Two more responses to the question about cofibrant objects....DMD _________________________________________________________ Subject: Re: another response and conf update Date: Tue, 22 Apr 2003 23:31:13 -0400 (EDT) From: Tom Goodwillie > > 1) A new kind of cheat: > Take your favorite bicomplete category and choose both fibrations and > weak equivalences as any morphisms. I know the intersection of the > three distinguished classes in a model structure must be the > isomorphisms, so the cofibrations are forced to be the isomorphisms. > Unless I'm missing something, this is a model structure and the only > cofibrant object is the initial object (only defined up to isomorphism > anyway). Hence, any bicomplete category with different initial and > final objects should yield an example. > That reminds me of a curious pattern I've noticed: Let C be a category. Let us call a pair (L,R) of classes of morphisms in C a "matched pair" if the elements of L are precisely the morphisms of C that have the left lifting property w.r.t. R and vice versa. Call it a "well matched pair" if in addition every morphism f of C can be factored f=pi, i in L and p in R. This comes up because if C is a closed model category then it has two well matched pairs: (L',R) = (trivial cofibrations, fibrations) (L,R') = (cofibrations, trivial fibrations). If every morphism is a weak equivalence, then these two pairs are equal. Here is the curious pattern: When I start making examples of matched pairs in various categories, a surprising number of them turn out to be well matched. Question: Is there any reason for this? -- Tom Goodwillie Exercise for the idle mathematician: (1) Let C be the category of sets. (a) Find all the matched pairs. (There are six. They are all well matched.) (b) Find all closed model structures. (There are nine. In eight of the nine cases at least one of the three classes {weak equivalences, cofibrations, fibrations} has all the morphisms.) (2) Repeat for the category of based sets. ______________________________________________________ Subject: question about cofibrant terminal objects Date: Wed, 23 Apr 2003 09:30:43 -0400 (Eastern Daylight Time) From: "Nicholas J. Kuhn" My goodness, I'm a bit confused about the complex and unnatural answers being sent in to the list, in answer to an example of a model category in which the terminal object is not cofibrant. Isn't the following example totally standard, and, indeed, presumably motivated the whole concept of model categories in the first place?... Let G be a finite group. The standard model category structure on topological spaces induces a standard structure on G-spaces. A single point is the terminal object, and its cofibrant replacement is typically denoted by EG. Nick Kuhn