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)?
- cofibrations and fibrations
- cofibrant objects and weak equivalences
- cofibrant objects and fibrations
- 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.
- true
- ?
- true
- false
Best Answer
This is just a flash answer without enough thought:
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.
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.
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.