Two more responses to the question about cofibrant objects....DMD
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Subject: Re: another response and conf update
Date: Tue, 22 Apr 2003 23:31:13 -0400 (EDT)
From: Tom Goodwillie <tomg@math.brown.edu>
>
> 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.
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Subject: question about cofibrant terminal objects
Date: Wed, 23 Apr 2003 09:30:43 -0400 (Eastern Daylight Time)
From: "Nicholas J. Kuhn" <njk4x@tetra.mail.virginia.edu>

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