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It is easy enough to construct the separable closure $k^{sep}$ of a field $k$, which then has $\pi_0(k^{sep}) = 0$ (profinite set of connected components), $\pi_1(k^{sep}) = 0$ (there are no nontrivial finite étale extensions of $k^{sep}$).

Let $B$ be a connected scheme. How do I construct a smooth (or formally smooth) map $E \rightarrow B$ such that $E$ has weakly contractible étale homotopy type?

This would be like a universal cover, but for the higher étale homotopy groups.

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  • $\begingroup$ What sort of an object are you expecting $E$ to be? In the topological setting already the $3$-connected cover of $S^3$ isn't a manifold (it's the string group $\text{String}(3)$) so I imagine it's too much to expect $E$ to be a scheme (as opposed to an ind-scheme or something stranger). $\endgroup$
    – Qiaochu Yuan
    Commented Sep 21, 2020 at 19:51
  • $\begingroup$ @QiaochuYuan. Could it still be a higher topos? My goal was to construct the loop space of a scheme by taking $E \times_B E$, but I did not foresee this problem. $\endgroup$
    – user30211
    Commented Sep 21, 2020 at 19:59
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    $\begingroup$ Depending on what you want out of a loop space you can do the following instead. You can construct the based loop space $\Omega X$ of a based space from its free loop space $LX$ by fixing a point on the circle, considering the corresponding projection $LX \to X$, and taking a (homotopy) fiber. There's an analogous construction in AG: the (derived) free loop space $LX$ of a scheme (or stack, derived stack, etc.) is the (derived) fiber product $X \times_{X \times X} X$, where the two maps $X \to X \times X$ are both the diagonal $\Delta$. This is a global version of Hochschild homology. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 21, 2020 at 20:04
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    $\begingroup$ Maybe you want a loop space that shifts all the etale homotopy up one degree though; I don't think this construction will have that property but I have no idea. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 21, 2020 at 20:07
  • $\begingroup$ Thanks Qiaochu, you're a legend $\endgroup$
    – user30211
    Commented Sep 21, 2020 at 20:23

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