7:37 AM
This is a pretty important technical paper, imo: arxiv.org/pdf/2003.11757.pdf . I don't know how into this stuff the other people here are, but I've been waiting to read this for a while.
2

2 hours later…
9:30 AM
@S.carmeli I have spent some time now looking at Lurie's Warning. Would it be possible for you to explain why equivalence on the stalk implies $\infty$-connectedness? Also, how do you see that $\mathbf{Z}/p^k\mathbf{Z} \mapsto \mathcal{S}(\mathbf{R}/p^k\mathbf{Z}, M(p))$ satisfies hyperdescent?

9:47 AM
@AdrianClough The fact that infinity connectedness is an equivalence of the stalk follows from the analogous statement for sheaves of groups (by definition and the fact that f^*\pi_i=\pi_i f^*). But sheaves of groups on continuous Z_p-sets are equivalent to groups with continuous Z_p action. This is a "classical" fact (still, should I explain this?).
For the second statement, look at the map s:BZ--> BZ_p. The formula I have written for the hypersheafification is s_s^. It is then hypercomplete because BZ is hypercomplete and the right adjoint in a geometric morphism preserve hypercompleteness (hypercompletion is a pointed functor on infinity topoi...). To see that this is actually the hypercompletion you need to use the fact that M(p) is p-torsion, so that the stalk of s_*s^*M(p) is M(p)
Indeed, you have to check that the canonical map M(p)--> map(R/p^kZ,M(p)) embedding the constant maps induces equivalence on the colimit over k. This follows by comparing the homotopy groups and using the fact that the transision maps are of degree.
p
thats explains anything?

3 hours later…
12:49 PM
@SaalHardali related to your question: a map HZ[1]-->Pic(C) is called in Hopkins-Lurie paper on the Brauer group a "polarization" I think.