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Oct 31 at 19:32 vote accept Tim Campion
Sep 19 at 21:11 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Aug 20 at 17:35 answer added Brendan Murphy timeline score: 3
Oct 6, 2023 at 13:46 comment added Tyler Lawson @TimCampion The mod-$p$ functor is given by tensoring with $\Bbb F_p$, equivalent to the perfect complex $\Bbb Z \to \Bbb Z$ in degrees 0 and 1. This complex is dualizable with dual $\Bbb F_p[-1]$, and so $\Bbb F_p \otimes M \simeq map(\Bbb F_p[-1], M)$. This makes the left adjoint tensoring with $\Bbb F_p[-1] = map(\Bbb F_p, \Bbb Z)$.
Sep 30, 2023 at 7:34 comment added Drew Heard Theorem 2.30 (and its proof) of arxiv.org/pdf/1507.06869.pdf are relevant
Sep 29, 2023 at 21:26 comment added Z. M The mod p functor is a finite colimit (the cofiber of multiplication by p), thus preserves (co)limits.
Sep 29, 2023 at 20:52 comment added LSpice TeX note: ${}^\to_\leftarrow$ {}^\to_\leftarrow is available without any need for tricky constructs as $\rightleftarrows$ \rightleftarrows. I edited accordingly.
Sep 29, 2023 at 20:52 history edited LSpice CC BY-SA 4.0
Oops, switched the arrows
Sep 29, 2023 at 20:36 history edited LSpice CC BY-SA 4.0
TeX; titles of papers
Sep 29, 2023 at 19:57 history edited Tim Campion CC BY-SA 4.0
deleted 36 characters in body
Sep 29, 2023 at 19:53 comment added Tim Campion Why does the mod p functor preserve limits? What is its left adjoint?
Sep 29, 2023 at 19:06 comment added Marc Hoyois I agree with Z. M. Moreover, the mod $p$ functor preserves both limits and colimits, so it is both monadic and comonadic.
Sep 29, 2023 at 19:04 comment added Z. M I think that this is a direct consequence of Beck's monadicity theorem (or more precisely, Lurie's generalization): the mod p functor is a left adjoint, conservative, and preserves totalizations of cosimplicial objects.
Sep 29, 2023 at 18:45 history asked Tim Campion CC BY-SA 4.0