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It is easy to prove that a model structure is determined by the following classes of maps (determined = two model structures with the mentioned classes in common are equal).

  • cofibrations and weak equivalences
  • fibrations and weak equivalences

The second statement follows immediately from the first by duality.

What about the following classes of maps/objects (A short argument would be very helpful)?

  1. cofibrations and fibrations
  2. cofibrant objects and weak equivalences
  3. cofibrant objects and fibrations
  4. cofibrant objects and fibrant objects

I think each of these classes determine the structure respectively. For the last one I suppose that one has to use framings but I cannot see how to do it.

Edit: Thank you all for the illuminative answers.

  1. true
  2. ?
  3. true
  4. false
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8  
Mike Shulman answered 2: it's false. The same category can have two distinct model structures such that every map is a weak equivalence and every object is cofibrant. Simple example: the category of sets, with (cofibrations, fibrations) being (1) (all maps, bijections), or (2) (injections, surjections). –  Tom Goodwillie Jun 27 '10 at 18:34
    
For what it's worth, the consensus seems to be: 1. easily seen to be true 2. false for trivial reasons 3. nontrivially true 4. easily seen to be false –  Tim Campion May 19 '12 at 9:04
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3 Answers 3

up vote 8 down vote accepted

This is just a flash answer without enough thought:

  1. Cofibrations determine trivial fibrations (by lifting) and fibrations determine trivial cofibrations. Any weak equivalence is a composite of a trivial cofibration and a trivial fibration. So cofibrations and fibrations determine the model structure.

  2. Cofibrant objects and fibrations (or weak equivalences) should NOT determine the model structure, or so my intuition says. If you know the model structure is left proper, maybe. You might be tempted to argue something like: fibrations determine trivial cofibrations, so we know all trivial cofibrations between cofibrant objects, so we ought to know all weak equivalences between cofibrant objects by something like Ken Brown's Lemma. But I bet this does not in fact work. Even if it does, I'm not sure what to do next. I'd be tempted to look for a counterexample.

  3. Cofibrant objects and fibrant objects surely must not determine the model structure. It must not be too hard to come up with a counterexample for this, but I need to think about it.

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For 3, cofibrant objects and fibrant objects do not even determine weak equivalences. For instance, Cat(=small categories) admits model structures with all objects fibrant and cofibrant, with either equivalences or isomorphisms as the weak equivs. –  Charles Rezk Jun 26 '10 at 21:32
    
Thank you for the example (This is point 4. not 3.). –  roger123 Jun 27 '10 at 18:09
7  
To clarify for anyone reading this: the second item in this response turns out to be wrong. As pointed out by other comments and answers, it's a theorem of Joyal that cofibrants + fibrations DO determine the model structure (although it seems to be more common to state the dual: cofibrations + fibrants determine...). An updated link to catlab proof: ncatlab.org/joyalscatlab/show/Model+categories#determination2 –  Tim Campion May 19 '12 at 8:52
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Among the nine model structures on the category of sets, there are:

Two in which cofibrant=empty and every set is fibrant.

Two in which cofibrant=empty and fibrant={empty or singleton}.

One in which every set is cofibrant and fibrant=nonempty.

One in which every set is cofibrant and fibrant=singleton.

One in which every set is cofibrant and fibrant={empty or singleton}.

Two in which every set is both cofibrant and fibrant.

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But you do not say that here are exactly nine model structures on the category of sets, right? –  user2146 Jul 9 '10 at 13:04
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I did mean to assert that there are exactly nine model structures on the category of sets. I did not say why this is true. It is a pleasant exercise to find them all. A recommended preliminary exercise is to find all six pairs (C,F) of classes of morphisms in the category of sets such C is determined by left lifting w.r.t. F and F is determined by right lifting w.r.t. C. –  Tom Goodwillie Jul 9 '10 at 14:18
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Wow! Thanks Tom. I just finished verifying your claim for myself. This is such a great exercise. Thank you so much! –  Chris Schommer-Pries Aug 12 '10 at 3:52
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I'm glad you enjoyed it. If you feel like doing more of the same, I recommend based sets. I don't know any serious use for any of this. –  Tom Goodwillie Aug 12 '10 at 11:27
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I think it's interesting for two reasons. First there's the pedagogic value. When I teach a class which covers model categories, I'll definitely include this exercise. Second, this may be the first step in classifying model structures on more complicated categories. Example: you can look at the forgetful functor from Cat to the Set of objects. This functor and its fully-faithful left adjoint both preserve colimits and limits. Hence you to transfer any model structure on Cat to one on Set, and hopefully to classify model structures on Cat relative to those on Set. Perhaps on presheaf cats too? –  Chris Schommer-Pries Aug 12 '10 at 13:33
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Mark's answer explains why (1) cofibrations and fibrations do determine the model structure, and Charles' and Tom's examples show that (4) cofibrant objects and fibrant objects do not.

For (2), any weak factorization system (L,R) on a category determines a model structure in which L is the cofibrations, R is the fibrations, and all maps are weak equivalences. Thus, it suffices to find a category C admitting two wfs (L,R) and (L',R') with the same cofibrant objects, which is pretty easy; take for instance C=Set with (all functions, isos) and (monos, epis).

Finally, for (3) it is true, surprisingly, that the cofibrant objects and the fibrations determine a model structure. In fact, merely the cofibrant+fibrant objects together with the fibrations also determine the model structure. This observation is due to Joyal and can be found on his Catlab (currently Proposition 4).

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Note: On Joyal's page, proposition 4 is not the same as lemma 4. That confused me a bit when I first clicked through. –  Harry Gindi Jun 27 '10 at 6:11
    
Yeah, that's an annoyance of the way Instiki does theorem/proposition/lemma numbering. –  Mike Shulman Jun 27 '10 at 16:51
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