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Hi, I have a question on weak-equivalences of spectra.

More precisely, I wonder whether filtered colimits of weak-equivalences of spectra are again weak-equivalences of spectra. Here, spectra are in the sense of Bousfield-Friedlander, i.e. a sequence of pointed simplicial sets $(E_0, E_1, \cdots, )$ with the morphisms $\sigma_n : S^1 \wedge E_n \to E_{n+1}$ for all $n \geq 0$. Of course, weak-equivalences of spectra are by definition the stable weak-equivalences.

After some efforts, I was able to find from a webpage that, if the spectra in question are all Kan spectra (spectra whose all levels are Kan simplicial sets), then this is indeed true. But, I have no idea how to proceed in general, partly because I do not think spectra are cofibrant in general.

I was informally told by someone that, in general, if we are working with a ``combinatorial'' model category, then such questions are likely be true. But, is it true that the category of spectra is combinatorial, and if that is the case, then does the main question hold?

I am fine with working with only Kan spectra if this question turns out to be negative, but since I didn't do my basic graduate school studies in algebraic topology, I hoped to know the answers for this question first for I thought the question might be easy for experts in the field. Thank you.

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    $\begingroup$ I don't have my copy of Hirschhorn handy, but an answer is also given there which is on the cofibrant side (whereas Fernando Muro's answer is on the fibrant side). Look in there about lambda sequences. There's something in Chapter 17 (it might be 17.4, but that's stretching my memory) about how you can take a cofibrant approximation to a lambda sequence and if a certain condition is satisfied then the colimit of that sequence of weak equivalences is a weak equivalence. Since this is phrased in model category generality, you might be able to find something more concrete for spectra $\endgroup$ Dec 23, 2012 at 5:45

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The answer is yes, but the reason is technical.

The reason is that, if I understand well, you're asking whether weak equivalences are closed under arbitrary filtered colimits. These are $\aleph_0$-filtered colimits, and there is a hierarchy of degrees of filtration parametrized by all infinite regular cardinals $\alpha$, being $\alpha=\aleph_0$ the first one.

In general combinatorial model categories, you know that there exists a big enough regular cardinal $\lambda$ such weak equivalences are closed under $\alpha$-filtered colimits for all $\alpha\geq \lambda$, but not necessarily for $\alpha<\lambda$.

The category of spectra of simplicial sets is locally $\aleph_0$-presentable and has sets of generating (trivial) cofibrations with $\aleph_0$-presentable sources. Hence you can take $\lambda=\aleph_0$.

See section 7 of

D. Dugger, Combinatorial model categories have presentations, Adv. Math. 164 (2001), 177–201,

and the appendix of

S. Schwede, Stable homotopy of algebraic theories, Topology 40 (2001), 1–41.

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  • $\begingroup$ I have one more question. From the references you mentioned, I found all informations I needed, except that the category of Bousfield-Friedlander spectra is locally $\alpha_0$-presentable, etc. Is this a standard fact, or an easy to prove statement? Or if there is a reference would you mind letting me know? $\endgroup$ Dec 22, 2012 at 3:57
  • $\begingroup$ The above $\alpha_0$ should be $\aleph_0$. I don't know how to correct a comment.. Sorry. $\endgroup$ Dec 22, 2012 at 3:58
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    $\begingroup$ In general, a stable model category is $\kappa$-presentable if it has a $\kappa$-compact generator, i.e. an object $X$ such that $[X,Y] = 0$ only if $Y$ is the zero object. In this case we're good because spheres detect when an object is trivial. See, e.g., Lurie's Higher Algebra Cor. 1.4.4.2. (though I'm sure some search involving model categories would yield a result in more traditional language) $\endgroup$ Dec 22, 2012 at 12:07
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    $\begingroup$ @Jinhyun it is in Schwede's paper under the equivalent name of locally finitely presentable $\endgroup$ Dec 22, 2012 at 14:47

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