# On a quasi-separated assumption in a lemma for the homotopy exact sequence of the etale fundamental group

Background:

I've seen two versions of the homotopy exact sequence for etale fundamental groups. One from Stacks:

Stacks 0BTX: Let $$k$$ be a field with algebraic closure $$\overline{k}$$. Let $$X$$ be a quasi-compact and quasi-separated scheme over $$k$$. If the base change $$X_{\overline{k}}$$ is connected, then there is a short exact sequence $$1\to \pi_1(X_{\overline{k}}) \to \pi_1(X)\to \pi_1(\operatorname{Spec} k) \to 1$$ of profinite topological groups.

And one from SGA:

SGA I, Theoreme 6.1 of chapter IX: Suppose $$S$$ is the spectrum of an Artinian ring $$A$$ with residue field $$k$$, $$\overline{k}$$ an algebraic closure of $$k$$, $$X$$ a $$S$$-scheme, $$X_0=X\times_A k$$, $$\overline{X}_0=X\times_A \overline{k}$$, $$\overline{a}$$ a geometric point of $$\overline{X}$$, $$a$$ the image in $$X$$, and $$b$$ the image in $$S$$. We suppose that $$X_0$$ is quasi-compact and geometrically connected over $$k$$ (N.B. if $$X$$ is proper over $$S$$, this means that $$H^0(X_0,\mathcal{O}_{X_0})$$ is an artinian local ring whose residue field is radicial over $$k$$). Then the canonical sequence of homomorphisms $$1\to \pi_1(\overline{X}_0,\overline{a})\to \pi_1(X,a)\to \pi_1(S,b) \to 1$$ is exact, and we have $$\pi_1(S,b)\stackrel{\sim}{\leftarrow}\pi_1(k,\overline{k})= Gal(\overline{k},k).$$

A key step in both of these proofs is the following lemma:

Lemma: Let $$X$$ be quasi-compact and geometrically connected. If we have a finite etale cover $$\overline{Y}$$ of $$\overline{X}=X\times_k \operatorname{Spec}\overline{k}$$, then it comes from a finite etale cover of $$X\times_k \operatorname{Spec} K$$, where $$k\subset K$$ is a finite extension.

Stacks adds the assumption that $$X$$ is quasi-separated, and SGA omits this assumption. In the case when $$X$$ is assumed quasi-separated, I think I understand how to show this lemma and I can even write down a recipe for producing the extension we need: using the fact that $$X$$ is quasi-compact and quasi-separated in combination with the definition of etale morphisms as locally of finite presentation, we can pick a finite affine open cover $$U_i=\operatorname{Spec} A_i$$ of $$X$$ and get a finite affine open cover $$\overline{U_i}=U_i\times_k \operatorname{Spec} \overline{k}$$ of $$\overline{Y}$$ by spectra of rings of the form $$(A_i\otimes_k \overline{k})[x_1,\cdots,x_n]/(f_1,\cdots,f_m)$$ which gives us a finite list of coefficients from $$\overline{k}$$ needed to define the $$\overline{U_i}$$. Covering $$U_i\cap U_j$$ with a finite number of open affines $$U_{ijk}$$ since $$X$$ is quasi-separated, we see that there's a finite number of coefficients from $$\overline{k}$$ necessary to define the maps $$\overline{U_{ijk}}\to \overline{U_i}$$ and $$\overline{U_{ijk}}\to \overline{U_j}$$, and we can apply the same trick to get a finite list of coefficients of $$\overline{k}$$ needed to define the data we use to patch together the $$\overline{U_i}$$ into $$\overline{Y}$$. We end up with a finite list of elements of $$\overline{k}$$ which are enough to specify all the data needed to put together $$\overline{Y}$$, and we can define our cover over a finite extension of $$k$$ containing all these elements.

Question: How can I prove the lemma when $$X$$ is not quasi-separated? SGA leaves the proof of the lemma to the reader, and it appears to me that my strategy fails without the quasi-separated hypothesis (some $$U_i\cap U_j$$ could fail to be quasi-compact and then I would have to deal with a potentially infinite list of elements of $$\overline{k}$$). Stacks' copy of the lemma seems to rely on quasi-separatedness in an essential way, and I don't see how to remove it.

This is more a comment than an answer: a few years back, in 2011, while working with some friends on SGA1, we also found out that we could not prove this statement without the hypothesis that $$X$$ is quasi-separated. Our question: Is this hypothesis simply missing in SGA1 ? reached Michel Raynaud and his answer was reported to be something like: Probably, but this is not very interesting.
• Thank you - I think this more or less resolves my issue (I don't need to prove this in the case that $X$ fails to be quasi-separated, and it does seem reasonable to me that it's a minor error in SGA), but I'm going to leave the question open for a bit longer just in case a more definitive answer should happen to come along. – KReiser Jun 28 at 23:33