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This is a kind of followup question to this old one.

Let $\mathbb{Q}$ and $\mathbb{R}$ have their usual (Euclidean) topology, and let $\beta\mathbb{Q}$ and $\beta\mathbb{R}$ stand for their respective Stone-Čech compactification. The inclusion $\mathbb{Q} \to \mathbb{R}$ induces a continuous map $h\,\colon \beta\mathbb{Q} \to \beta\mathbb{R}$ (by functoriality). Without thinking, I had casually assumed that $h$ was injective (because “$\mathbb{R}$ is larger than $\mathbb{Q}$ so surely its Stone-Čech compactification must be larger as well” — of course the reasoning is stupid because the S.-Č. compactification of $\mathbb{R}\cup\{\pm\infty\}$ certainly isn't larger than that of $\mathbb{R}$); however, this is not true. Indeed:

Proposition: $h$ is surjective, and it is not injective.

Proof: The image of $h$ is compact because $\beta\mathbb{Q}$ is compact, so it is closed in $\beta\mathbb{R}$; but it contains $\mathbb{Q}$ (which we can unproblematically identify as a subspace of all spaces involved), which is dense in $\mathbb{R}$, which is dense in $\beta\mathbb{R}$: this shows that $h$ is surjective. As for non-injectivity: let $x$ be an irrational number; the sets $C_+ := \{r\in\mathbb{Q} : r>x\}$ and $C_- := \{r\in\mathbb{Q} : r<x\}$ are closed in $\mathbb{Q}$ (because $r>x$ is equivalent to $r\geq x$), and, in fact, they are zero-sets (e.g., $C_+$ is the zero-set of the function $\max(x-r,0)$); since they are disjoint in $\mathbb{Q}$, their closures in $\beta\mathbb{Q}$ are disjoint (cf. Gillman & Jerison, Rings of Continuous Functions, theorem 6.5(III)); but their closures in $\mathbb{R}$ certainly aren't disjoint, so neither are their closures in $\beta\mathbb{R}$: so $h$ cannot be injective (and more precisely, there are points of the closures of $C_+$ and $C_-$ in $\beta\mathbb{Q}$ which map to $x \in \beta\mathbb{R}$ even though they are disjoint in $\beta\mathbb{Q}$). ∎

I find this… difficult to picture intuitively. The best I can do is imagine that $\beta\mathbb{R}$ is a real line with an impossibly complicated mess at both ends, whereas $\beta\mathbb{Q}$ has an impossibly complicated mess at every gap between rationals, and the map $h$ squashes these messes to real numbers. But I suppose strange things must happen at the ends, too. Also, as pointed out in an answer to the old question I linked to, Stannett, “The Stone-Čech Compactification of the Rational World”, Glasgow Math. J. 30 (1988) 181–188, proves that $\beta\mathbb{Q}\setminus\mathbb{Q}$ is separable (whereas $\beta\mathbb{R}\setminus\mathbb{R}$ isn't).

Anyway, I hope I will be forgiven for asking a very vague—

Question: what other interesting properties can be shown of $h$ (or perhaps of its fibers) that might help visualize it?

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  • $\begingroup$ I think your intuitive picture is completely right. When the Stone-Čech compactification functor "sees" a sequence trying not to have a convergent subsequence, it's going to try to compactify in every possible way. So we get a huge mess near every irrational number when we take the Stone-Čech compactification of $\mathbb{Q}$, all of which just maps to that irrational number in the Stone-Čech compactification of $\mathbb{R}$. But $\mathbb{Q}$ and $\mathbb{R}$ has the same mess at the ends. I don't think this is saying anything you don't already see, hence not an answer! $\endgroup$
    – hunter
    Commented Aug 23 at 13:41
  • $\begingroup$ @hunter I should probably have made it clear that what I mostly can't visualize is what the fibers of $h$ above the points at infinity are like (i.e., how the two different messes at infinity compare). But I don't have a precise question either. $\endgroup$
    – Gro-Tsen
    Commented Aug 23 at 13:55
  • $\begingroup$ Maybe one question is whether $\beta \mathbb{Q} \setminus (\beta\iota)^{-1}(\mathbb{R})$ is homeomorphic (via $\beta\iota$) to $\beta \mathbb{R} \setminus \mathbb{R}$, where $\beta\iota$ is the map on s-c compactifications induced by the inclusion $\iota: \mathbb{Q} \to \mathbb{R}$. $\endgroup$
    – hunter
    Commented Aug 23 at 14:15
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    $\begingroup$ That $h$ is not injective, also follows (more quickly but perhaps less intuitive than your proof), from the fact that h is continuous, closed, surjective, but not a homeomorphism. (For instance, $\beta \mathbb R$ is connected, but $\beta \mathbb Q$ isn't.) $\endgroup$
    – Ulli
    Commented Aug 24 at 9:15

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