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Let $\mathbb Q^\omega_0:=\{(x_i)_{i\in\omega}\in\mathbb Q^\omega:\exists n\in\omega\;\forall m\ge n\;\;x_m=0\}$ and observe that $\mathbb Q^\omega_0$ is a countable dense set in $\mathbb R^\omega$ (endowed with Tychonoff product topology).

Question. Is there a continuous map $f:\mathbb R^\omega\to\mathbb R^\omega$ such that $f^{-1}(\mathbb Q^\omega)=\mathbb Q^\omega_0$?

Remark. For any countable set $A$ in a finite-dimensional metrizable separable space $X$ there exists a topological embedding $f:X\to\mathbb R^\omega$ such that $f^{-1}(\mathbb Q^\omega)=A$.

Indeed, for $n=2\dim(X)+2$ we can find a topological embedding $g:X\to\mathbb R^n$ such that the image $g(X)$ does not intersect the set $\mathbb Q^n$. Since the set $\mathbb Q^n\cup g(A)$ is countable and dense in $\mathbb R^n$, there exists a homeomorphism $h:\mathbb R^n\to\mathbb R^n$ such that $h^{-1}(\mathbb Q^n)=\mathbb Q^n\cup g(A)$. Consider the embedding $e:\mathbb R^n\to\mathbb R^\omega$, $e:(x_1,\dots,x_n)\mapsto(x_1,\dots,x_n,0,0,\dots)$, and observe that $e^{-1}(\mathbb Q^\omega)=\mathbb Q^n$. Then the map $f=e\circ h\circ g:X\to\mathbb R^\omega$ is a required topological embedding with $f^{-1}(\mathbb Q^\omega)=A$.

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