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Let $X$ be a spectrum. In various places, I have encountered the statement that $$ X \simeq \text{hocolim}_n \Sigma^{\infty-n}X_n. $$ I was wondering how this homotopy colimit is defined, and why we have such an equivalence.

The particular model for spectra I use is the one defined in Adams' blue book.

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  • $\begingroup$ If you have a countable sequence of maps in a triangulated category, you can build an object which you can meaningfully call the homotopy colimit. Taking this approach in the setting of symmetric spectra, you can read a proof of the result in Proposition II.5.12 in Schwede's Symmetric Spectra book project. $\endgroup$
    – Bruno Stonek
    Commented Dec 8, 2021 at 12:08

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If you use CW spectra, as in Adams' blue book, then you can write this as a (strict) colimit of an increasing sequence of spectra. Namely, let $X$ be a CW spectrum and denote by $\underline{X_n}$ the CW spectrum that has $(\underline{X_n})_i=X_i$ for $i\leq n$ and $(\underline{X_n})_{n+i}=\Sigma^iX_n$ for $i>0$ (which is an explicit model for $\Sigma^{\infty - n}X$). Then we have a sequence of inclusions of subspectra $\underline{X_0} \to \cdots \to \underline{X_n} \to \cdots$ whose union is $X$. This sequential colimit is also a homotopy colimit, since the maps are cellular inclusions of CW spectra.

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  • $\begingroup$ If we start with a CW spectrum $X$, so that $\Sigma X_i \hookrightarrow X_{i+1}$ is the inclusion of subcomplex, then the version of $\underline{X_n}$ I wrote down is also a CW spectrum in Adams' sense, no? $\endgroup$
    – Manuel Araújo
    Commented May 1 at 10:33
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    $\begingroup$ Yes, you are right. I'll delete my irrelevant comment. $\endgroup$
    – John Rognes
    Commented May 1 at 10:58

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