Hurewicz versus Wadge hierarchy of zero-dimensional Borel sets?

Question

Given two subsets $A,B$ of the Cantor cube $2^\omega$ we write $A\le_W B$ (resp. $A\le_H B$) if there is a continuous (and injective) function $f:2^\omega\to 2^\omega$ such that $f^{-1}(B)=A$. The relation $\le_W $ is the well-known (and well-studied) Wadge reducibility relation. The relation $\le_H$ is called the Hurewicz reducibility to make associations with some known Hurewicz-type tests (proved by Hurewicz, Louveau, Saint Raymond and others).

Problem 1. What is known about the relation of Hurewicz reducibility between Borel subsets of the Cantor cube? Is it equivalent to the Wadge reducibility for sufficiently complex Borel subsets of $2^\omega$?

More precisely: Assume that $A,B\subset 2^\omega$ are two Borel subsets of complexity (say) $A,B\notin \Sigma^0_3\cap \Pi^0_3$. Does $A\le_W B$ imply $A\le_HB$?

By a result of Louveau and Saint Raymond (1987) this implication is true if $A,B$ belong to the class $\Gamma\setminus \check\Gamma=\{C\in\Gamma:2^\omega\setminus C\notin\Gamma\}$ for some class $\Gamma$ of Borel subsets of $2^\omega$ which is closed under intersections with $G_\delta$-sets and unions with $F_\sigma$-sets in $2^\omega$. So, for any $\xi\ge 3$ and any subsets $A,B\in\Pi^0_\xi\setminus \Sigma^0_\xi$ we get $A\le_H B$ and $B\le_H A$.

Let us say that two topological spaces $X,Y$ are Hurewicz equivalent if each of them admits a closed topological embedding into the other. It is clear that two subsets $A,B\subset 2^\omega$ are Hurewicz equivalent if $A\le_H B$ and $B\le _H A$. The converse is not true: the Cantor cube contains two homeomorphic dense $G_\delta$-sets $A,B$ such that $A\not\le_H B$.

Problem 2. Assume that $A,B\subset 2^\omega$ are two Borel subsets of sifficiently hight Borel complexity (say $A,B\notin \Sigma^0_3\cap \Pi^0_3$). Assume that $A,B$ are Hurewicz equivalent. Is $A\le_H B$?

Problem 3. Classify Borel subsets of $2^\omega$ up to Hurewicz equivalence. Is it true that for each Borel subset $B\subset 2^\omega$ of sufficiently hight Borel complexity (say $B\notin\Pi^0_2\cup \Sigma^0_2$) the Wadge class $W(A)=\{B\subset 2^\omega:B\le_W A,\;A\le_W B\}$ of $A$ contains only finitely many spaces (better, a unique space) that are pairwise Hurewicz non-equivalent?

Answer 1

After some thoughts I realized that the answers to Problems 1 and 3 are negative. Namely, each limit Wadge class contains infinitely many Hurewicz non-equivalent spaces. Indeed, take a sequence $(U_n)_{n\in\omega}$ of pairwise disjoint clopen subsets of the Cantor cube $2^\omega$ that converge to some point $x_\infty$. In each set $U_n$ choose a Borel subset $A_n$ so that the sequence $(A_n)_{n\in\omega}$ is strictly increasing in the Warge hierarchy. Finally, consider the Borel set $A=\{x_\infty\}\cup\bigcup_{n\in\omega}A_n$. It can be shown that for every number $n=\{0,\dots,n-1\}\in\mathbb N$ endowed with the discrete topology the space $A\times n$ is Wadge equivalent to $A$. Nonetheless, for $m>n$ the space $A\times m$ admits no closed embedding into $A\times n$. So, the spaces $A\times n$, $n\in\mathbb N$, are pairwise Hurewicz non-equivalent.