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Let $X$ be a nice scheme (additional assumptions could be added), and let $Et(X)$ be its (Artin-Mazur) etale homotopy type. I am looking for a/the scheme $Y$ over $X$ whose etale homotopy type $Et(Y)$ will be the topological universal cover of $Et(X)$. By definition $Et(X)$ is the geometric realization of a simplicial set, and it was pointed out to me that if $R \rightarrow S$ is the universal cover of a simplicial set $S$, then the geometric realization $|R|$ is the topological universal cover of $|S|$.

What is the meaning of "Universal cover of a simplicial set"; Is there a reference for that, and also for the second assertion?. How could we apply this to find $Y$?

Edit: If we want to avoid the simplicial method. Suppose that there is a scheme $Y$ over $X$ such that

  1. The first etale homotopy group $\pi_1^{et}(Y)$ is trivial.

  2. For all $n \geq 2$ we have $\pi_n^{et}(Y) \simeq \pi_n^{et}(X)$.

Are these conditions sufficient to state that $Et(Y)$ is the topological universal cover of $Et(X)$?

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    $\begingroup$ Do you have a universal cover of the multiplicative group $\mathbb G_m$? The conditions that you listed seem to be sufficient if you replace the second by the stronger condition that the map $\pi_n^{\operatorname{ét}}(Y)\to\pi_n^{\operatorname{ét}}(X)$ induced by the map $Y\to X$ is an isomorphism for every $n>1$. However, I suspect that the existence is very rare. $\endgroup$
    – Z. M
    Commented Jul 20, 2021 at 9:39
  • $\begingroup$ @Z.M Thank you for your comment. Yes, the condition you mentioned is exactly the one i meant. $\endgroup$ Commented Jul 25, 2021 at 16:29

1 Answer 1

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Such an "étale universal cover" exists at least if $X$ is Noetherian and geometrically unibranch (and for all qcqs $X$ if one considers profinite étale homotopy types).

Background. I will regard the étale homotopy type of a scheme as an object in the $\infty$-category $\mathrm{Pro}(\mathcal S)$ of pro-spaces. In the $\infty$-category $\mathcal S$, the universal cover of a pointed space $(X,x)$ can be characterized as the initial object in the $\infty$-category of pointed $0$-truncated morphisms to $(X,x)$ (i.e., morphisms with discrete fibers). Similarly, one can define $0$-truncated morphisms in $\mathrm{Pro}(\mathcal S)$, and every pointed object admits a universal cover. References for the étale homotopy type and for $n$-truncated/$n$-connected morphisms in $\mathrm{Pro}(\mathcal S)$ are Sections E.2 and E.4.2 in Spectral Algebraic Geometry.

Let $\mathcal S_{<\infty}$ be the $\infty$-category of truncated spaces and let $\mathrm{Et}\colon \mathrm{Sch} \to \mathrm{Pro}(\mathcal S_{<\infty})$ be the protruncated étale homotopy type, which is the homotopy-coherent incarnation of the construction of Artin and Mazur. Under the equivalence $$\mathrm{Pro}(\mathcal S_{<\infty}) \simeq \mathrm{Fun}^\mathrm{acc,lex}(\mathcal S_{<\infty},\mathcal S)^\mathrm{op},$$ $\mathrm{Et}(X)$ is the functor $\mathcal S_{<\infty}\to\mathcal S$ sending a truncated space $K$ to the global sections $\Gamma(X_\mathrm{et},K)$ of the constant étale sheaf on $X$ with value $K$.

Construction. Let $X$ be a scheme with a geometric point $x\colon \operatorname{Spec} k\to X$, with $k$ separably closed. Consider the category $\mathrm{FEt}(X,x)$ of pointed finite étale covers of $X$, that is, factorizations of $x$ through a finite étale morphism $X'\to X$. This category has finite limits and in particular is cofiltered (it is also essentially small). Thus, the limit $\tilde X$ of the forgetful functor $$ \mathrm{FEt}(X,x) \to \mathrm{Sch}_{/X} $$ exists (note that $\tilde X$ depends on $x$, so the notation is abusive). This is a natural candidate for the "universal cover" of $(X,x)$.

Now, the statement that $\mathrm{Et}(\tilde X)\to\mathrm{Et}(X)$ is the universal cover of the pointed pro-space $(\mathrm{Et}(X),x)$ is equivalent to the following three conditions: 1) the map $\mathrm{Et}(\tilde X)\to\mathrm{Et}(X)$ is $0$-truncated, 2) $\mathrm{Et}(\tilde X)$ is connected, and 3) $\pi_1^\mathrm{et}(\tilde X,x)$ is trivial.

Results.

1) always holds if $X$ is quasi-compact and quasi-separated.

First, I claim that the functor $\mathrm{Et}$ preserves the limit of the diagram defining $\tilde X$. Under the above equivalence of $\infty$-categories, this is the statement that for $K$ a truncated space, $\Gamma((-)_\mathrm{et},K)$ transforms this limit into a colimit, which is a standard property of étale cohomology with respect to inverse limits of qcqs schemes. [Here it is important that $K$ is truncated, otherwise this may not be true.]

Then, if $p\colon X'\to X$ is finite étale, I claim that the morphism $\mathrm{Et}(p)\colon \mathrm{Et}(X')\to \mathrm{Et}(X)$ is $0$-truncated. In fact, it is the pullback of a morphism of groupoids $\pi\colon\Xi'\to\Xi$ with finite discrete fibers. To see this, note that the morphism of étale $\infty$-topoi induced by $p$ is itself the pullback of such a morphism $\pi$. The point is then that for any space $K$, $\Gamma(X'_\mathrm{et}, K)$ is $\Gamma(X_\mathrm{et},p_*K)$ and $p_*K$ is a locally constant sheaf on $X$ in a strong sense: it is the pullback of the sheaf $\pi_*K$ on $\Xi$ (the fact that the fibers of $\pi$ are finite spaces is used here, to commute the pushforward with the pullback). One can thus apply Proposition 2.15 in Higher Galois theory to compute $\Gamma(X_\mathrm{et},p_*K)$ in terms of the étale homotopy type of $X$. Unpacking this formula gives $\mathrm{Et}(X')=\mathrm{Et}(X)\times_\Xi\Xi'$. [Here, $K$ need not be truncated, so $p$ induces a $0$-truncated morphism on actual étale homotopy types, and $X$ need not even be qcqs.]

ETA: If $X$ is locally noetherian, another proof of this claim is Lemma 2.1 in Schmidt-Stix, Anabelian geometry with etale homotopy types.

Finally, since $0$-truncated morphisms are stable under limits, $\mathrm{Et}(\tilde X)\to\mathrm{Et}(X)$ is also $0$-truncated.

2) also holds if $X$ is qcqs. In this case $\tilde X$ is connected, as any clopen subset of $\tilde X$ lifts to a clopen subset of some finite étale $X'\to X$.

3) holds if we assume moreover that the pro-group $\pi_1^\mathrm{et}(X,x)$ is profinite, e.g., $X$ is Noetherian and geometrically unibranch. Then $\pi_1^\mathrm{et}(\tilde X,x)$ is also profinite by 1), so 3) is equivalent to the statement that every finite étale cover of $\tilde X$ is trivial, which holds by construction.

Note that the additional assumption for 3) is not needed if we pass to the profinite completions. That is, if $X$ is qcqs, then $\mathrm{Et}(\tilde X)^\wedge \to \mathrm{Et}(X)^\wedge$ is the universal cover of the profinite étale homotopy type $(\mathrm{Et}(X)^\wedge,x)$.

Example. Let $X=\mathbb G_m$ over a field $k$ of characteristic zero, pointed at $1$ by an algebraic closure $\bar k$ of $k$. Then $\tilde X=\operatorname{Spec}\bar k[t^{\pm 1}][t^{1/n}, n\geq 2]$.

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  • $\begingroup$ Thank you for your detailed answer. Could we translate the meaning of a n-truncated morphism $Et(Y) \rightarrow Et(X)$ by $\pi_p(Et(Y)) \simeq \pi_p(Et(X))$ for all $p \geq n+2$? $\endgroup$ Commented Jul 25, 2021 at 21:19
  • $\begingroup$ Not quite, $n$-truncated means iso on $\pi_p$ for $p\geq n+2$ and mono on $\pi_{n+1}$. However, for pro-spaces one cannot quite define them this way because pro-spaces may not have enough base points (eg, the set of points could be empty). The actual definition is: a morphism in $\mathrm{Pro}(\mathcal S)$ is $n$-truncated if it is a cofiltered limit of $n$-truncated morphisms in $\mathcal S$. $\endgroup$ Commented Jul 26, 2021 at 5:58
  • $\begingroup$ I guess that the subtlety of $\operatorname{Pro}(\mathcal S)$ disappears if you pass to the profinite completion? $\endgroup$
    – Z. M
    Commented Jul 26, 2021 at 11:10
  • $\begingroup$ @Z.M Indeed profinite space are much better behaved. In particular a morphism of profinite spaces is $n$-truncated iff it is so after passing to the limit, this is Prop. E.4.6.1 in Spectral Algebraic Geometry. In order to be able to check $n$-truncativity using homotopy pro-groups, I would guess that $\pi_0$ being profinite is enough, which is certainly the case for $\mathrm{Et}$ of a qcqs scheme. $\endgroup$ Commented Jul 26, 2021 at 12:50

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