Proof We fix a countable base $\mathcal B$ of H which is closed under finite unions, and denote by f the restriction $\overline f|X$ of any $\overline f\in E(H)$ . For every $y\in \overline Y$ , there is a map $\alpha _y:E(H)\to \overline {\mathbb R}$ , $\alpha _y(\overline f)=\overline {\varphi (f)}(y)$ . Since $\varphi $ is uniformly continuous, so is the map $\beta _y:E_p(X)\to \mathbb R$ , $\beta _y(f)=\varphi (f)(y)$ . Let $H=\bigcup _kH_k$ be the union of an increasing sequence $\{H_k\}$ of compact sets. Following Krupski [Reference Krupski12], for every $y\in \overline Y$ and every $p,k\in \mathbb N$ , we define the families

$$ \begin{align*}\mathcal A^k(y)=\{K\subset H_k:K{~}\mbox{is closed and}{~}a(y,K)<\infty\}\end{align*} $$

and

$$ \begin{align*}\mathcal A_p^k(y)=\{K\subset H_k:K{~}\mbox{is closed and}{~}a(y,K)\leq p\},\end{align*} $$

where

$$ \begin{align*}a(y,K)=\sup\{|\alpha_y(\overline f)-\alpha_y(\overline g)|:\overline f,\overline g\in E(H), |\overline f(x)-\overline g(x)|<1{~}\forall x\in K\}.\end{align*} $$

Possibly, some or both of the values $\alpha _y(\overline f),\alpha _y(\overline g)$ from the definition of $a(y,K)$ could be $\pm \infty $ . That’s why we use the following agreements:

1. (2.4) $\infty +\infty =\infty , \infty -\infty =-\infty +\infty =0, -\infty -\infty =-\infty .$

Note that $a(y,\varnothing )=\infty $ since $\varphi $ is surjective.

Using that $E(X)$ and $E(H)$ are $QS$ -algebras on X and H, and following the arguments from Krupski’s paper [Reference Krupski12] (see also the proofs of [Reference Gul’ko7, Proposition 1.4] and [Reference Marciszewski and Pelant16, Proposition 3.1]), one can establish the following claims (for the sake of completeness, we provide the proofs).

This claim follows from the proof of [Reference Krupski12, Proposition 2.1]. Indeed, since $\varphi $ is uniformly continuous there is $p\in \mathbb N$ and a finite set $K\subset X$ such that if $f,g\in E(X)$ and $|f(x)-g(x)|<1/p$ for every $x\in K$ , then $|\alpha _y(\overline f)-\alpha _y(\overline g)|=|\varphi (f)(y)-\varphi (g)(y)|<1$ . Take arbitrary $\overline f,\overline g\in E(H)$ with $|\overline f(x)-\overline g(x)|<1$ for every $x\in K$ and consider the functions $\overline f_m=\overline f+ \frac {m}{p}(\overline g-\overline f)\in E(H)$ for each $m=0,1,\ldots ,p$ . Obviously $|f_m(x)-f_{m+1}(x)|<1/p$ for all $x\in K$ , so $|\alpha _y(\overline f_m)-\alpha _y(\overline f_{m+1})|<1$ . Consequently, $|\alpha _y(\overline f)-\alpha _y(\overline g)|\leq \sum _{m=0}^{p-1}|\alpha _y(\overline f_m)-\alpha _y(\overline f_{m+1})|<p$ . Because K is finite, there is $k\in \mathbb N$ with $K\subset H_k$ . Hence, $K\in \mathcal A_p^k(y)$ .

Consider the sets $Y_p^k=\{y\in \overline Y:\mathcal A_p^k(y)\neq \varnothing \}$ , $p,k\in \mathbb N$ .

We use the proof of [Reference Krupski12, Lemma 2.2]. Suppose $y\not \in Y_p^k$ . Since $y\in Y_p^k$ iff $H_k\in \mathcal A_p^k(y)$ , $H_k\not \in \mathcal A_p^k(y)$ . So, there exist $\overline f,\overline g\in E(H)$ with $|\overline f(x)-\overline g(x)|<1$ for all $x\in H_k$ and $|\alpha _y(\overline f)-\alpha _y(\overline g)|>p$ . Then, $V=\{z\in \overline Y: |\alpha _z(\overline f)-\alpha _z(\overline g)|>p\}$ is a neighborhood of y with $V\cap Y_p^k=\varnothing $ .

Following the proof of [Reference Krupski12, Lemma 2.3], we first show that the set $Z=\{(y,K)\in Y_p^k\times [H_k]^{\leq q}:K\in \mathcal A_p^k(y)\}$ is closed in $Y_p^k\times [H_k]^{\leq q}$ , where $[H_k]^{\leq q}$ denotes the space of all subsets $K\subset H_k$ of cardinality $\leq q$ endowed with the Vietoris topology. Indeed, if $(y,K)\in Y_p^k\times [H_k]^{\leq q}\backslash Z$ , then $K\not \in \mathcal A_p^k(y)$ . Hence, $a(y,K)>p$ and there are $\overline f,\overline g\in E(H)$ such that $|\overline f(x)-\overline g(x)|<1$ for all $x\in K$ and $|\alpha _y(\overline f)-\alpha _y(\overline g)|>p$ . Let $U=\{z\in Y_p^k: |\alpha _z(\overline f)-\alpha _z(\overline g)|>p\}$ and $V=\{x\in H_k:|\overline f(x)-\overline g(x)|<1\}$ . The set $U\times <V>$ is a neighborhood of $(y,K)$ in $Y_p^k\times [H_k]^{\leq q}$ disjoint from Z (here $<V>=\{F\in [H_k]^{\leq q}: F\subset V\}$ ). Since $Y_p^k\times [H_k]^{\leq q}$ is compact and $Y_{p,q}^k$ is the image of Z under the projection $Y_p^k\times [H_k]^{\leq q}\to Y_p^k$ , $Y_{p,q}^k$ is closed in $Y_p^k$ .

For every k, let $Y_k=\bigcup _{p,q}Y_{p,q}^k$ . Obviously, $Y_k\subset \{y\in \overline Y:\mathcal A^k(y)\neq \varnothing \}$ . Since ${H_k\subset H_{k+1}}$ for all k, the sequence $\{Y_k\}$ is increasing. It may happen that $Y_k=\varnothing $ for some k, but Claim 1 implies that $Y\subset \bigcup _{k}Y_{k}$ .

We follow the proof of [Reference Krupski12, Lemma 2.5] to show that $K_1\cap K_2\in \mathcal A^k(y)$ for any $K_1,K_2\in \mathcal A^k(y)$ . Let $\overline f,\overline g\in E(H)$ with $|\overline f(x)-\overline g(x)|<1$ for all $x\in K_1\cap K_2$ and $U=\{x\in H:|\overline f(x)-\overline g(x)|<1\}$ . Take an open set W in H containing $K_1$ with $W\cap K_2\subset U$ and choose $\overline u\in E(H)$ such that $\overline u|K_1=1$ , $\overline u|(H\backslash W)=0$ and $\overline u(x)\in [0,1]$ for all $x\in H$ (see condition (2.3)). Then, $\overline h=\overline u\cdot (\overline f-\overline g)+\overline g\in E(H)$ , $\overline h|K_1=\overline f|K_1$ , $\overline h|(K_2\backslash W)=\overline g|(K_2\backslash W)$ , and $|\overline h(x)-\overline g(x)|<1$ for $x\in K_2$ . Since $K_1\in \mathcal A^k(y)$ and $\overline h|K_1=\overline f|K_1$ , we have $|\alpha _y(\overline f)-\alpha _y(\overline h)|\leq a(y,K_1)<\infty $ . Similarly, $K_2\in \mathcal A^k(y)$ and $|\overline h(x)-\overline g(x)|<1$ for $x\in K_2$ imply $|\alpha _y(\overline h)-\alpha _y(\overline g)|\leq a(y,K_2)<\infty $ . Therefore,

$$ \begin{align*}|\alpha_y(\overline f)-\alpha_y(\overline g)|\leq |\alpha_y(\overline f)-\alpha_y(\overline h)|+|\alpha_y(\overline h)-\alpha_y(\overline g)|\leq a(y,K_1)+a(y,K_2).\end{align*} $$

Note that the last inequality is true if some of $\alpha _y(\overline f), \alpha _y(\overline h), \alpha _y(\overline g)$ are $\pm \infty $ . Indeed, if $\alpha _y(\overline f)=\pm \infty $ , then $|\alpha _y(\overline f)-\alpha _y(\overline h)|<\infty $ implies $\alpha _y(\overline h)=\pm \infty $ . Consequently, ${\alpha _y(\overline g)=\pm \infty} $ because $|\alpha _y(\overline h)-\alpha _y(\overline g)|<\infty $ . Similarly, if $\alpha _y(\overline h)=\pm \infty $ or $\alpha _y(\overline g)=\pm \infty $ , then the other two are also $\pm \infty $ . Hence, $a(y,K_1\cap K_2)\leq a(y,K_1)+a(y,K_2)$ , which means that $K_1\cap K_2\neq \varnothing $ (otherwise, $a(y,K_1\cap K_2)=\infty $ ) and $K_1\cap K_2\in \mathcal A^k(y)$ .

Since each family $\mathcal A^k(y)$ , $y\in Y_k$ , consists of compact subsets of $H_k$ , $K(y,k)=\bigcap \mathcal A^k(y)$ is nonempty and compact.

Let $y\in Y_k$ . We already observed that $K(y,k)$ is compact and nonempty. Since $y\in Y_{p,q}^k$ for some $p,q$ , $\mathcal A^k(y)$ contains finite sets. Hence, $K(y,k)$ is also finite and $K(y,k)\in \mathcal A^k(y)$ because it is an intersection of finitely many elements of $\mathcal A^k(y)$ . If $y\in Y$ , then by Claim 1, there is k such that $\mathcal A^k(y)$ contains a finite subset of X. Since $K(y,k)$ is the minimal element of $\mathcal A^k(y)$ , it is also a subset of X.

Following [Reference Gul’ko7], for every k, we define $M^k(p,1)=Y_{p,1}^k$ and $M^k(p,q)=Y_{p,q}^k\backslash Y_{2p,q-1}^k$ if $q\geq 2$ .

Since $M^k(p,q)\subset Y^k_{p,q}\subset Y_k$ , $\bigcup \{M^k(p,q):p,q=1,2,\ldots \}\subset Y_k$ . If $y\in Y_k$ , then $K(y,k)\in \mathcal A^k(y)$ is a finite subset of $H_k$ . Assume $|K(y,k)|=q$ and $a(y,K(y,k))\leq p$ for some $p,q$ . So, $y\in Y_{p,q}^k$ . Moreover, $y\not \in Y_{2p,q-1}^k$ , otherwise, there would be $K\in \mathcal A^k(y)$ with $a(y,K)\leq 2p$ and $|K|\leq q-1$ . The last inequality contradicts the minimality of $K(y,k)$ . Hence, $y\in M^k(p,q)$ which shows that $Y_k=\bigcup \{M^k(p,q):p,q=1,2,\ldots \}$ .

Suppose $y\in M^k(p,q)$ . Then, there exists a set $K\in \mathcal A^k(y)$ with $a(y,K)\leq p$ and $|K|\leq q$ . Since, $y\not \in Y_{2p,q-1}^k$ , $|K|=q$ . If there exists another $K'\in \mathcal A^k(y)$ with $a(y,K')\leq p$ and $|K'|=q$ , then $K\cap K'\neq \varnothing $ , $|K\cap K'|\leq q-1$ and, by Claim 4, $a(y,K\cap K')\leq a(y,K)+ a(y,K')\leq 2p$ . This means that $y\in Y_{2p,q-1}^k$ , a contradiction. Hence, there exists a unique $K_{kp}(y)\in \mathcal A^k(y)$ such that $a(y,K_{kp}(y))\leq p$ and $|K_{kp}(y)|=q$ .

For every q, let $[H_k]^q$ denote the set of all q-points subsets of $H_k$ endowed with the Vietoris topology.

Because $K_{kp}(y)\subset H_k$ consists of q points for all $y\in M^k(p,q)$ , it suffices to show that if $K_{kp}(y)\cap U\neq \varnothing $ for some open $U\subset H$ , then there is a neighborhood V of y in $\overline Y$ with $K_{kp}(z)\cap U\neq \varnothing $ for all $z\in V\cap M^k(p,q)$ . We can assume that $K_{kp}(y)\cap U$ contains exactly one point $x_0$ .

Let $q\geq 2$ , so $K_{kp}(y)=\{x_0,x_1,\ldots ,x_{q-1}\}$ . Since $y\not \in Y_{2p,q-1}^k$ we have $a(y,K)>2p$ , where $K=\{x_1,\ldots ,x_{q-1}\}$ . Hence, there are $\overline f, \overline g\in E(H)$ such that $|\overline f(x)-\overline g(x)|<1$ for all $x\in K$ and $|\alpha _y(\overline f)-\alpha _y(\overline g)|>2p$ . The last inequality implies $\overline f(x_0)\neq \overline g(x_0)$ , otherwise, $a(y,K_{kp}(y))$ would be greater than $2p$ (recall that $y\in M^k(p,q)$ implies $a(y,K_{kp}(y))\leq p$ ). So, at least one of the numbers $\overline f(x_0), \overline g(x_0)$ is not zero. Without loss of generality, we can assume that $\overline f(x_0)>0$ , and let r be a rational number with $\frac {-1+\delta }{\overline f(x_0)}<r<\frac {1+\delta }{\overline f(x_0)}$ , where $\delta =\overline f(x_0)-\overline g(x_0)$ . Then, $-1<(1-r)\overline f(x_0)-\overline g(x_0)<1$ , and choose $\overline h_1\in E(H)$ such that $\overline h_1(x_0)=r$ and $\overline h_1(x)=0$ for all $x\not \in U$ . Consider the function $\overline h=(1-\overline h_1)\overline f$ . Clearly, $\overline h(x_0)=(1-r)\overline f(x_0)$ and $\overline h(x)=\overline f(x)$ if $x\not \in U$ . Hence, $\overline h\in E(H)$ and $|\overline h(x)-\overline g(x)|<1$ for all $x\in K_{kp}(y)$ . This implies $|\alpha _y(\overline h)-\alpha _y(\overline g)|\leq p$ . Then,

$$ \begin{align*}|\alpha_y(\overline f)-\alpha_y(\overline h)|\geq |\alpha_y(\overline f)-\alpha_y(\overline g)|-|\alpha_y(\overline h)-\alpha_y(\overline g)|>2p-p=p.\end{align*} $$

Observe that it is not possible $\alpha _y(\overline f)=\alpha _y(\overline h)=\pm \infty $ because $|\alpha _y(\overline h)-\alpha _y(\overline g)|\leq p$ would imply $\alpha _y(\overline g)=\pm \infty $ . Then, $|\alpha _y(\overline f)-\alpha _y(\overline g)|=0$ , a contradiction.

The set $V=\{z\in \overline Y:|\alpha _z(\overline f)-\alpha _z(\overline h)|>p\}$ is a neighborhood of y. Since $\overline h(x)=\overline f(x)$ for all $x\not \in U$ , $K_{kp}(z)\cap U=\varnothing $ for some $z\in V\cap M^k(p,q)$ would imply $|\alpha _z(\overline h)-\alpha _z(\overline f)|\leq p$ , a contradiction. Therefore, $K_{kp}(z)\cap U\neq \varnothing $ for $z\in V\cap M^k(p,q)$ .

If $q=1$ , then $K_{kp}(y)=\{x_0\}$ and $K(y,k)=K_{kp}(y)$ . So, $H_k\backslash U\not \in \mathcal A^k(y)$ (otherwise, $K(y,k)\subset H_k\backslash U$ ). Hence, there exist $\overline f,\overline g\in E(H)$ such that $|\overline f(x)-\overline g(x)|<1$ for all $x\in H_k\backslash U$ and $|\alpha _y(\overline f)-\alpha _y(\overline g)|>2p$ . Define $\overline h\in E(H)$ as in the previous case and use the same arguments to complete the proof.

Since $Y_{p,q}^k$ are compact subsets of $\overline Y$ , each $M^k(p,q)$ is a countable union of compact subsets $\{F_n^k(p,q):n=1,2,\ldots \}$ of $\overline Y$ . So, by Claim 6, $Y_k=\bigcup \{F_n^k(p,q):n,p,q=1,2,\ldots \}$ . According to Claim 7, all maps $\Phi _{kpq}^n=\Phi _{kpq}|F_n^k(p,q):F_n^k(p,q)\to [H_k]^q$ are continuous. Moreover, since $Y\subset \bigcup _kY_k$ , $Y\subset \bigcup \{F_n^k(p,q):n,p,q,k=1,2,\ldots \}$ .

We follow the arguments from the proof of [Reference Gorak, Krupski and Marciszewski6, Theorem 4.2]. Fix $z\in F_n^k(p,q)$ for some $n,p,q, k$ and let $A=\{y\in F_n^k(p,q):K_{kp}(y)=K_{kp}(z)\}$ . Since $\Phi _{kpq}^n$ is a perfect map, A is compact. Suppose A is infinite, so it contains a convergent sequence $S=\{y_m\}$ of distinct points. Because $E(\overline Y)$ contains a $QS$ -algebra $\Gamma $ on $\overline Y$ , for every $y_m$ there exist its neighborhood $U_m$ in $\overline Y$ and a function $\overline g_m\in \Gamma $ , $\overline g_m:\overline Y\to [0,2p]$ such that: $U_m\cap S=\{y_m\}$ , $\overline g_m(y_m)=2p$ , and $\overline g_m(y)=0$ for all $y\not \in U_m$ . Since $\varphi $ is c-good, for each m, there is $f_m\in E(X)$ with $\varphi (f_m)=\overline g_m|Y=g_m$ and $||f_m||\leq c||g_m||$ . So, $||\overline f_m||\leq 2pc$ , $m=1,2,\ldots $ and the sequence $\{\overline f_m\}$ is contained in the compact set $[-2pc,2pc]^{H}$ . Hence, $\{\overline f_m\}$ has an accumulation point in $[-2pc,2pc]^{H}$ . This implies the existence of $i\neq j$ such that $|\overline f_i(x)-\overline f_j(x)|<1$ for all $x\in K_{kp}(z)$ . Consequently, since $K_{kp}(y_j)=K_{kp}(z)$ , $|\alpha _{y_j}(\overline f_j)-\alpha _{y_j}(\overline f_i)|\leq p$ . On the other hand, $\alpha _{y_j}(\overline f_j)=\overline {\varphi (f_j)}(y_j)=\overline g_j(y_j)=2p$ and $\alpha _{y_j}(\overline f_i)=\overline {\varphi (f_i)}(y_j)=\overline g_i(y_j)=0$ , so $|\alpha _{y_j}(\overline f_j)-\alpha _{y_j}(\overline f_i)|=2p$ , a contradiction.

Now, we can complete the proof of Proposition 2.1. Suppose H has a property $\mathcal P$ satisfying conditions (a)–(c). Then so does $H_k^q$ for each $k,q$ because $H_k^q$ is closed in $H^q$ . We claim that the space $[H_k]^q$ also has the property $\mathcal P$ . Indeed, let $\mathcal C$ be a countable base of $H_k$ . For every q-tuple $(U_1,\ldots ,U_q)$ of elements of $\mathcal C$ with pairwise disjoint closures, the closed set

$$ \begin{align*}W(U_1,\ldots, U_q) =\{\{x_1,\ldots,x_q\}:x_i\in\overline U_i, i=1,\ldots,q\}\subset [H_k]^q\end{align*} $$

is homeomorphic to the closed subset $\overline U_1\times \cdots \times \overline U_q$ of $H_k^q$ . Hence, $W(U_1,\ldots ,U_q)\in \mathcal P$ . Clearly, the space $[H_k]^q$ can be covered by countably many sets of the form $W(U_1,\ldots , U_q)$ , therefore, it belongs to $\mathcal P$ . Finally, since the maps $\Phi _{kpq}^n:F_n^k(p,q)\to [H_k]^q$ are perfect and have finite fibers, each $F_n^k(p,q)$ has the property $\mathcal P$ . Therefore, by condition (b), $Y_\infty =\bigcup \{F_n^k(p,q):n,p,q,k=1,2,\ldots \}$ has the property $\mathcal P$ .

It remains to show that all powers of $Y_\infty $ also have the property $\mathcal P$ . Simplifying the notations, we observed that $Y_\infty =\bigcup _{m=1}^\infty F_m$ , where every $F_m$ is a compact set admitting a map $\Phi _m$ with finite fibers onto a compact subset of $H^m$ . Then, for every k, we have

$$ \begin{align*}Y_\infty^k=\bigcup_{(m_1,m_2,\ldots,m_k)}F_{m_1}\times F_{m_2}\times\cdots\times F_{m_k}.\end{align*} $$

Consequently, $\prod _{i=1}^k\Phi _{m_i}:\prod _{i=1}^kF_{m_i}\to H^{m_1+m_2+\cdots +m_k}$ is a map which fibers are products of k-many finite sets. Hence, the fibers of $\prod _{i=1}^k\Phi _{m_i}$ are finite. Because $H^{m_1+m_2+\cdots +m_k}\in \mathcal P$ , so is $\prod _{i=1}^kF_{m_i}$ . Finally, by property $(b)$ , $Y_\infty ^k\in \mathcal P$ .

Proof Let $X_0$ be a metrizable compactification of X and for every $f\in E(X)$ denote by $Z_f$ the closure of $f(X)$ in $\mathbb R$ . Consider the diagonal product h of the maps j and $\triangle \{f:f\in E(X)\}$ , where $j:X\hookrightarrow X_0$ is the embedding. Then, the closure $X_1$ of $h(X)$ in the product $X_0\times \prod _{f\in E(X)}Z_f$ is a compactification of X such that every $f\in E(X)$ can be extended over $X_1$ . Let $\theta :\beta X\to X_1$ be the map witnessing that $\beta X$ is a compactification of X larger than $X_1$ . Since $\dim \beta X=0$ , by the Mardešić factorization theorem [Reference Engelking2, Theorem 3.4.1] there is a metrizable compactum $\overline X$ and maps $\nu :\beta X\to \overline X$ and $\eta :\overline X\to X_1$ such that $\dim \overline X=0$ and $\theta =\eta \circ \nu $ . Evidently, $\nu |i(X)$ is a homeomorphism, where $i:X\hookrightarrow \beta X$ is the embedding, so $\overline X$ is a compactification of X. Because every $f\in E(X)$ is extendable to a function $\overline f:X_1\to \mathbb R$ , the composition $\overline f\circ \eta $ is an extension of f over $\overline X$ .

Proof The proof is similar to the proof of Lemma 3.1. The only difference is that for every $f\in E(X)$ , we consider $Z_f$ to be the closure of $f(X)$ in $\overline {\mathbb R}$ .

Proof of Theorem 1.1

Let $T:D_p(X)\to D_p(Y)$ be a uniformly continuous c-good surjection. Everywhere below, for $f\in C^*(X)$ , let $\overline f\in C(\beta X)$ be its extension; similarly, if $g\in C(Y)$ then $\overline g\in C(\beta Y,\overline {\mathbb R})$ is the extension of g. According to $(3.1)$ , it suffices to prove that for every $h\in \mathcal F_{Y}$ , there is $h_0\in \mathcal F_{Y}$ such that $\dim h_0(Y)=0$ and ${h_0\succ h}$ . So, we fix $h\in \mathcal F_{Y}$ and let $\overline h:\beta Y\to \overline {h(Y)}$ be an extension of h, where $\overline {h(Y)}$ is a metrizable compactification of $h(Y)$ . We will construct by induction two sequences $\{\Psi _n\}_{n\geq 1}\subset C(\beta X)$ and $\{\Phi _n\}_{n\geq 1}\subset C(\beta Y,\overline {\mathbb R})$ of countable sets, countable $QS$ -algebras $\mathcal A_n$ on $(\triangle \Psi _n)(\beta X)$ and countable $QS$ -algebras $\Lambda _{n}$ on $Y_n'=(\triangle \Phi _n')(\beta Y)$ , where $\Phi _n'=\{\overline {T(f)}:\overline f\in \Psi _n\}$ , satisfying the following conditions for every $n\geq 1$ :

1. (3.5) $\Phi _1\subset C(\beta Y)$ is admissible and $\triangle \Phi _1\succ \overline h$ ;

2. (3.6) $\Phi _n\subset \Phi _{n+1}=\Phi _n'\cup \{\lambda \circ (\triangle \Phi _n'):\lambda \in \Lambda _{n}\}$ ;

3. (3.7) each $\Psi _n$ is admissible, $\dim (\triangle \Psi _n)(\beta X)=0$ and $\Psi _n\subset \Psi _{n+1}$ ;

4. (3.8) $\mathcal A_n$ is a $QS$ -algebra on $(\triangle \Psi _n)(\beta X)$ satisfying condition (2.3);

5. (3.9) $\Lambda _{n+1}$ contains $\{\lambda \circ \delta _{n}:\lambda \in \Lambda _{n}\}$ , where $\delta _{n}:Y^{\prime }_{n+1}\to Y_{n}'$ is the surjective map generated by the inclusion $\Phi _{n}'\subset \Phi _{n+1}'$ ;

6. (3.10) for every $\overline {g}\in \Phi _n\cap C(\beta Y),$ there is $\overline f_g\in \Psi _{n}$ with $||f_g||\leq c||g||$ and $T(f_g)=g$ .

Since $\overline h(\beta Y)$ is a separable metrizable space, by $(3.3)$ , there is a countable admissible set $\Phi _1\subset C(\beta Y)$ with $\triangle \Phi _1\succ \overline h$ . Choose a countable set $\Psi _1'\subset C(\beta X)$ such that for every $\overline g\in \Phi _1$ , there is $\overline f_g\in \Psi _1'$ with $||f_g||\leq c||g||$ and $T(f_g)=g$ . Next, use $(3.2)$ to find countable $\Theta _1\subset C(\beta X)$ containing $\Psi _1'$ such that $\dim (\triangle \Theta _1)(\beta X)=0$ . Finally, by $(3.3)$ , we can extend $\Theta _1$ to a countable admissible set $\Psi _1\subset C(\beta X)$ such that $(\triangle \Psi _1)(\beta X)$ is homeomorphic to $(\triangle \Theta _1)(\beta X)$ and the $QS$ -algebra $\mathcal A_1=\pi (\Psi _1)$ on $(\triangle \Psi _1)(\beta X)$ satisfies condition (2.3). Evidently, $\dim (\triangle \Psi _1)(\beta X)=0$ .

Suppose $\Phi _k$ and $\Psi _k$ are already constructed for all $k\leq n$ . Then, $\Phi _{n}'=\{\overline {T(f)}:\overline f\in \Psi _n\}$ is a countable set in $C(\beta Y,\overline {\mathbb R})$ . Because $\Psi _{n-1}\subset \Psi _n$ , $\Phi _{n-1}'\subset \Phi _n'$ . So, there is a surjective map $\delta _{n-1}: Y_n'\to \ Y_{n-1}'$ (see $(3.9))$ . Since $\triangle \Phi _n'\succ \triangle \Phi _{n-1}'$ , we have $\delta _{n-1}(y)=\triangle \Phi _{n-1}'((\triangle \Phi _n')^{-1}(y))$ for all $y\in Y_n'$ . Choose a countable $QS$ -algebra $\Lambda _{n}$ on $Y_n'$ containing the family $\{\lambda \circ \delta _{n-1}:\lambda \in \Lambda _{n-1}\}$ and let $\Phi _{n+1}=\Phi _n'\cup \{\lambda \circ (\triangle \Phi _n'):\lambda \in \Lambda _{n}\}$ . Next, take a countable set $\Psi _{n+1}'\subset C(\beta X)$ containing $\Psi _n$ such that for every $\overline {g}\in \Phi _{n+1}\cap C(\beta Y),$ there is $\overline f_g\in \Psi _{n+1}'$ with $||f_g||\leq c||g||$ and $T(f_g)=g$ . Then, by $(3.2),$ there is countable $\Theta _{n+1}\subset C(\beta X)$ containing $\Psi _{n+1}'$ with $\dim (\triangle \Theta _{n+1})(\beta X)=0$ . Finally, according to $(3.3)$ , we extend $\Theta _{n+1}$ to a countable admissible set $\Psi _{n+1}\subset C(\beta X)$ such that $(\triangle \Psi _{n+1})(\beta X)$ is homeomorphic to $(\triangle \Theta _{n+1})(\beta X)$ and the $QS$ -algebra $\mathcal A_{n+1}=\pi (\Psi _{n+1})$ on $(\triangle \Psi _{n+1})(\beta X)$ satisfies condition (2.3). This completes the induction.

Let $\Psi =\bigcup _n\Psi _n$ , $X_0=(\triangle \Psi )(X)$ , $\overline X_n=(\triangle \Psi _n)(\beta X),$ and $\overline X_0=(\triangle \Psi )(\beta X)$ . Similarly, let $\Phi =\bigcup _n\Phi _n$ , $Y_0=h_0(Y),$ and $\overline Y_0=(\triangle \Phi )(\beta Y)$ , where $h_0=(\triangle \Phi )|Y$ . Both $\Psi $ and $\Phi $ are countable and $\Psi $ is an admissible subset of $C(\beta X)$ (see $(3.4))$ . Hence, the family $E(\overline X_0)=\{\pi _{\overline f}:\overline f\in \Psi \}$ is a countable $QS$ -algebra on $\overline X_0$ . Moreover, the family $E(Y_0)=\{\pi _{\overline g}|Y_0:\overline g\in \Phi \}$ is extendable over $\overline Y_0$ . Since $\Psi _n\subset \Psi _{n+1}$ for every n, there are maps $\theta _n^{n+1}:\overline X_{n+1}\to \overline X_n$ . Because $\Psi =\bigcup _n\Psi _n$ , the space $\overline X_0$ is the limit of the inverse sequence $S_X=\{\overline X_n,\theta _n^{n+1}\}$ with $\dim \overline X_n=0$ for all n. Hence, $\overline X_0$ is also zero-dimensional (see [Reference Engelking2, Theorem 3.4.11]. Observe also that $E(\overline X_0)=\bigcup _{n=1}^\infty \{h\circ \theta _n:h\in \mathcal A_n\}$ .

Let us show that $E(\overline X_0)$ ia a $QS$ -algebra satisfying condition (2.3). Because $E(\overline X_0)$ is the union of the increasing sequence of the families $\{h\circ \theta _n:h\in \mathcal A_n\}$ and each $\mathcal A_n$ is a $QS$ -algebra, $E(\overline X_0)$ is closed under sums, multiplications, and multiplications by rational numbers. It remains to show that for every $x\in \overline X_0$ and every its neighborhood $U\subset \overline X_0$ , there is $\overline f\in E(\overline X_0)$ such that $\overline f(x)=1$ and $\overline f(\overline X_0\backslash U)=0$ . But that follows from the proof of the more general condition (2.3). To show that $E(\overline X_0)$ satisfies $(2.3)$ , take a compact set $K\subset \overline X_0$ and an open set $W\subset \overline X_0$ containing K. Because $\overline X_0$ is the limit of the inverse sequence $S_X=\{\overline X_n,\theta _n^{n+1}\}$ , there are n, a compact set $K_n\subset \overline X_n$ and an open set $W_n\subset \overline X_n$ containing $K_n$ such that $K\subset \theta _n^{-1}(K_n)\subset \theta _n^{-1}(W_n)\subset W$ , where ${\theta _n:\overline X_0\to \overline X_n}$ denotes the n-th projection in $S_X$ . Then, according to condition (3.8), there is $f_n\in \mathcal A_n$ with $f_n|K_n=1$ , $f_n|(\overline X_n\backslash W_n)=0$ and $f_n:\overline X_n\to [0,1]$ . Finally, observe that the function $\overline f=f_n\circ \theta _n:\overline X_0\to [0,1]$ belongs to $E(\overline X_0)$ and $\overline f$ satisfies the conditions $\overline f|K=1$ and $\overline f|(\overline X_0\backslash W)=0$ .

It follows from the construction that $\Phi =\{\overline {T(f)}:\overline f\in \Psi \}$ and for every $\overline g\in \Phi \cap C(\beta Y),$ there is $\overline f_g\in \Psi $ with $T(f_g)=g$ and $||f_g||\leq c||g||$ . Observe that for all $\overline f\in \Psi $ and $\overline g\in \Phi ,$ we have $f=\pi _{\overline f}\circ (\triangle \Psi )|X$ and $g=\pi _{\overline g}\circ (\triangle \Phi )|Y$ . Therefore, there is a surjective map $\varphi :E_p(X_0)\to E_p(Y_0)$ defined by $\varphi (\pi _f)=\pi _{T(f)}$ , where $\pi _f$ and $\pi _g$ denote, respectively, the functions $\pi _{\overline f}|X_0$ and $\pi _{\overline g}|Y_0$ . Moreover, $||f||=||\pi _f||$ and ${||g||=||\pi _g||}$ for all $\overline f\in \Psi $ and $\overline g\in \Phi \cap C(\beta Y)$ . This implies that $\varphi $ is a c-good surjection.

Let’s show that $\varphi $ is uniformly continuous. Suppose

$$ \begin{align*}V=\{\pi_g:|\pi_g(y_i)|<\varepsilon{~}\forall i\leq k\}\end{align*} $$

is a neighborhood of the zero function in $E_p(Y_0)$ . Take points $\overline y_i\in Y$ with $h_0(\overline y_i)=y_i$ , $i=1,2,\ldots ,k$ , and let $\widetilde V=\{g\in D_p(Y):|g(\overline y_i)|<\varepsilon {~}\forall i\leq k\}$ . Since T is uniformly continuous, there is a neighborhood

$$ \begin{align*}\widetilde U=\{f\in D_p(X):|f(\overline x_j)|<\delta{~}\forall j\leq m\}\end{align*} $$

of the zero function in $D_p(X)$ such that $f-f'\in \widetilde U$ implies $T(f)-T(f')\in \widetilde V$ for all $f,f'\in D_p(X)$ . Let $x_j=(\triangle \Psi )(\overline x_j)$ and

$$ \begin{align*}U=\{\pi_f:|\pi_f(x_j)|<\delta{~}\forall j\leq m\}.\end{align*} $$

Obviously, $\pi _f-\pi _{f'}\in U$ implies $f-f'\in \widetilde U$ . Hence, $T(f)-T(f')\in \widetilde V$ , which yields $\varphi (\pi _f)-\varphi (\pi _{f'})\in V$ .

Finally, we can show that $E(\overline Y_0)$ contains a $QS$ -algebra on $\overline Y_0$ . Since $\Lambda _n\subset C(Y_n')$ is a $QS$ -algebra on $Y_n'$ , it separates the points and the closed sets in $Y_n'$ . So, $Y_n'$ is homeomorphic to $(\triangle \Lambda _n)(Y_n')$ . This implies that $Y_{n+1}=(\triangle \Phi _{n+1})(\beta Y)$ is homeomorphic to $Y_n'$ . Therefore, $\Lambda _n$ can be considered as a $QS$ -algebra on $Y_{n+1}$ . On the other hand, $\overline Y_0$ is the limit of the inverse sequence $S_Y=\{Y_{n+1},\gamma _{n+1}^{n+2},n\geq 1\}$ , where $\gamma _{n+1}^{n+2}:Y_{n+2}\to Y_{n+1}$ is the surjective map generated by the inclusion $\Phi _{n+1}\subset \Phi _{n+2}$ . According to condition (3.9), we can also assume that $\Lambda _{n+1}$ contains the family $\{\lambda \circ \gamma _{n+1}^{n+2}:\lambda \in \Lambda _n\}$ . Denote by $\gamma _{n+1}:\overline Y_0\to Y_{n+1}$ the projections in $S_Y$ , and let $\Gamma _n=\{\lambda \circ \gamma _{n+1}:\lambda \in \Lambda _n\}$ . Then, $\{\Gamma _n\}_{n\geq 1}$ is an increasing sequence of countable families and $\Gamma _n\subset E(\overline Y_0)=\{\pi _{\overline g}:\overline g\in \Phi \}$ for every n. We claim that $\Gamma =\bigcup _n\Gamma _n$ is a $QS$ -algebra on $\overline Y_0$ . Indeed, $\Gamma $ is closed under addition, multiplication, and multiplication by rational numbers. Because for every $y\in \overline Y_0$ and its neighborhood $V\subset \overline Y_0$ , there is n and an open set $V_n\subset Y_{n+1}$ such that $y_n=\gamma _{n+1}(y)\in V_n$ and $\gamma _{n+1}^{-1}(V_n)\subset V$ , there exists $\lambda \in \Lambda _n$ with $\lambda (y_n)=1$ and $\lambda (Y_{n+1}\backslash V_n)=0$ . Then, $g=\lambda \circ \gamma _{n+1}\in \Gamma $ , $g(y)=1,$ and $g(\overline Y_0\backslash V)=0$ .

To prove Theorem 1.1, we apply Proposition 2.1 with $H=\overline X_0$ to find a $\sigma $ -compact set $Y_\infty \subset \overline Y_0$ containing $Y_0$ with $\dim Y_\infty =0$ . Therefore, by [Reference Engelking2, Proposition 1.2.2], $\dim Y_0=0$ .

Proof By the Closed Graph Theorem, T considered as a map between the Banach spaces $C^*_u(X)$ and $C^*_u(Y)$ , both equipped with the sup-norm, is continuous. Then, T induced a linear isomorphism $T_0$ between $C^*_u(X)/K$ and $C^*_u(Y)$ , where K is the kernel of T. So, for every $g\in C_u^*(Y),$ we have $||T_0^{-1}(g)||\leq ||T_0^{-1}||\cdot ||g||$ . Because

$$ \begin{align*}||T_0^{-1}(g)||=\inf\{||f-h||:h\in K\},\end{align*} $$

where $f\in C_u^*(X)$ with $T(f)=g$ , there exists $h_g\in K$ such that $||f-h_g||\leq 2||T_0^{-1}(g)||$ . Hence, $||f-h_g||\leq 2||T_0^{-1}||\cdot ||g||$ . Since $T(f-h_g)=T(f)=g$ , we obtain that T is c-good with $c=2||T_0^{-1}||$ .

Proof of Theorem 1.3

Following the proof of Theorem 1.1, we construct two sequences $\{\Psi _n\}_{n\geq 1}\subset C(\beta X)$ and $\{\Phi _n\}_{n\geq 1}\subset C(\beta Y,\overline {\mathbb R})$ of countable sets and countable $QS$ -algebras $\mathcal A_n$ on $\triangle \Psi _n(\beta X)$ and $\Lambda _{n}$ on $Y_n'=(\triangle \Phi _n')(\beta Y)$ satisfying the conditions (3.5)–(3.10) except $(3.7)$ . Because X and Y are separable metrizable spaces, we can choose countable sets $\Psi _1$ and $\Phi _1$ such that $(\triangle \Phi _1)|Y$ and $(\triangle \Psi _1)|X$ are homeomorphisms.

Then, following the notations from the proof of Theorem 1.1, we have that $X_0$ and $Y_0$ are homeomorphic to X and Y, respectively. Moreover, there exists a uniformly continuous c-good surjection $\varphi : E_p(X_0)\to E_p(Y_0)$ such that $E(\overline X_0)$ is a $QS$ -algebra on $\overline X_0$ satisfying condition (2.3), $E(\overline Y_0)\subset C(\overline Y_0,\overline {\mathbb R})$ and $E(\overline Y_0)$ contains a countable $QS$ -algebra $\Gamma $ on $\overline Y_0$ . According to the proof of Proposition 2.1 with $H=X_0$ , there is a $\sigma $ -compact set $Y_\infty \subset \overline Y_0$ containing $Y_0$ which a countable union of closed sets $F\subset Y_\infty $ such that F admits a map with finite fibers into a finite power of $X_0$ . Let $Y_0=\bigcup _m Y_m$ with each $Y_m$ being compact. Then, $Y_0$ can also be represented as a countable union of compact sets each admitting a map with finite powers into a finite power of $X_0$ . Now, we apply the following fact which was actually used in the proof of Proposition 2.1: Assume all powers of a $\sigma $ -compact metrizable space P have a property $\mathcal P$ satisfying conditions (a)–(c). If a metrizable space Z is the union of countably many compact sets $Z_n$ such that each $Z_n$ admits a map with finite fibers into a finite power $P^{k_n}$ , then all finite powers of Z also have the property $\mathcal P$ . Since all finite powers of $X_0$ have the property $\mathcal P$ , where $\mathcal P$ is either weakly infinite-dimensionality or the $(m-C)$ -space property, by mentioned above fact, all finite powers of $Y_0$ are either weakly infinite-dimensional or have the $(m-C)$ -space property.

Proof Let $E(\overline X)=\{\overline f:f\in E(X)\}$ . Every $y\in \overline Y$ generates a map $l_y:E(\overline X)\to \overline {\mathbb R}$ , $l_y(\overline f)=\overline {\varphi (f)}(y)$ . Assuming the equalities from $(2.4)$ , for any $\overline f_1,\overline f_2\in E(\overline X)$ and $x\in \overline X$ , we can write $\overline f_1(x)+\overline f_2(x)$ but not always $\overline f_1+\overline f_2=\overline {f_1+f_2}$ . Also, if $\overline f_1+\overline f_2\in E(\overline X)$ , it is possible that $l_y(\overline f_1+\overline f_2)\neq l_y(\overline f_1)+l_y(\overline f_2)$ . If $\lambda \in \mathbb R\backslash \{0\}$ and $\overline {\lambda \cdot f}\in E(\overline X)$ , then $l_y(\overline {\lambda \cdot f})=\lambda \cdot l_y(\overline f)$ (here, $\lambda \cdot (\pm \infty )=\pm \infty $ if $\lambda>0$ and $\lambda \cdot (\pm \infty )=\mp \infty $ if $\lambda <0$ ). In case $\lambda =0$ , we have $l_y(\overline {0\cdot f})=0$ . More general, if $\overline h\in C(\overline X)$ such that $h\cdot f\in E(X)$ for some $f\in E(X)$ , then $l_y(\overline {h\cdot f})$ is well defined.

The support of $l_y$ , $y\in \overline Y$ , is defined to be the set $supp(l_y)$ of all $x\in \overline X$ satisfying the following condition [Reference Valov and Vuma22]: for every neighborhood $U\subset \overline X$ of $x,$ there is $f\in E(X)$ such that $\overline f(\overline X\backslash U)=0$ and $l_y(\overline f)\neq 0$ . Obviously, $supp(l_y)$ is closed in $\overline X$ .

Indeed, because the set of real-valued functions from $E(\overline Y)$ is dense in $C_p(\overline Y)$ , for every $y\in \overline Y,$ there is $g\in E(\overline Y)$ with $g(y)\neq 0$ . Then, take $f_y\in E(X)$ such that $\varphi (f_y)=g|Y$ , so $l_y(\overline f_y)=g(y)$ . If $supp(l_y)=\varnothing $ , every $x\in \overline X$ has a neighborhood $V_x$ such that $l_y(\overline f)=0$ for any $f\in E(X)$ with $\overline f(\overline X\backslash V_x)=0$ . Passing to smaller neighborhoods, we can assume that $V_x\in \mathcal B$ for all x. Hence, there is a finite open cover $\gamma =\{V_{x_1},\ldots ,V_{x_k}\}$ of $\overline X$ and a partition of unity $\mu =\{\overline h_1,\ldots ,\overline h_k\}\subset E(\overline X)$ subordinated to $\gamma $ . Then, $f_y\cdot h_i\in E(X)$ and $\overline {f_y\cdot h_i}|(\overline X\backslash V_{x_i})=0,$ which implies $l_y(\overline {f_y\cdot h_i})=0$ for all i. Because $\mu $ is a partition of unity, $\overline f_y=\sum _{i=1}^k\overline {f_y\cdot h_i}$ . So, $l_y(\overline f_y)=\sum _{i=1}^kl_y(\overline {f_y\cdot h_i})=0$ , a contradiction.

First, let $\overline f(U)=0$ for some $\overline f\in E(\overline X)$ . We can assume that U is a finite union of open sets $V_i$ from $\mathcal B$ , $i=1,\ldots ,k$ . Every $x\in \overline X\backslash U$ has a neighborhood $V_x$ such that $l_y(\overline g)=0$ for any $g\in E(X)$ with $\overline g(\overline X\backslash V_x)=0$ . As in the proof of Claim 9, we can assume that $V_x\in \mathcal B$ , and choose another neighborhood $U_x\in \mathcal B$ with $\overline U_x\subset V_x$ . Take a finite open cover $\{U_{x_1},\ldots ,U_{x_m}\}$ of $\overline X\backslash U$ . Then, $\gamma =\{V_1,\ldots ,V_k,U_{x_1},\ldots ,U_{x_m}\}$ is an open cover of $\overline X$ consisting of elements from $\mathcal B$ . Hence, there exits a partition of unity $\{\overline h_1,\ldots ,\overline h_k,\overline \theta _1,\ldots ,\overline \theta _m\}\subset E(\overline X)$ subordinated to $\gamma $ . Then, $h_i\cdot f, \theta _j\cdot f\in E(X)$ for all $i,j$ and $f=\sum _{i=1}^kh_i\cdot f+\sum _{j=1}^m\theta _j\cdot f$ . Take a sequence $\{y_n\}\subset Y$ with $\lim _n y_n=y$ . So, $\varphi (f)(y_n)=\sum _{i=1}^k\varphi (h_i\cdot f)(y_n)+\sum _{j=1}^m\varphi (\theta _j\cdot f)(y_n)$ . Observe that $(h_i\cdot f)(x)=0$ for all $x\in X$ , so $\varphi (h_i\cdot f)$ is the zero function on Y and $\varphi (h_i\cdot f)(y_n)=0$ , $i=1,\ldots ,k$ . Hence, $\varphi (f)(y_n)=\sum _{j=1}^m\varphi (\theta _j\cdot f)(y_n)$ . On the other hand, $(\theta _j\cdot f)(x)=0$ for all $x\in X\backslash U_{x_j}$ . So, $(\overline {\theta _j\cdot f})(x)=0$ for all $x\in \overline X\backslash V_{x_j}$ because $\overline X\backslash V_{x_j}\subset \overline {X\backslash U_{x_j}}$ . This implies that $l_y(\overline {\theta _j\cdot f})=0$ for all $j=1,\ldots ,m$ . Since, ${l_y(\overline {\theta _j\cdot f})=\lim _n\varphi (\theta _j\cdot f)(y_n)=0}$ , we have

$$ \begin{align*}l_y(\overline f)=\lim_n\varphi(f)(y_n)=\lim_n\sum_{j=1}^m\varphi(\theta_j\cdot f)(y_n)=0.\end{align*} $$

Suppose now that $\overline f|U=\overline g|U$ for some $f,g\in E(X)$ . Then, $f(x)-g(x)=0$ for all $x\in U\cap X$ . Consequently, $(\overline {f-g})(x)=0$ for all $x\in U$ and, according to the previous paragraph, $l_y(\overline {f-g})=0$ . Hence, for every sequence $\{y_n\}\subset Y$ converging to $y,$ we have

$$ \begin{align*}l_y(\overline{f-g})=\lim_n\varphi(f-g)(y_n)=\lim_n\varphi(f)(y_n)-\lim_n\varphi(g)(y_n)=l_y(\overline f)-l_y(\overline g).\end{align*} $$

Let $x_0\in supp(l_{y_0})\cap U$ and $\overline f\in E(\overline X)$ be such that $\overline f(\overline X\backslash W)=0$ and $l_{y_0}(\overline f)\neq 0$ , where W is a neighborhood of $x_0$ with $\overline W\subset U$ . Assuming the claim is not true, we can find a sequence $\{y_n\}\subset \overline Y$ converging to $y_0$ such that $supp(l_{y_n})\cap U=\varnothing $ for every n. Since $\overline X\backslash \overline W$ is a neighborhood of each $supp(l_{y_n})$ , by Claim 10, $l_{y_n}(\overline f)=0$ . Because $\lim _n l_{y_n}(\overline f)=l_{y_0}(\overline f)$ , we have $l_{y_0}(\overline f)=0$ , a contradiction.

Following the notations from the proof of Proposition 2.1, for every $y\in \overline Y,$ we define

$$ \begin{align*}a(y)=\sup\{|l_y(\overline f)|:\overline f\in E(\overline X)\text{ and }|\overline f(x)|<1{~} \forall x\in supp(l_y)\}.\end{align*} $$

Let $\widetilde \varphi :LE_p(X)\to LE_p(Y)$ be the continuous extension of $\varphi $ . Then, every $l_y$ , $y\in Y$ , can be continuously extended to a linear functional $\widetilde l_y:LE_p(X)\to \mathbb R$ , $\widetilde l_y(\sum _{i=1}^k\lambda _i\cdot f_i)=\sum _{i=1}^k\lambda _i\cdot l_y(f_i)$ . Since $\widetilde \varphi $ is uniformly continuous, according to the proof of Claim 1 from Proposition 2.1, for every $y\in Y,$ there exists a finite set $K=\{x_1(y),x_2(y),\ldots ,x_q(y)\}\subset X$ such that

$$ \begin{align*}\sup\{|\widetilde l_y(f)|:f\in LE(X)\text{ and }|f(x)|<1{~} \forall x\in K\}<\infty.\end{align*} $$

Because $\widetilde l_y$ is linear, we can show that $\widetilde l_y(g)=0$ for any $g\in LE(X)$ with $g(x_i(y))=0$ , $i=1,\ldots ,q$ . Since $E(X)$ is a $QS$ -algebra, for every $i,$ there is a function $g_i\in E(X)$ such that $g_i(x_i(y))=1$ and $g_i(x_j(y))=0$ with $j\neq i$ . Now, for every $f\in E(X),$ the function $g=f-\sum _{i=1}^qg_i\cdot f(x_i(y))$ belongs to $LE(X)$ and $g(x_i(y))=0$ for all i. So, $\widetilde l_y(g)=0$ and $l_y(f)=\sum _{i=1}^q\lambda _i(y)f(x_i(y))$ , where $\lambda _i(y)=l_y(g_i)$ . Because $y\in Y$ , we can also write

$$ \begin{align*}l_y(\overline f)=\sum_{i=1}^q\lambda_i(y)\overline f(x_i(y))\end{align*} $$

with $\lambda _i(y)=l_y(\overline g_i)$ . Note that each $\lambda _i(y)$ is a real number because $l_y(g_i)=\varphi (g_i)(y)$ . The last equality is valid for all $\overline f\in E(\overline X)$ and shows that $supp(l_y)=\{x_i(y):\lambda _i(y)\neq 0\}\subset K$ . Note that, by Claim 9, $supp(l_y)\neq \varnothing $ .

To complete the proof of Claim 12, assume that $supp(l_y)=K$ , and for every natural k take a function $f_k\in E(X)$ with $f_k(x_i(y))=\varepsilon _i(1-1/k)$ , where $\varepsilon _i=1$ if $\lambda _i(y)>0$ and $\varepsilon _i=-1$ if $\lambda _i(y)<0$ . Clearly, $|f_k(x_i(y))|<1$ for all $i,k$ and $\lim _kl_y(\overline f_k)=\sum _{i=1}^q|\lambda _i(y)|$ . Hence, $\sum _{i=1}^q|\lambda _i(y)|\leq a(y)$ . The reverse inequality $a(y)\leq \sum _{i=1}^q|\lambda _i(y)|$ follows from $l_y(\overline f)=\sum _{i=1}^q\lambda _i(y)\overline f(x_i(y))$ , $\overline f\in E(\overline X)$ .

For every $p,q\in \mathbb N,$ let $Y_{p,q}=\{y\in \overline Y:|supp(l_y)|\leq q{~}\mbox {and}{~}a(y)\leq p\}$ .

Let $\{y_n\}$ be a sequence in $Y_{p,q}$ converging to $y\in \overline Y$ . Suppose $y\not \in Y_{p,q}$ . Then either $supp(l_y)$ contains at least $q+1$ points or $a(y)>p$ . If $supp(l_y)$ contains at least $q+1$ points $x_1,x_2,\ldots ,x_{q+1}$ , we choose disjoint neighborhoods $O_i$ of $x_i$ , $i=1,\ldots ,q+1$ . By Claim 11, there is a neighborhood V of y such that $supp(l_z)\cap O_i\neq \varnothing $ for all i and $z\in V$ . This implies that $supp(l_{y_n})$ has at least $q+1$ points for infinitely many n’s, a contradiction.

If $y\not \in Y_{p,q}$ and $a(y)>p$ , then there exists $\overline f\in E(\overline X)$ such that $|\overline f(x)|<1$ for all $x\in supp(l_y)$ and $|l_y(\overline f)|>p$ . Take a neighborhood U of $supp(l_y)$ with $U\subset \{x\in \overline X:|\overline f(x)|<1\}$ . Choose another neighborhood W of $supp(l_y)$ with $\overline W\subset U$ . Next, there is $h\in E(X)$ such that $\overline h(\overline W)=1$ , $\overline h(\overline X\backslash U)=0,$ and $h(x)\in [0,1]$ for all $x\in X$ . Then, $\overline g=\overline {h\cdot f}\in E(\overline X)$ and $|\overline g(x)|<1$ for all $x\in \overline X$ . Moreover, $\overline g|W=\overline f|W$ . So, by Claim 10, $|l_y(\overline g)|=|l_y(\overline f)|>p$ . Therefore, $V=\{z\in \overline Y:|l_z(\overline g)|>p\}$ is a neighborhood of y in $\overline Y$ with $V\cap Y_{p,q}=\varnothing $ , a contradiction.

According to Claim 12, for every $y\in Y,$ then there exist $p,q\in \mathbb N$ and real numbers $\lambda _i(y)$ such that for all $\overline f\in E(\overline X),$ we have $l_y(\overline f)=\sum _{i=1}^q\lambda _i(y)\overline f(x_i(y))$ with $a(y)=\sum _{i=1}^q|\lambda _i(y)|\leq p$ , where $\{x_1(y),\ldots ,x_q(y)\}$ is the support $supp(l_y)$ . Hence, $Y\subset \bigcup \{Y_{p,q}:p,q\in \mathbb N\}$ . For every $p\geq 1$ and $q\geq 2,$ we define

$$ \begin{align*}M(p,1)=Y_{p,1}\text{ and }M(p,q)=Y_{p,q}\backslash Y_{2p,q-1}.\end{align*} $$

Some of the sets $M(p,q)$ could be empty, but $Y\subset \bigcup \{M(p,q):p,q=1,2,\ldots \}$ . Indeed, by Claim 12, there exist $p,q$ with $|supp(l_y)|=q$ and $a(y)\leq p$ . Then, $y\in M(p,q)$ . Since each $Y_{p,q}$ is closed, $M(p,q)=\bigcup _{n=1}^\infty F_n'(p,q)$ such that each $F_n'(p,q)$ is a compact subset of $\overline Y$ . We define $F_n(p,q)=\overline {Y\cap F_n'(p,q)}$ . Then, $Y\subset \bigcup \{F_n(p,q):n,p,q=1,2,\ldots \}$ . Obviously, $supp(l_y)$ consists of q different points for any $y\in M(p,q)$ . So, we have a map $S_{p,q}:M(p,q)\to [\overline X]^q$ , $S_{p,q}(y)=supp(l_y)$ . According to Claim 11, $S_{p,q}$ is continuous when $[\overline X]^q$ is equipped with the Vietoris topology. For every $y\in M(p,q),$ let $S_{p,q}(y)=\{x_i(y)\}_{i=1}^q$ . Everywhere below, we consider the restriction $S_{p,q}|F_n(p,q)$ and for every $z\in F_n(p,q)$ denote by $A(z)=\{y\in F_n(p,q):S_{p,q}(y)=S_{p,q}(z)\}$ the fiber $S_{p,q}^{-1}(S_{p,q}(z))$ generated by z. Since $F_n(p,q)$ is compact and $S_{p,q}$ is continuous, each $A(z)$ is a compact subset of $F_n(p,q)$ .

Choose neighborhoods $O_i$ of $x_i(z)$ in $\overline X$ with disjoint closures and functions $\overline g_i\in E(\overline X)$ with $\overline g_i|O_i=1$ and $\overline g_i|O_j=0$ if $j\neq i$ (this can be done by choosing $g_i\in E(X)$ with $g_i(O_i\cap X)=1$ and $g_i(O_j\cap X)=0$ when $j\neq i$ ). According to the proof of Claim 12, for every $y\in A(z)\cap Y$ and the real numbers $\lambda _i(y)=l_y(\overline g_i),$ we have $l_y(\overline f)=\sum _{i=1}^{q}\lambda _i(y)\overline f(x_i(z))$ for all $\overline f\in E(\overline X)$ . Let’s show this is true for all $y\in A(z)$ . So, fix $y\in A(z)\backslash Y$ and take a sequence $\{y_m\}\subset Y\cap F_n(p,q)$ converging to y. Then, by Claim 12, $S_{p,q}(y_m)=\{x_i(y_m)\}_{i=1}^q\subset X$ and there are real numbers $\{\lambda _i(y_m)\}_{i=1}^q$ such that $l_{y_m}(\overline f)=\sum _{i=1}^q\lambda _i(y_m)\overline f(x_i(y_m))$ for all $\overline f\in E(\overline X)$ . On the other hand, since the map $S_{p,q}$ is continuous, each sequence $\{x_i(y_m)\}_{m=1}^\infty $ converges to $x_i(y)=x_i(z)$ , $i=1,\ldots ,q$ . So, we can assume that $\{x_i(y_m)\}_{m=1}^\infty \subset O_i$ . Consequently, $l_{y_m}(\overline g_i)=\lambda _i(y_m)$ and $\lim _m\lambda _i(y_m)=l_y(\overline g_i)$ . Since $\{y_m\}\subset Y\cap Y_{p,q}$ , $\sum _{i=1}^q|\lambda _i(y_m)|\leq p$ for each m (see Claim 12). Hence, $\sum _{i=1}^q|l_y(\overline g_i)|\leq p$ . Denoting $\lambda _i(y)=l_y(\overline g_i)$ , we obtain $\sum _{i=1}^q|\lambda _i(y)|\leq p$ . The last inequality means that all $\lambda _i(y)$ are real numbers. Since $\lim _m\overline f(x_i(y_m))=\overline f(x_i(y))$ for all $\overline f\in E(\overline X)$ and each $\overline f(x_i(y))$ is a real number (recall that $z\in Y$ , so by Claim 12, $x_i(z)=x_i(y)\in X$ ), we have $l_y(\overline f)=\lim l_{y_m}(\overline f)=\sum _{i=1}^q\lambda _i(y)\overline f(x_i(y))$ .

Finally, the equality $\lambda _i(y)=\overline {\varphi (g_i)}(y)$ implies that $\lambda _i$ are continuous on $A(z)$ .

Following the previous notations, for every $h\in C(S_{p,q}(z))$ and $y\in A(z),$ we define $\varphi _z(h)(y)=\sum _{i=1}^q\lambda _i(y)h(x_i(z))$ . Because $\lambda _i$ are continuous real-valued functions on $A(z)$ , so is each $\varphi _z(h)$ . Continuity of $\varphi _z$ with respect to the pointwise convergence topology is obvious. Let’s show that $\varphi _z(C(S_{p,q}(z)))$ is dense in $C_p(A(z))$ . Indeed, take $\theta \in C_p(A(z))$ and its neighborhood $V\subset C_p(A(z))$ . Then extend $\theta $ to a function $\overline \theta \in C(\overline Y)$ . Because the set of real-valued elements of $E_p(\overline Y)$ is dense in $C_p(\overline Y)$ , there is $\overline g\in E_p(\overline Y)$ with $\overline g|A(z)\in V$ , so $\overline g(y)\in \mathbb R$ for all $y\in \overline Y$ . Next, choose $f\in E(X)$ such that $\varphi (f)=g$ . Since $z\in Y$ , each $x_i(z)\in X$ . So, all $\overline f(x_i(z))$ are real numbers. Then, $h=\overline f|S_{p,q}(z)\in C(S_{p,q}(z))$ and, according to Claim 14, we have $\varphi _z(h)=\overline g|A(z)$ .

Since $|S_{p,q}(z)|=q$ , $C_p(S_{p,q}(z))$ is isomorphic to $\mathbb R^q$ . Hence, Claim 15 together with basic facts from linear algebra imply that $|A(z)|\leq q$ , in particular, $A(z)$ is zero-dimensional.

Now, we can complete the proof of Proposition 4.1. As in the proof of Proposition 2.1, using that $\overline X^q\in \mathcal P$ , we can show that $[\overline X]^q\in \mathcal P$ . Therefore, condition (c’) implies that each $F_n(p,q)$ has also the property $\mathcal P$ , and so does $Y_n(p,q)=F_n(p,q)\cap Y$ . Finally, since Y is the union of its closed subsets $Y_n(p,q)$ , we conclude that $Y\in \mathcal P$ .

Proof Let $\phi '=\triangle \Phi '$ . Since $\Phi '$ is countable, $\phi '(\beta Y)$ is a metrizable compactum. Hence, by [Reference Gul’ko7, Proposition 1.2], there is a countable $QS$ -algebra $E\subset C(\phi '(\beta Y))$ . Let $\Phi =\Phi '\cup \{g\circ \phi ':g\in E\}$ . Since the functions of E separate the points and closed subsets of $\phi '(\beta Y)$ , $(\triangle \Phi )(\beta Y)$ is homeomorphic to $\phi '(\beta Y)$ . Since E is a $QS$ -algebra on $\phi '(\beta Y)$ , E is a dense subset of $C_p(\phi '(\beta Y))$ . Clearly, E is a subset of $E_\Phi $ and consists of real-valued functions.