# Homotopy groups of filtered homotopy limits

Let $X$ be the homotopy limit of a filtered system of simplicial sets $X_i$. When are the morphisms $\pi_j(X)\to \varprojlim \pi_j(X_i)$ surjective for all $j\ge 0$? This seems to be no problem when the system is countable since then there is a short exact sequence from section IX.3 of A. K. Bousfield and D. M. Kan, Homotopy limits, completions and localizations, Lecture Notes in Mathematics 304, Springer, 1972. What about uncountable limits?

Also, how could one translate the corresponding result into a computation of morphisms in a simplicial (proper?) model category?

• I don't understand the last part of your question. What is the translation of your question for the more general setting of a simplicial proper model category ? – Philippe Gaucher Dec 21 '13 at 8:32

Check out page 34 of May-Ponto "More Concise Algebraic Topology" for a more modern treatment of the homotopy limit of a sequence of spaces. In particular, Proposition 2.2.9 is exactly the surjection you're asking about. Their proof is to realize $X$ as a homotopy equalizer, then apply Proposition 2.2.7, which is a general statement about homotopy equalizers. So it seems this proof generalizes completely to the non-countable case. You simply have to realize the homotopy limit of an uncountable sequence as a homotopy equalizer and like May and Ponto we should set $Y = \prod X_\alpha$ and look at the homotopy equalizer of $id_Y$ and $\prod f_\alpha$ where the $f_\alpha:X_{\alpha+1}\to X_\alpha$ are the maps in the system. Incidentally, it's not that uncommon for statements from before 1990 or so to have only been proven for the countable case, but often very similar proofs work far more generally. A good example is Neeman's book on triangulated categories, which introduces the notion of a well-generated triangulated category.
One difference with the classical situation is that instead of working with $[S^n,X]$ one works with the simplicial mapping space. Another is that it's very hard for a model category to say that a map is surjective, but it can say something is a weak equivalence. The theorem you started with doesn't just give the surjection; it identifies the kernel as a $lim^1$ term. So it can be phrased as looking for an isomorphism. The closest thing in Hirschhorn I can find is Theorem 18.7.4 and it's corollaries (which are stated for hocolim but have natural dual versions for holim). There's a lot of stuff in Hirschhorn, so if this doesn't sound like the model category version of the statement for Top it's very possible a different result of his will be a better fit. I encourage people to leave comments if they think there's a better model category analogue.
• I think what you state in the first paragraph will not work, and is in fact very specific to countable limits. For limits along uncountable ordinals, you will probably need to consider more maps than $f_\alpha:X_{\alpha+1}\to X_\alpha$, as there are elements without a predecessor. As such, you might not be able to express them as homotopy equalizers. In fact, that would be rather surprising, as the higher derived functors $\operatorname{lim}^i$, $i>1$, of the limit of abelian groups along an uncountable ordinal are in general non-zero. This is unlike the countable case, where they are zero. – Ricardo Andrade Dec 21 '13 at 17:47
• To the best of my limited knowledge, all we can do in the general situation of the question is to try applying the fringed Bousfield-Kan spectral sequence for the homotopy groups of the homotopy limit. The $E_2$ term of that spectral sequence involves the higher derived functors of limit. Thus, the above cited work on the derived functors of limit might be helpful. – Ricardo Andrade Dec 21 '13 at 17:50
• Hi Ricardo. Thanks for your comments; you raise some interesting points. Sorry for the delayed response. Christmas and all that. First, I agree that one step in what I wrote seemed sketchy, and that was taking an uncountable product of maps. Still, I'm not sure if I buy your argument about limit ordinals or not; morally, it feels like a limit containing all the $f_\alpha$ will get the limit maps for free. More concerning to me is your comment about lim$^i$. Can you think of an explicit example of an uncountable sequence of maps with a nonzero lim$^2$ term? That would definitely convince me. – David White Dec 24 '13 at 12:38
• Dear @David: No worries! Just to clarify my first objection, it was essentially the following. If you apply your homotopy equalizer construction to a limit along the ordinal sum $\omega + \omega$, none of the maps you consider connects the first copy of $\omega$ to the second one. So the homotopy equalizer you get is actually the product of the two homotopy limits along each copy of $\omega$. In the case of a countable limit, this objection can certainly be circumvented (by taking a cofinal copy of $\omega$), but I fail to see how to avoid this problem for uncountable limits. – Ricardo Andrade Dec 24 '13 at 18:13