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Apr 29, 2013 at 8:14 vote accept Philippe Gaucher
Jun 28, 2012 at 21:53 comment added Philippe Gaucher Actually, there is a slightly shorter argument: Marc Olschok already proves in his PhD that, in this case, all objects of the left determined one are fibrant. So the left determined one has the same cofibration and the same fibrant objects than Quillen's, so the two model structures are equal.
Jun 28, 2012 at 21:48 comment added Philippe Gaucher I mean: every trivial cofibration of the left determined one is a trivial cofibration of Quillen's, so dually, every fibrant of the Quillen's is fibrant in the left determined one, i.e. all spaces are fibrant in the left determined one. BTW, the same argument also proves that with Vopenka's principle, every combinatorial model category such that all objects are fibrant is left determined, fact that I had never noticed before asking this question.
Jun 28, 2012 at 17:56 comment added David White Thanks for posting the answer. It's nice to get these things resolved, even when the OP has to answer his own question. One question: when you say that the left determined model structure has more fibrant objects, you mean more than the Quillen model structure on $\Delta$-generated spaces, right? What do the fibrant objects there look like?
Jun 28, 2012 at 7:06 history answered Philippe Gaucher CC BY-SA 3.0