I found an easy proof that the (levelwise) homotopy limit of a pointwise equivalence of finite diagrams of orthogonal spectra is an equivalence, without assuming that the spectra in the diagrams are fibrant. This makes me a bit nervous. Does anyone know if this is in fact true?
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1$\begingroup$ Problems might begin with the question: How do you form a levelwise homotopy limit of spectra? Is then the map $\Sigma X_n \to X_{n+1}$ still well-defined? So I assume, when you write equivalence that you mean stable equivalence? Can you write down your proof? $\endgroup$– Lennart MeierCommented Sep 20, 2013 at 16:03
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$\begingroup$ I'm using the Bousfield-Kan formula to define homotopy limits. Since the cotensored structure of spectra over simplicial sets is levelwise, the homotopy limits turn out to be levelwise (and well defined). The proof uses only that stable equivalences are $\pi_\ast$-equivalences, and that directed homotopy colimits and finite homotopy limits commute in spaces. $\endgroup$– Emanuele DottoCommented Sep 20, 2013 at 17:08
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$\begingroup$ Does not the Bousfield-Kan formula presuppose that the objects are fibrant? $\endgroup$– Lennart MeierCommented Sep 20, 2013 at 18:00
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$\begingroup$ It makes sense even without having a model structure $\endgroup$– Emanuele DottoCommented Sep 20, 2013 at 18:36
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2$\begingroup$ You can define whatever you want. But the Bousfield-Kan construction is in general not homotopy invariant (and therefore does not agree with I would call the homotopy limit), neither in simplicial sets nor in spectra. See, for example the Bousfield-Kan book XI.5.6, Section 2.3 of math.mit.edu/conferences/talbot/2007/tmfproc/Chapter07/… or the Hirschhorn book, Thm 18.5.2 (most precise). $\endgroup$– Lennart MeierCommented Sep 20, 2013 at 19:46
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It's perhaps a little strange to answer this question after three and a half years, but I've thought about this before too and couldn't resist posting. If the category $C$ indexing your diagram is finite in the sense that the classifying space $BC$ is a finite CW complex (or equivalently $C$ has finitely many composable strings of morphisms) then the Bousfield-Kan homotopy limit preserves stable equivalences, even if you don't make the spectra in the diagram fibrant first.
The first way to see this is to argue that the levels of your spectra can be made cofibrant first without changing the homotopy type of the levels or their homotopy limits. Then you observe that fibrant replacement can be achieved by a sequential colimit of loopspaces of the levels. But the kind of finite homotopy limit discussed above commutes with filtered homotopy colimits, so you're done.
A second argument is by induction up the coskeletal filtration of the cosimplicial object that defines the homotopy limit, see for instance section 4 of these notes for a space-level version. There are only finitely many such levels by our assumption on $C$. You end up only needing that homotopy pullback constructions and finite product constructions preserve stable equivalences of spectra, even if you don't make the spectra fibrant before plugging them in. The first is true because homotopy pullback squares are always equivalent to homotopy pushout squares without any point-set assumptions, and the second is true because a finite product commutes with the sequential colimit that defines the homotopy groups of a spectrum.
Having finitely many morphisms in $C$ isn't good enough. A counterexample is when $C$ has one object and morphisms $\mathbb Z/2$, and your spectrum is the sphere spectrum in the category of prespectra (or symmetric or orthogonal spectra), so the $n$th level is the space $S^n$. The homotopy limit over $C$ doesn't commute with fibrant replacement. If I remember correctly, the verification of this uses both the Segal conjecture and the Sullivan conjecture.
