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The ambiguity inherent in defining the absolute Galois group $G_\mathbb{Q}$ - that it is determined only up to inner automorphisms - arises from the fact that one has to choose an algebraic closure of $\mathbb{Q}$ and there are many choices. Viewing the Galois group as fundamental group, choosing an algebraic closure amounts to choosing a base point with respect to which the fundamental group is defined.

This naturally leads to: what is the set of all algebraic closures of $\mathbb{Q}$? Is there a meaningful sense in which it forms a topological space? What are its topological properties? Is it path-connected? Is it $K(G_\mathbb{Q}, 1)$?

The answer I feel is obvious, but I'm too obtuse to see it.

A tangentially related question: is there a good example of a situation/theorem where working systematically with fundamental groupoid rather than fundamental group has proved crucial to the argument?

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    $\begingroup$ The collection of all algebraic closures naturally forms a groupoid, to get a "space of algebraic closures" you could take the nerve of that. Since the groupoid is connected, that is going to give you a $K(G_\mathbb{Q}, 1)$. $\endgroup$ Commented Apr 12, 2019 at 6:20
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    $\begingroup$ As a more advanced version of what Achim Krause said, you can also take the profinite space given by the étale homotopy type of $\mathrm{Spec}\,k$ (which is also a $K(G_\mathbb{Q},1)$, although in a slightly more refined way that remembers the topology of $G_\mathbb{Q}$) $\endgroup$ Commented Apr 12, 2019 at 7:19
  • $\begingroup$ To answer your last question, I don't know what "crucial" means here, but there is a groupoid version of the Seifert-Van Kampen theorem, which allows one to compute the fundamental group of $S^1$, while the group version doesn't allow that. Similarly, you can compute the fundamental group of a quotient through groupoid techniques, and you can't always with the group. See Brown's Topology and groupoids $\endgroup$ Commented Apr 12, 2019 at 9:17

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