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Apr 19, 2022 at 21:17 answer added M L timeline score: 2
Apr 19, 2022 at 21:02 comment added user40276 @PiotrAchinger I've edited the question and corrected a mistake. I said profinite étale $\pi_1$ when I've actually meant étale $\pi_1$ for the map with dense image. In order to get the usual SGA1 $\pi_1$ you need to take profinite completion of the shape.
Apr 19, 2022 at 21:00 comment added Michael Hardy $$\begin{align} & pro-ét \\ {} \\ & \text{pro-ét} \end{align}$$ Above one sees what some parts of this posting looked like before and after the edits by LSpice. In particular, note that a hyphen looks different from a minus sign.
Apr 19, 2022 at 20:58 history edited Michael Hardy CC BY-SA 4.0
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Apr 19, 2022 at 20:56 history edited LSpice CC BY-SA 4.0
`\DeclareMathOperator`; `\text`; names of references; minor typos; deleted "thanks"
Apr 19, 2022 at 20:53 comment added Piotr Achinger Sorry, I deleted my comment because I could not edit it. Let me repeat here my comments so that the above makes sense: 1) the map with dense image goes from proetale $\pi_1$ to etale $\pi_1$; in fact there is a third prodiscrete group $\pi_1^{\rm SGA3}$ and maps $\pi_1^{\rm proet}\to \pi_1^{\rm SGA3}\to \pi_1^{\rm et}$ which are respectively the prodiscrete and profinite completion. 2) Artin-Mazur observed that $\pi_1^{\rm SGA3}$ is the $\pi_1$ of the etale homotopy type. Does that imply that $\Pi_\infty(X_{\rm et}$ and $\Pi_\infty(X_{\rm proet})$ are different bc they have different $\pi_1$?
Apr 19, 2022 at 20:48 history edited user40276 CC BY-SA 4.0
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Apr 19, 2022 at 20:44 comment added user40276 @PiotrAchinger I'm not well-versed too. I've included the reference (Remark 7.4.12 in BS). What's the étale $\pi_1$ in your comment? I think you mean profinite $\pi_1$. If so, you need to take the profinite completion of the shape or take the finite étale topoi. Am I misunderstanding anything?
Apr 19, 2022 at 20:40 history edited user40276 CC BY-SA 4.0
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Apr 19, 2022 at 20:05 history asked user40276 CC BY-SA 4.0