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Jul 27, 2021 at 12:57 history edited Marc Hoyois CC BY-SA 4.0
Added reference to Schmidt-Stix
Jul 26, 2021 at 12:50 comment added Marc Hoyois @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.
Jul 26, 2021 at 11:10 comment added Z. M I guess that the subtlety of $\operatorname{Pro}(\mathcal S)$ disappears if you pass to the profinite completion?
Jul 26, 2021 at 5:58 comment added Marc Hoyois 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 $\mathcal S$.
Jul 25, 2021 at 21:19 comment added Moutand Mohammed 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 $p \geq n+2$?
Jul 25, 2021 at 16:22 vote accept Moutand Mohammed
Jul 25, 2021 at 16:22 vote accept Moutand Mohammed
Jul 25, 2021 at 16:22
Jul 22, 2021 at 5:52 history answered Marc Hoyois CC BY-SA 4.0