Proof Obviously, this is true for a countable set A, so we suppose that A is uncountable. Let $f^{-1}$ be the inverse of f. Using that P is C-embedded in X and any continuous function on X depends on countably many coordinates (see [Reference Engelking3, Reference Mibu8]), we construct by induction sequences of countable sets $B(n)\subset A$ and maps $f_{B(2n-1)}:P_{B(2n)}\to P_{B(2n-1)}$ and $g_{B(2n)}:P_{B(2n+1)}\to P_{B(2n)}$ such that:

• • $B(1)=C$ , $B(n)\subset B(n+1)$ .

• • $\displaystyle \pi _{B(2n-1)}\circ f=f_{B(2n-1)}\circ \pi _{B(2n)}$ .

• • $\displaystyle \pi _{B(2n)}\circ f^{-1}=g_{B(2n)}\circ \pi _{B(2n+1)}$ .

Then $B=\bigcup _{n=1}^{\infty } B(n)$ is countable and the equality $\pi _B(x)=\pi _B(y)$ implies $\pi _B(f(x))=\pi _B(f(y))$ and $\pi _B(f^{-1}(x))=\pi _B(f^{-1}(y))$ for all $x,y\in P$ . Since $P_B$ is compact, there exist maps $f_B:P_B\to P_B$ and $g_B:P_B\to P_B$ with $\displaystyle \pi _B\circ f=f_B\circ \pi _B$ , $\displaystyle \pi _B\circ f^{-1}=g_B\circ \pi _B$ , and $f_B\circ g_B$ is the identity on $P_B$ . Then $f_B$ is an autohomeomorphism of $P_B$ .

Proof Since $(\{z\}\times X_{A\backslash C})\cap P$ is $\tau $ -negligible in $\{z\}\times X_{A\backslash C}$ for every $z\in P_C$ , the cardinality of $A\backslash C$ is at least $\tau $ . Suppose that $P_{A\backslash C}=X_{A\backslash C}$ . Passing to a subset of P, we may assume that the projection $\pi _{A\backslash C}$ restricted to P is an irreducible map onto $X_{A\backslash C}$ . Denote this map by f and fix $z\in P_C$ . Because f is irreducible, we have

$$ \begin{align*} \pi\chi(f((\{z\}\times X_{A\backslash C})\cap P),X_{A\backslash C})\leq\pi\chi((\{z\}\times X_{A\backslash C})\cap P,P). \end{align*} $$

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 $ . This, according to [Reference Mednikov6, Lemma 6], means that $f((\{z\}\times X_{A\backslash C})\cap P)$ is not $\tau $ -negligible in $X_{A\backslash C}$ . Since $f((\{z\}\times X_{A\backslash C})\cap P)$ is homeomorphic to $(\{z\}\times X_{A\backslash C})\cap P$ and $X_{A\backslash C}$ is homeomorphic to $\{z\}\times X_{A\backslash C}$ , $(\{z\}\times X_{A\backslash C})\cap P$ is not $\tau $ -negligible in $\{z\}\times X_{A\backslash C}$ , a contradiction.

Proof Let $\Gamma \subset A$ be a set of cardinality $<\tau $ . Since P is a $\tau $ -negligible set in X, so are the sets $P(z)=(\{z\}\times X_{A\backslash \Gamma })\cap P$ for all $z\in P_{\Gamma }$ . This implies that each $P(z)$ is $\tau $ -negligible in $\{z\}\times X_{A\backslash \Gamma }$ . Otherwise, $P(z^*)$ would contain a closed $G_{\lambda }$ -set in $\{z^{*}\}\times X_{A\backslash \Gamma }$ for some $z^*\in P_{\Gamma }$ and $\lambda <\tau $ . Because $\{z^*\}\times X_{A\backslash \Gamma }$ is $G_{\mu }$ -set in X, where $\mu $ is the cardinality of $\Gamma $ , $P(z^*)$ contains a $G_{\lambda '}$ -subset of X with $\lambda '=\max \{\lambda ,\mu \}<\tau $ , a contradiction.

Using the above observation, 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:

• • $C_1\subset A\backslash C$ .

• • $C_{n+1}\subset A\backslash \bigcup _{k\leq n}C\cup C_k$ .

• • $P_{C_n}\neq X_{C_n}$ for all n.

Indeed, suppose that we already constructed the sets $C_k$ , $k=1,2,\ldots ,n$ . Since the cardinality of $C_n'=\bigcup _{k\leq n}C\cup C_k$ is $<\tau $ , $(\{z\}\times X_{A\backslash C_n'})\cap P$ is $\tau $ -negligible in $\{z\}\times X_{A\backslash C_n'}$ for every $z\in P_{C_n'}$ . Then, by Lemma 2.2, $P_{A\backslash C_n'}\neq X_{A\backslash C_n'}$ . Hence, we can choose a finite set $C_{n+1}\subset A\backslash C_n'$ and an open set $V\subset X_{C_{n+1}}$ such that $V\times X_{A\backslash (C_n'\cup C_{n+1})}$ is disjoint from $P_{A\backslash C_n'}$ . This implies $P_{C_{n+1}}\neq X_{C_{n+1}}$ .

One can show that $B=\bigcup _{n\geq 1}C\cup C_n$ is the required set.

If $f:P\to P$ 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:

• • $C_1\subset A\backslash C$ ;

• • $C_{n+1}\subset A\backslash \bigcup _{k\leq n}C\cup C_k'$ , where $C_k'=\bigcup _{\alpha \in C_k}B(\alpha )$ ;

• • $P_{C_n}\neq X_{C_n}$ for all n.

Then $B=\bigcup _{n\geq 1}C\cup C_n'$ is f-admissible and satisfies the required conditions.

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 $(x,y)\in P_{B}\times X_{A\backslash B}$ such that for each $x\in P_{B}$ , the map $\varphi _x$ , defined by $\varphi _x(y)=h(x,y)$ , belongs to ${\mathcal H}(X_{A\backslash B})$ . So, we have a map $\varphi :P_{B}\to {\mathcal H}(X_{A\backslash B})$ (see [Reference Engelking4, Theorem 3.4.9]). Because ${\mathcal H}(X_{A\backslash B})$ is a complete separable metric space, it is an absolute extensor for zero-dimensional compacta (for example, this follows from Michael’s zero-dimensional selection theorem [Reference Michael9]). Hence, $\varphi $ can be extended to a map $\widetilde \varphi : X_B\to {\mathcal H}(X_{A\backslash B})$ . Define $\widetilde h:X\to X_{A\backslash B}$ , $\widetilde h(x,y)=\widetilde \varphi (x)(y)$ , where $(x,y)\in X_B\times X_{A\backslash B}$ . Finally, ${\widetilde f}(x,y)=({\widetilde f}_B,\widetilde h(x,y))$ provides a homeomorphism in ${\mathcal H}(X)$ extending f such that ${\widetilde f}_B\circ \pi _B=\pi _B\circ {\widetilde f}$ .

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 $c\in \pi _X^{-1}(x)\cap P'$ . Since $f|(\pi _X^{-1}(x)\cap P')$ is a homeomorphism between the nowhere dense subsets $\pi _{\mathfrak {C}}((\{x\}\times \mathfrak {C})\cap P')$ and $\pi _{\mathfrak {C}}((\{g(x)\}\times \mathfrak {C})\cap P')$ of $\mathfrak {C}$ , Knaster–Reichbach’s theorem [Reference Knaster and Reichbach5] cited above yields a homeomorphism $h_x\in \mathcal {H}(\mathfrak {C})$ extending $f|(\pi _X^{-1}(x)\cap P')$ . Hence, $\Phi (x)\neq \varnothing $ for all $x\in X$ . 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 ${\widetilde f}:X\times \mathfrak {C}\to X\times \mathfrak {C}$ , defined by ${\widetilde f}(x,c)=(g(x),\phi (x)(c))$ , is the required homeomorphism extending f. Therefore, according to Michael’s [Reference Michael9] zero-dimensional selection theorem, it suffices to show that $\Phi $ is lower semi-continuous, i.e., the set $\{x\in X:\Phi (x)\cap W\neq \varnothing \}$ is open in X for any open $W\subset {\mathcal H}(\mathfrak {C})$ .

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 $\{h\in {\mathcal H}(\mathfrak {C}): h(U_i)\subset V_i, i=1,2,\dots ,k\}$ , 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 $\widetilde U_i, \widetilde V_i\subset X\times \mathfrak {C}$ such that:

1. (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. (2) $O(x^*)$ can be chosen so small that $f(\widetilde U_i\cap P')\subset \widetilde V_i\cap P'$ .

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, $\widetilde U_i(x)\cap P'$ and $\widetilde V_i(x)\cap P'$ are nowhere dense sets in $\widetilde U_i(x)$ and $\widetilde V_i(x)$ , respectively, and $f^x_i=f|(\widetilde U_i(x)\cap P')$ is a homeomorphism between $\widetilde U_i(x)\cap P'$ and $\widetilde V_i(x)\cap P'$ . Hence, by Knaster–Reichbach’s theorem [Reference Knaster and Reichbach5], for every i, there exists a homeomorphism ${\widetilde f}^x_i:\widetilde U_i(x)\to \widetilde V_i(x)$ 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 $\pi _X^{-1}(g(x))$ , respectively, the homeomorphisms ${\widetilde f}^x_i$ , $i=1,2,\ldots ,k$ , provide a homeomorphism $h_x'$ between $\pi _X^{-1}(x)$ and $\pi _X^{-1}(g(x))$ extending $f|\pi _X^{-1}(x)\cap P'$ . Then the equality $h_x(c)=h_x'(x,c)$ , $c\in \mathfrak {C}$ , defines a homeomorphism $h_x\in \mathcal {H}(\mathfrak {C})$ with $h_x\in \Phi (x)\cap W$ . Therefore, $\Phi $ is lower semi-continuous.

Proof of Theorem 1.2

We identify $D^{\tau }$ with $D^A$ , where A is a set of cardinality $\tau $ . We already observed that the theorem is true when A is countable. So, let $A=\{\alpha :\alpha <\omega (\tau )\}$ be uncountable. Let show that the proof is reduced to the case of one negligible subset $P\subset D^A$ and an autohomeomorphism $f\in {\mathcal H}(P)$ . Indeed, take two disjoint copies X and Y of $D^A$ with $P\subset X$ and $K\subset Y$ , and let $Q=P\biguplus K$ be the disjoint union of P and K. Obviously, $X\biguplus Y$ is homeomorphic to $D^A$ , Q is negligible in $X\biguplus Y$ , and $f\biguplus f^{-1}$ is an autohomeomorphism of Q. Suppose that $f\biguplus f^{-1}$ can be extended to a homeomorphism $F:X\biguplus Y\to X\biguplus Y$ . Choose two clopen neighborhoods $X'$ and $Y'$ of $P$ and $K$ in $X$ and $Y$ , respectively, with $X\backslash X'\neq\varnothing\neq Y\backslash Y'$ such that $F(X')=Y'$ . Then there is a homeomorphism $G:X\backslash X'\to Y\backslash Y'$ . Hence, $F|X'$ and G provide a homeomorphism ${\widetilde f}:X\to Y$ extending f. Therefore, we can suppose that we have one negligible subset P of $D^A$ and an autohomeomorphism $f\in {\mathcal H}(P)$ .

We identify $D^A$ with $X=\mathfrak {C}^A$ and take a functionally open set $V(P)$ in X which is dense in $X\setminus P$ . Because every continuous function on X depends on countably many coordinates, we can choose a countable set $C\subset A$ such that $\pi _C^{-1}(\pi _C(V(P)))=V(P)$ . Hence, $P_B$ is a nowhere dense subset of $X_B$ for any set $B\subset A$ containing C. Next, using Proposition 2.3, we can cover A by an increasing transfinite family $\{A(\alpha ):\alpha <\omega (\tau )\}$ and find homeomorphisms $f_{\alpha }\in {\mathcal H}(P_{A(\alpha )})$ satisfying the following conditions:

1. (3) $A(1)$ is countable and the cardinality of each $A(\alpha )$ is less than $\tau $ .

2. (4) $A_{\alpha }=\bigcup _{\beta <\alpha }A(\beta )$ if $\alpha $ is a limit ordinal.

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

4. (6) $\displaystyle \pi _{A(\alpha )}\circ f=f_{\alpha }\circ \pi _{A(\alpha )}$ .

5. (7) Each $P_{A(\alpha +1)\backslash A(\alpha )}$ is a nowhere dense set in $X_{A(\alpha +1)\backslash A(\alpha )}$ .

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

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

The proof is by transfinite induction. The first extension ${\widetilde f}_1$ exists by Knaster–Reichbach’s theorem [Reference Knaster and Reichbach5] because $P_{A(1)}$ is nowhere dense in $X_{A(1)}$ . If ${\widetilde f}_{\alpha }$ is already defined for all $\alpha <\beta $ , where $\beta $ is a limit ordinal, then item (4) 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 space $P_{A(\alpha )}\times X_{A(\alpha +1)\backslash A(\alpha )}$ , the set $P'=P_{A(\alpha +1)}\subset P_{A(\alpha )}\times X_{A(\alpha +1)\backslash A(\alpha )}$ , and the homeomorphisms $f_{\alpha +1}$ , $f_{\alpha }$ . For any $x\in P_{A(\alpha )}$ , consider the set

$$ \begin{align*} P'(x)=P'\cap(\{x\}\times X_{A(\alpha+1)\backslash A(\alpha)}). \end{align*} $$

Item (7) yields that $\pi _{A(\alpha +1)\backslash A(\alpha )}(P'(x))$ is nowhere dense in $X_{A(\alpha +1)\backslash A(\alpha )}$ for every $x\in P_{A(\alpha )}$ . Therefore, by Lemma 3.1, the homeomorphism $f_{\alpha +1}$ can be extended to a homeomorphism

$$ \begin{align*} f_{\alpha+1}':P_{A(\alpha)}\times X_{A(\alpha+1)\backslash A(\alpha)}\to P_{A(\alpha)}\times X_{A(\alpha+1)\backslash A(\alpha)} \end{align*} $$

such that $\pi _{A(\alpha )}\circ f_{\alpha +1}'=f_{\alpha }\circ \pi _{A(\alpha )}$ . Finally, by Lemma 2.4, there is a homeomorphism ${\widetilde f}_{\alpha +1}\in {\mathcal H}(\mathfrak {C}^{A(\alpha +1)})$ satisfying condition (8).