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Let $\varphi:\mathbb R^n \to \mathbb R^n$ be just some continuous function.

If the image of $\varphi$ happens to contain $\mathbb Q^n$, does it follow that in fact all of $\mathbb R^n$ is contained in the image as well?

No, it does not. For instance, the map given by $$(x,y)\overset{\varphi}{\longmapsto} (xy-1+\pi, x^2(xy-1)+y)$$ has as its image every point in $\mathbb R^2$ except for $(\pi,0)$.

But now suppose instead that $\mathbb Q^n$ is actually contained in the image of just $\mathbb Q^n$. That is, instead of requiring only that $$\mathbb Q^n \subseteq \varphi[\mathbb R^n]\phantom{.}$$ we require that in fact $$\mathbb Q^n \subseteq \varphi[\mathbb Q^n].$$

Question. Does it then follow that $\varphi[\mathbb R^n]=\mathbb R^n$, or is there a counterexample?

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    $\begingroup$ Yes for $n=1$. No for $n\ge 2$. For instance, for $n=2$ take the complex exponential plus an irrational ($f:z\mapsto\exp(z)+z_0$), and precompose with a self-homeomorphism of $\mathbf{R}^2$ mapping $\mathbf{Q}^2$ onto $f^{-1}(\mathbf{Q}^2)$. $\endgroup$
    – YCor
    Commented Jun 25, 2021 at 12:15
  • $\begingroup$ @YCor : How is such a self-homeomorphism constructed? $\endgroup$
    – Iosif Pinelis
    Commented Jun 25, 2021 at 12:56
  • $\begingroup$ @IosifPinelis this inverse image is dense, and $\mathrm{Homeo}(\mathbf{R}^n)$ acts transitively on dense countable subsets of $\mathbf{R}^n$ (I think the latter fact appears somewhere on MO). $\endgroup$
    – YCor
    Commented Jun 25, 2021 at 13:02
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    $\begingroup$ @YCor : Thank you. I have indeed found this answer: mathoverflow.net/a/281159/36721 . $\endgroup$
    – Iosif Pinelis
    Commented Jun 25, 2021 at 13:23
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    $\begingroup$ @YCor, since that (together with @‍IosifPinelis's reference) seems to be a complete answer, maybe it could be posted as such? $\endgroup$
    – LSpice
    Commented Jun 25, 2021 at 13:47

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A counterexample for $n=2$ is the map $\varphi(x,y) = (x,(x^2-2)y)$. Each point $(r,s)\in\mathbb{Q}^2$ is the image of $\left(r,\frac{s}{r^2-2}\right)\in\mathbb{Q}^2$, but e.g. $(\sqrt{2},1)\not\in\varphi(\mathbb{R}^2)$.

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    $\begingroup$ For $n\ge 2$ then take $(x_1,\dots,x_n)\mapsto (x_1(x_n^2-2),x_2,\dots,x_n)$, whose inverse as birational map is $(x_1,\dots,x_n)\mapsto (x_1/(x_n^2-2),x_2,\dots,x_n)$ $\endgroup$
    – YCor
    Commented Jun 25, 2021 at 18:38
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    $\begingroup$ Fabulous! I was secretly wondering about polynomial maps, so this is extra useful for me. Thank you! $\endgroup$
    – Louis Deaett
    Commented Jun 26, 2021 at 0:17

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