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Nov 4, 2021 at 5:19 comment added Dustin Clausen Sorry, i just noticed your latest comment now. Since S is discrete, the only nuclear modules are discrete. You don't get a Verdier sequence when you restrict to the nuclear full subcategory.
Apr 13, 2021 at 9:00 comment added Z. M Wait, what about the kernel $Z$ of $D((S,\mathbb Z)_\blacksquare)\to D(S_\infty)$? It seems to me that for any compact idempotent in a stable symmetric monoidal $\infty$-category $\mathcal C$, we have an associated Verdier exact sequence (probably split exact?) $\mathcal C'\to\mathcal C\to\mathcal C''$ with $\mathcal C\to\mathcal C''$ being symmetric monoidal with ideal $\mathcal C'$? I suppose that this construction passes to nuclear objects and the full subcategory of $Z$ spanned by nuclear objects is what you want?
Apr 12, 2021 at 13:18 comment added Dustin Clausen Exactly right, it doesn't preserve nuclearity
Apr 12, 2021 at 12:44 comment added Z. M I twisted myself by thinking about the subcategory $D(S_\infty)\subseteq D((S,\mathbb Z)_\blacksquare)$ and was wondering its quotient, but this inclusion does not preserve nuclearity.
Apr 12, 2021 at 12:17 vote accept Z. M
Apr 12, 2021 at 10:49 comment added Dustin Clausen It's a map of ring objects in solid $\mathbb{Z}$-modules, and everything can be done in that context. $R=\mathbb{Z}$ here.
Apr 12, 2021 at 10:47 comment added Dustin Clausen Yes, the map is $K(S) \rightarrow K(S_\infty)$, exactly as you say. Where is the confusion?
Apr 12, 2021 at 10:43 comment added Z. M I am a bit confused by the direction of the map you depicted. If I am not mistaken, the natural map is $S\to S_\infty$, and then I guess that the map should look like $K(S)\to K(S_\infty)$? Maybe I am mistaken, but $D(S_\infty)$ looks like a reflective subcategory of $D((S,\mathbb Z)_\blacksquare)$?
Apr 12, 2021 at 10:30 comment added Dustin Clausen Right, this K-theory is disrete, but you can promote it to a condensed spectrum by replacing the rings with internal hom from an extr. disc. profinite T to them to define the T-valued points of the condensed K-theory spectrum. About the Morita-invariance question, I don't follow waht you're asking. About compactly supported topological K-theory being the K-theory of a category, it's also not clear to me, but I think it should be.
Apr 10, 2021 at 22:17 comment added Z. M The continuous K-theory in question is discrete and so is the proposed K-theory? I don't know the ring of functions near the boundary in general, but does that matter, in view of the Morita-invariance? By the way, I don't know whether the compactly supported topological K-theory is the K-theory of some category.
Apr 10, 2021 at 21:28 history answered Dustin Clausen CC BY-SA 4.0