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Let $\mathbb P$ denote the space of irrationals. Is there a continuous bijection (one-to-one and onto) $f:\mathbb P\to \mathbb Q ^\omega$ that maps each closed subset of $\mathbb P$ to a $G_\delta$-subset of $\mathbb Q ^\omega$?

Remark 1. Suppose that $f:\mathbb P\to \mathbb Q ^\omega$ is a continuous bijection mapping closed sets to $G_{\delta}$ sets. Then $f^{-1}:\mathbb Q^\omega\to \mathbb P$ is a Baire class $1$ functions, i.e. $(f^{-1})^{-1}(U)=f(U)$ is an $F_{\sigma}$-subset of $\mathbb Q ^\omega$ for every open set $U\subseteq \mathbb P$. By Theorem 4.1 in the reference below, either there are countably many sets $X_n\subseteq \mathbb Q ^\omega$ such that $\mathbb Q ^\omega=\bigcup \{X_n:n<\omega\}$ and $f^{-1}\restriction X_n$ is continuous, or $f^{-1}$ contains Pawlikowski's function $P:(\omega+1)^\omega\to \omega^\omega$.

Remark 2. While trying to solve this problem, I discovered an example involving complete Erdos space $\mathfrak E_c$. There is a continuous bijection $f:\mathfrak E_c^\omega\to \mathbb Q ^\omega$ which maps closed sets to $G_{\delta}$ sets and such that $f^{-1}$ is not a countable union of continuous functions. So by Theorem 4.1 it must contain $P$. I proved that all similar examples, including the one in @Arno's answer also must contain $P$.

My feeling now is that my question probably has a positive answer, although the zero-dimensionality of $\mathbb P$ makes things interesting.

Solecki, Sławomir, Decomposing Borel sets and functions and the structure of Baire class 1 functions, J. Am. Math. Soc. 11, No. 3, 521-550 (1998). ZBL0899.03034.

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    $\begingroup$ Is $\mathbb{P}$ the set of irrationals, or something else? $\endgroup$ – Nate Eldredge Aug 9 '20 at 23:14
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The "canonical" continuous bijection works. We start by observing that $\mathbb{P}$ is homeomorphic to $\mathbb{N}^\omega$. We pick some bijection $\tau : \mathbb{N} \to \mathbb{Q}$, which is trivially continuous, and has a Baire class 1 inverse. We can then lift $\tau$ to obtain a continuous bijection $\tau^\omega : \mathbb{N}^\omega \to \mathbb{Q}^\omega$ with Baire class 1 inverse $(\tau^\omega)^{-1}$. As $(\tau^\omega)^{-1}$ is Baire class 1, the preimage of a closed set under it is $\Pi^0_2$, hence $\tau$ maps closed sets to $\Pi^0_2$-sets as desired.

D.S. Lipham gave some more details in the comments for checking that the inverse is Baire class 1. We can directly show that $\tau^\omega$ maps open sets to $F_\sigma$-sets. Each basic open subset of $\mathbb{N}^\omega$ maps to a product of $F_\sigma$-subsets of $\mathbb{Q}$ whose factors are eventually all of $\mathbb{Q}^\omega$. Hence, the image is $F_\sigma$ in $\mathbb{Q}^\omega$. Each open subset of $\mathbb{N}^\omega$ is a countabe union of basic open sets, so its image is a countable union of $F_\sigma$-sets.

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  • $\begingroup$ @D.S.Lipham Any sequence of rationals appears in the range as the image of the sequence of the indices of those rationals. $\endgroup$ – Arno Aug 10 '20 at 21:55
  • $\begingroup$ I see. Looks correct then. $\endgroup$ – D.S. Lipham Aug 10 '20 at 22:26
  • $\begingroup$ To check that the inverse is Baire class 1, we should show $\tau ^\omega$ maps open sets to $F_\sigma$-sets. Each basic open subset of $\mathbb N^\omega$ maps to a product of $F_{\sigma}$-subsets of $\mathbb Q$ whose factors are eventually all of $\mathbb Q ^\omega$. So that image is $F_{\sigma}$ in $\mathbb Q ^\omega$. Each open subset of $\mathbb N ^\omega$ is a countably union of basic open sets, so its image is a countable union of $F_{\sigma}$-sets. $\endgroup$ – D.S. Lipham Aug 10 '20 at 22:55

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