4
$\begingroup$

Let $\mathcal{M}$ be a pointed model category and $\mathcal{C}$ a class of maps in $\mathcal{M}$ for which the left Bousfield localization ${\rm L}_{\mathcal{C}}\mathcal{M}$ exists (see Hirschhorn, Model categories and their localizations, Ch. 3).

Assume $F\to E\to B$ is a fibre sequence and all three objects are fibrant (in $\mathcal{M}$). If $F, B$ are fibrant in ${\rm L}_{\mathcal{C}}\mathcal{M}$ (i.e. they are $\mathcal{C}$-local, Hirschhorn, def. 3.1.4), do we have that $E$ is also $\mathcal{C}$-local? Or can we replace this fibre sequence (in $\mathcal{M}$) by $F\to E'\to B$ with $E'$ $\mathcal{C}$-local?

$\endgroup$
  • 2
    $\begingroup$ If $\mathcal M$ is stable, then comparing the fiber sequences $Map(X,F)\to Map(X,E)\to Map(X,B)$ to $Map(A,F)\to Map(A,E)\to Map(A,B)$ for any $A\to X$ in $\mathcal C$, one gets the desired result from the five-lemma. Otherwise, I'm not sure what happens; a bit of digging around found no clear statement of a five lemma type result for maps of fibrations of spaces or pointed spaces which are not necessarily connected, as in this case the mapping spaces cannot possibly be. $\endgroup$ – Kevin Carlson Sep 14 at 20:51
  • $\begingroup$ You are right. Unfortunately, my main concern is the non-stable case. I guess mostly probably the answer is negative, as in your comparing of fiber sequences (I think it comes down to result about sSet). $\endgroup$ – Lao-tzu Sep 14 at 20:55
  • 2
    $\begingroup$ I do think the result needs to be modified, since certainly a map of fibrations of pointed but not necessarily connected simplicial sets which is an equivalence on base and fiber has no reason to be an equivalence on the total space. But I believe the analogous statement is true for a map of fibrations of non-pointed simplicial sets which is an equivalence on the base and on every choice of fiber. $\endgroup$ – Kevin Carlson Sep 14 at 21:03
  • $\begingroup$ What you said is true, that follows from the fibrewise characterization of being homotopy cartesian in sSet. There is indeed a five lemma type result in Hovey's book: (dual of ) Proposition 6.5.3. But that is about the fiber. $\endgroup$ – Lao-tzu Sep 14 at 21:09
4
$\begingroup$

The idea you're looking for is called "fibrewise localization". It's defined in Dror Farjoun's book "Cellular Spaces, Null Spaces, and Homotopy Localization", and also in Hirschhorn's book (since you clearly have a copy of the latter, I'll use references from there).

Hirschhorn's Definition 6.1.1 defines the fiberwise localization of a map $p: E\to B$, of pointed spaces, as a factorization $p = q\circ i$ where $q$ is a fibration, and for every $b\in B$, the induced map on homotopy fibers $HFib_b(p) \to HFib_b(q)$ is a $\mathcal{C}$-local equivalence. In general, you construct such localizations by localizing the fibers. To be fibrant in the fiberwise localization model structure means that $p$ is a fibration and that all fibers are $\mathcal{C}$-local. When the fiberwise localization exists, it tells you how to replace your $F\to E\to B$ by the weakly equivalent $F\to E' \to B$.

Unfortunately, Hirschhorn's 6.1.4 proves that fiberwise localization DOES NOT EXIST in the category of pointed spaces. Further conditions are needed, as Kevin Carlson pointed out. So the general answer to your question has to be "no", or else fiberwise localization would exist for free.

One place that gives such conditions is Chataur and Scherer, "Fiberwise localization and the Cube Theorem", Theorem 4.3 (note that the "cube axiom" is required). In the situation of your question, $F$ is already local, so $\eta: F\to L_C(F)$ can be taken to be the identity. The proof of 4.3 shows that $E'$ (there denoted $\overline{E}$) is constructed as a localization of $E$, so is local, as you wanted.

$\endgroup$
  • $\begingroup$ Thank you very much! I didn't check carefully Chataur and Scherer's paper, but I think in their Theorem 4.3, it gives a weak equivalence $E\to E'$ in the localised category, but I want a weak equivalence $E\to E'$ in $\mathcal{M}$. I was in some specific situation, and I think I found some way around the desired result above now. Anyway, having your definitive answer is really very nice. $\endgroup$ – Lao-tzu Sep 15 at 18:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.