Timeline for When is a stable $\infty$-category the stabilization of an $\infty$-topos?
Current License: CC BY-SA 4.0
17 events
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| Nov 10, 2022 at 22:16 | comment | added | user40276 | Is even the $1$-categorial version of that known (for $1$-topos and Grothendieck abelian categories)? | |
| Nov 10, 2022 at 21:47 | answer | added | Tim Campion | timeline score: 3 | |
| Jul 8, 2022 at 20:09 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Jul 8, 2022 at 13:14 | comment | added | Tim Campion | @PeterLeFanuLumsdaine that seems reasonable. I’m just following the usage of Joyal et al. And I’m the oo-categorical setting, still nobody knows what a “logical morphism “ should be. | |
| Jul 8, 2022 at 8:12 | comment | added | Peter LeFanu Lumsdaine | A side note, but calling that category of infinity-toposes “logoi” seems peculiar — the established use of “logos” I’m familiar with is typically to distinguish elementary (infinity-)toposes and logical morphisms, by contrast with Grothendieck (infinity-)toposes and geometric morphisms; or else something closely analogous to this, distinguishing the logical category from the geometric. | |
| Jul 8, 2022 at 0:08 | answer | added | Doron Grossman-Naples | timeline score: 7 | |
| Jul 5, 2022 at 13:55 | comment | added | Bbb | Ah thanks @TimCampion indeed I was being silly. Although trying to reconstruct X from the stable operad structure coming from differentiation as I suggested above could be interesting as well. | |
| Jul 5, 2022 at 3:43 | comment | added | Tim Campion | @Bbb the cartesian monoidal structure on an additive category is not closed, but the monoidal structure I’m talking about is closed, so they are very different. The cartesian monoidal structure on the topos induces a very much non cartesian monoidal structure on the stabilization. Eg this is exactly how the smash product of spectra is induced by the cartesian product of spaces | |
| Jul 4, 2022 at 22:59 | comment | added | Bbb | @TimCampion but that wouldn’t look anything like the smash product right? I might be confused but wouldn’t that give you Sp(X) with its cartesian monoidal structure? | |
| Jul 4, 2022 at 22:54 | comment | added | Tim Campion | @Bbb sure but stabilization is a symmetric monoidal functor from PrL to stprl so it does turn any presentable Cartesian closed category into a symmetric monoidal closed stable category | |
| Jul 4, 2022 at 22:41 | comment | added | Bbb | @TimCampion stabilizing the cartesian product doesn’t give a symmetric monoidal struture in general - this “multiplicative structure” from differentiating the cartesian product only exhibits Sp(X) as an operad. This is discussed in HA 6.2.4 and also in the thesis of Heuts. | |
| Jul 4, 2022 at 14:55 | comment | added | Tim Campion | I think there's always a symmetric monoidal "smash product" on $Stab(\mathcal X)$ arising from the cartesian product on $\mathcal X$. It may make sense to take such a symmetric monoidal structure as part of the input when trying to reconstruct a topos $\mathcal X$. | |
| Jul 3, 2022 at 21:14 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Jul 3, 2022 at 21:13 | comment | added | Tim Campion | @MaximeRamzi Oh you're right -- it's rather that the right adjoint itself has a right adjoint. Hm... so Question 3 seems to be a dead end | |
| Jul 3, 2022 at 21:12 | comment | added | Maxime Ramzi | I don't think $Stab$ has a left adjoint, does it ? For instance, it does not preserve the pullback $Spaces \times_{CMon} CGrp = 0$ (along the free functor and the inclusion) Also, its right adjoint $R$ is the forgetful functor, so $R\mathcal A$ is never a topos unless $\mathcal A$ is, which I think happens iff $\mathcal A = 0$. | |
| Jul 3, 2022 at 20:29 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Jul 3, 2022 at 20:23 | history | asked | Tim Campion | CC BY-SA 4.0 |