Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Weak equivalences in the standard model structure on simplicial sets are allegedly closed under transfinite composition.

What's a reference for that?

share|improve this question
2  
More specifics would be helpful. Do you mean that you have a transfinite composite of maps which are weak equivalences and want to show that all the objects are equivalent to the colimit? The homotopy colimit? Do you mean that you have two transfinite composites with a map between them that is a levelwise weak equivalence, and you want to show the colimits are weakly equivalent? The homotopy colimits? –  Tyler Lawson Jan 23 '10 at 18:20
    
Sorry if this was't clear: I want to know if the cocone component on the first object of the colimiting cocone that defines the transfinite composition is a weak equivalence. That's the morphism that is the "transfinite composite" of a transfinite sequence of weak equivalences. –  Urs Schreiber Jan 23 '10 at 20:39

1 Answer 1

up vote 7 down vote accepted

I don't have a complete reference (and like Tyler, I don't know exactly what result you want). But here are some observations:

  • there is a functor $\mathrm{Ex}^\infty$, which replaces a simplicial set with a weakly equivalent fibrant replacement, and which commutes with filtered colimits. (See ch. 3 of Goerss-Jardine.)

  • if you have a transfinite composition(s) in which all the simplicial sets are fibrant, it is straightforward to understand their behaviour with respect to weak equivalences, using the formula for simplicial homotopy groups, which gives the right homotopy groups for Kan complexes; (in particular, simplicial homotopy groups commute with filtered colimits in this setting.) [Added later:] in particular, in any transfinite composition of weak equivalences between Kan complexes, the cocone componenent of the first object (i.e., the map from the first object to the colimit of the trasnfinite sequence) has to be a weak equivalence.

[Added:] These two facts taken together imply that a "transfinite composition" of weak equivalences is a weak equivalence.

share|improve this answer
    
Okay, thanks. So here is what I tried: we can check if we have a weak equivalence by homming into all possible fibrant objects and checking if that produces a WE each time. So I hom the colimit cocone that defines the transfinite composition into a Kan complex. The result is now a limiting cone over a diagram of fibrant objects. I want to know the the cone component over the first (or any) object is a weak equivalence. Okay, so I now hom simplicial spheres into this and then argue that... and here I am not sure. It's probably very easy, though. –  Urs Schreiber Jan 23 '10 at 20:43
    
@Urs. I can't see how to complete your argument. If you have a transfinite tower (aka, a cocone) of weak equivalences between fibrant objects, I see no general reason for the map from the limit to be a weak equivalence. –  Charles Rezk Jan 23 '10 at 20:52
    
Charles, now thanks for that updated version of your reply. Nice! I didn't think of/know if that respect for filtered colimits. With that it's clear. Thanks again! –  Urs Schreiber Jan 23 '10 at 23:10
    
@CharlesRezk How can I see that $\mbox{Ex}^\infty$ commutes with filtered colimits? I didn't see what you were referring to in Goerss--Jardine. –  Aaron Mazel-Gee Nov 4 at 20:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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