H}(\mathfrak{C})$. One can show that if $\Phi$
admits a continuous selection $\phi:X\to\mathcal{H}(\mathfrak{C})$, then the
map $\widetilde{f}:X\times\mathfrak{C}\to Y\times\mathfrak{C}$, defined by
$\widetilde{f}(x,c)=(g(x),\phi(x)(c))$, is the required homeomorphism
extending $f$. Therefore, according to Michael’s [7] zero-dimensional
selection theorem, it suffices to show that $\Phi$ is lower semi-continuous.

To prove that, let $x^{*}\in X$ be a fixed point and $h^{*}\in\Phi(x^{*})\cap
W$, where $W$ is open in $\mathcal{H}(\mathfrak{C})$. We can assume that $W$
is of the form $\\{h\in\mathcal{H}(\mathfrak{C}):h(U_{i})\subset
V_{i},i=1,2,..k\\}$, where $\\{U_{i}\\}_{i=1}^{k}$ is a clopen disjoint cover
of $\\{x^{*}\\}\times\mathfrak{C}$ and $\\{V_{i}\\}_{i=1}^{k}$ is a disjoint
clopen cover of $\\{g(x^{*})\\}\times\mathfrak{C}$. We extend the sets $U_{i}$
and $V_{i}$ to clopen sets $\widetilde{U}_{i}\subset X\times\mathfrak{C}$ and
$\widetilde{V}_{i}\subset Y\times\mathfrak{C}$ such that

  * (1)

$\widetilde{U}_{i}=O(x^{*})\times U_{i}$ and
$\widetilde{V}_{i}=g(O(x^{*}))\times V_{i}$, where $O(x^{*})$ is a clopen
neighborhood of $x^{*}$ in $X$;

  * (2)

$O(x^{*})$ can be chosen so small that $f(\widetilde{U}_{i}\cap
P^{\prime})\subset\widetilde{V}_{i}\cap K^{\prime}$.

We are going to show that for every $x\in O(x^{*})$ there exists
$h_{x}\in\Phi(x)\cap W$. We fix such $x$ and observe that all sets
$\widetilde{U}_{i}(x)=\widetilde{U}_{i}\cap(\\{x\\}\times\mathfrak{C})$ and
$\widetilde{V}_{i}(x)=\widetilde{V}_{i}\cap(\\{g(x)\\}\times\mathfrak{C})$ are
compact and perfect. Moreover, $P_{i}^{\prime}(x)=\widetilde{U}_{i}(x)\cap
P^{\prime}$ and $K_{i}^{\prime}(x)=\widetilde{V}_{i}\cap K^{\prime}$ are
nowhere dense sets in $\widetilde{U}_{i}(x)$ and $\widetilde{V}_{i}(x)$,
respectively, and $f^{x}_{i}=f|(\widetilde{U}_{i}(x)\cap P^{\prime})$ is a
homeomorphism between $P_{i}^{\prime}(x)$ and $K_{i}^{\prime}(x)$ for every
$i$. Hence, by [4], there exist homeomorphisms
$\widetilde{f}^{x}_{i}:\widetilde{U}_{i}(x)\to\widetilde{V}_{i}(x)$ extending
$f^{x}_{i}$. Because $\\{\widetilde{U}_{i}(x)\\}_{i=1}^{k}$ and
$\\{\widetilde{V}_{i}(x)\\}_{i=1}^{k}$ are disjoint clopen covers of
$\pi_{X}^{-1}(x)$ and $\pi_{Y}^{-1}(g(x))$, respectively, the homeomorphisms
$\widetilde{f}^{x}_{i}$, $i=1,2,..,k$, provide a homeomorphism
$h_{x}^{\prime}$ between $\pi_{X}^{-1}(x)$ and $\pi_{Y}^{-1}(g(x))$ extending
$f|\pi_{X}^{-1}(x)\cap P^{\prime}$. Then the equality
$h_{x}(c)=h_{x}^{\prime}(x,c)$, $c\in\mathfrak{C})$, defines a homeomorphism
from $\mathcal{H}(\mathfrak{C})$ with $h_{x}\in\Phi(x)\cap W$. Therefore,
$\Phi$ is lower semi-continuous. ∎

Proof of Theorem $4.1$. We identify $D^{\tau}$ with $D^{A}$, where $A$ is a
set of cardinality $\tau$. We first consider the case when $\lambda=\tau$. We
already observed that the theorem is true when $A$ is countable. So, let
$A=\\{\alpha:\alpha<\omega(\tau)\\}$ be uncountable and consider two copies of
$D^{A}=\mathfrak{C}^{A}$, one containing $P$ and the other containing $K$.
Each of the open sets $\mathfrak{C}^{A}\setminus P$ and
$\mathfrak{C}^{A}\setminus K$ contains dense functionally open sets, say
$V(P)$ and $V(K)$, and choose a countable set $C\subset A$ such that
$\pi_{C}^{-1}(\pi_{C}(V(P)))=V(P)$ and $\pi_{C}^{-1}(\pi_{C}(V(K)))=V(K)$.
Hence, $P_{B}$ and $K_{B}$ are nowhere dense subsets of $\mathfrak{C}^{B}$ for
any set $B\subset A$ containing $C$. Next, using Corollary 2.3, we can cover
$A$ by an increasing family $\\{A(\alpha):\alpha<\omega(\tau)\\}$ and find
homeomorphisms $f_{\alpha}:P_{A(\alpha}\to K_{A(\alpha)}$ satisfying the
following conditions:

  * (3)

$A_{\alpha}=\bigcup_{\beta<\alpha}A(\beta)$ if $\alpha$ is a limit ordinal and
the cardinality of each $A(\alpha)$ is less than $\tau$;

  * (4)

$A(\alpha+1)\backslash A(\alpha)$ is countable and $C\subset A(\alpha)$ for
all $\alpha$;

  * (5)

$\displaystyle\pi_{A(\alpha)}\circ f=f_{\alpha}\circ\pi_{A(\alpha)}$;

  * (6)

$\pi_{A(\alpha+1)\backslash A(\alpha)}(P)$ and $\pi_{A(\alpha+1)\backslash
A(\alpha)}(K)$ are nowhere dense sets in $\mathfrak{C}^{A(\alpha+1)\backslash
A(\alpha)}$.

It remains to prove that each $f_{\alpha}$ can be extended to a homeomorphism
$\widetilde{f}_{\alpha}\in\mathcal{H}(\mathfrak{C}^{A(\alpha)})$ such that

  * (7)

$\displaystyle\pi_{A(\alpha)}^{A(\alpha+1)}\circ\widetilde{f}_{\alpha+1}=\widetilde{f}_{\alpha}\circ\pi_{A(\alpha)}^{A(\alpha+1)}$.

The proof is by transfinite induction. The first extension $\widetilde{f}_{1}$
exists by [4] because $P_{A(1)}$ and $K_{A(1)}$ are nowhere dense subsets of
$\mathfrak{C}^{A(1)}$. If $\widetilde{f}_{\alpha}$ is already defined for all
$\alpha<\beta$, where $\beta$ is a limit ordinal, then item (3) implies the
existence of $\widetilde{f}_{\beta}$. Therefore, we need only to define
$\widetilde{f}_{\alpha+1}$ provided $\widetilde{f}_{\alpha}$ exists.

To this end, consider the spaces $X=P_{A(\alpha)}$ and $Y=K_{A(\alpha)}$, the
sets $P^{\prime}=P_{A(\alpha+1)}\subset
X\times\mathfrak{C}^{A(\alpha+1)\backslash A(\alpha)}$,
$K^{\prime}=K_{A(\alpha+1)}\subset Y\times\mathfrak{C}^{A(\alpha+1)\backslash
A(\alpha)}$ and the homeomorphisms $f_{\alpha+1}$, $f_{\alpha}$. For any $x\in
X$ and $y\in Y$ let

$P^{\prime}(x)=P^{\prime}\cap(\\{x\\}\times\mathfrak{C}^{A(\alpha+1)\backslash
A(\alpha)}){~{}}\mbox{and}{~{}}K^{\prime}(y)=K^{\prime}\cap(\\{y\\}\times\mathfrak{C}^{A(\alpha+1)\backslash
A(\alpha)}).$

Item (6) yields that $\pi_{A(\alpha+1)\backslash A(\alpha)}(P^{\prime}(x))$
and $\pi_{A(\alpha+1)\backslash A(\alpha)}(K^{\prime}(y))$ are nowhere dense
sets in $\mathfrak{C}^{A(\alpha+1)\backslash A(\alpha)}$ for every $x\in
P_{A(\alpha)}$ and $y\in K_{A(\alpha)}$. Therefore, by Lemma 4.2, the
homeomorphism $f_{\alpha+1}$ can be extended to a fiberwise homeomorphism

$f_{\alpha+1}^{\prime}:P_{A(\alpha)}\times\mathfrak{C}^{A(\alpha+1)\backslash
A(\alpha)}\to K_{A(\alpha)}\times\mathfrak{C}^{A(\alpha+1)\backslash
A(\alpha)}.$

Finally, by Corollary 3.5, there is a homeomorphism
$\widetilde{f}_{\alpha+1}\in\mathcal{H}(\mathfrak{C}^{A(\alpha+1)})$
satisfying condition (7).

Suppose now that $P$ and $K$ are $\lambda$-negligible in $\mathfrak{C}^{A}$
and $\lambda<\tau$. Then $A$ is uncountable and $P=P^{(\lambda)}$,
$K=K^{(\lambda)}$. So, by [2], there is a set $B\subset A$ of cardinality
$\lambda$ with $P=\pi_{B}^{-1}(P_{B})$ and $K=\pi_{B}^{-1}(K_{B})$. According
to Proposition 2.1, we can assume that $B$ is $f$-admissible. So, there exists
a homeomorphism $f_{B}:P_{B}\to K_{B}$ with $\pi_{B}\circ
f=f_{B}\circ\pi_{B}$. Moreover, the sets $P_{B}$ and $K_{B}$ are
$\lambda$-negligible in $\mathfrak{C}^{B}$. Hence, by the already established
case of Theorem 4.1, $f_{B}$ can be extended to a homeomorphism
$\widetilde{f}_{B}\in\mathcal{H}(\mathfrak{C}^{B})$. Then, Corollary 3.5
implies the existence of a homeomorphism
$\widetilde{f}\in\mathcal{H}(\mathfrak{C}^{A})$ extending $f$. $\Box$

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