Mapping $\mathbb P$ onto $\mathbb Q ^\omega$

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.

• Is $\mathbb{P}$ the set of irrationals, or something else? – Nate Eldredge Aug 9 '20 at 23:14

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.
• 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. – D.S. Lipham Aug 10 '20 at 22:55