What determines a model structure? 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

 A: 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.  
A: 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.
A: 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).
