Given the Chromatic Convergence Theorem, can we state this globally as some convergence in the category of categories? That is, we have the subcategories of finite spectra in each stable homotopy category of $E(n)$-localized spectra, say $\mathcal{L}_n^{fin}$. Is it correct that there should be "inclusion" functors $\mathcal{L}_n^{fin}\to\mathcal{L}_{n+1}^{fin}$? Can we say that if we take some kind of limit (maybe this should be phrased in terms of quasi-categories or something?) we get the homotopy category of finite spectra?
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You have inclusion and localization functors. The inclusion functors go in the direction you indicate (any $E(n)$-local spectrum is also $E(n+1)$-local), and the localization functors are left adjoints, goint in the oposite direction. In your heuristics, I would consider the localization functors because of the following reason. Given a spectrum $X$ the units of the adjunctions give rise to a tower $$\cdots\rightarrow L_{E(n+1)}X\longrightarrow L_{E(n)}X\rightarrow\cdots\rightarrow L_{E(1)}X\longrightarrow L_{E(0)}X=X_{\mathbb{Q}},$$ and the chromatic convergence theorem says that for $X$ a $p$-local finite spectrum, $X$ can be computed as the following homotopy inverse limit
Now we can use this imput for the computation of mapping spaces between two $p$-local finite spectra $X$ and $Y$,
If $\mathcal{L}_{n}^{fin}$ denotes the category of $E(n)$-local $p$-local finite spectra, simplicially enriched via mapping spaces (e.g. using the hammock localization), then the sequence of localization functors
can be regarded as a sequence of simplicial functors which are the identity on objects. Given two objects $X$ and $Y$, the formula above shows that the homotopy limit at the level of mapping spaces recovers morphisms in the category of $p$-local spectra. I guess you would now like to say that this means that |
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As Fernando showed, you do have a functor from the category of finite spectra to an appropriate limit of the categories of $E(n)$-local finite spectra, and this recovers the mapping space between any pair of finite spectra:
For each $n$, choose a finite type $n$ complex However, there is no finite complex $Y$ such that |
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