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Recall that a model category is a complete and cocomplete category with classes of morphisms called cofibrations, fibrations, and weak equivalences. These are closed under composition and satisfy certain axioms, such as lifting properties. Furthermore, according to Mark Hovey (but this apparently varies in the literature), any morphism must admit a functorial factorization into the composite of a cofibration and a trivial fibration (i.e. a fibration which is a weak equivalence), as well as a composite of a trivial cofibration and a fibration.

The standard examples of model categories are simplicial sets and chain complexes. Yet the words "fibration" and "cofibration" suggest not category theory but the topological homotopy lifting property. A morphism of topological spaces $f: X \to Y$ is called a (Hurewicz) fibration if whenever $p: T \times [0,1] \to Y$ is a map and $\widetilde{p}: T \to X$ is a map lifting $p|_{T \times \{0\}}$, $\widetilde{p}$ can be extended to a lifting of $p$. Cofibrations can be defined by an analogous homotopy extension property, and for Hausdorff spaces this is equivalent to being a deformation retract of a suitable neighborhood (defined as $\{x: u(x)<1\}$ where $u$ is a suitable continuous function).

So, although this is not explicitly stated anywhere I see, I take it that topological spaces with the usual notions of fibrations and cofibrations (with weak equivalences the homotopy equivalences) do indeed form a model category.

Question: Am I right in assuming this?

The axioms are slightly tricky to check in general. Topological spaces admit limits and colimits. I believe that the lifting properties of cofibrations with respect to trivial fibrations (or trivial cofibrations with respect to fibrations) follows directly from the definition of a cofibration (namely, if you can extend something up to homotopy, you can do it exactly for a cofibration).

I'm getting slightly confused on the functorial factorizations. Let $f: X \to Y$ be a morphism. We need a canonical way of factoring this (actually, two canonical ways). One way is to use the inclusion of $X$ in the mapping cylinder $M_f$ (which is a cofibration because $(M_f, X)$ is checked to be an NDR-pair). Moreover, $M_f \to Y$ is a homotopy equivalence. I don't see why this is a fibration though (it's true that the fibers have the same homotopy type at least).

Correction: As Aaron observes in the comments, there are easy counterexamples for when the map from the mapping cylinder is not a fibration. This means that another approach is needed to construct the functorial factorization of a map into a cofibration and a trivial fibration?

Could someone clarify?

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The Wikipedia article states that Top forms a model category with Serre fibrations and cofibrations, as well as with Hurewicz fibrations and cofibrations. (That's all I got.) –  Qiaochu Yuan Nov 29 '10 at 0:31
    
(Presumably the weak equivalences are the homotopy equivalences?) Anyway, I understand the one that gets used more often is the one for Serre fibrations (and where cofibrations are CW pairs or something like that), but I'm still used to thinking of Hurewicz fibrations. –  Akhil Mathew Nov 29 '10 at 0:47
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Are you sure that $M_f \rightarrow Y$ is a fibration? Here's (what seems to me to be) an easy counterexample. $f:S^1\rightarrow D^2$ is the inclusion, $G:S^1\times I\rightarrow D^2$ is the obvious nullhomotopy $G_t(x)=(1-t)x$, and $H_0:S^1\rightarrow M_f$ is the inclusion at the top of the cylinder. Then there's no $H_t$ covering $G_t$. –  Aaron Mazel-Gee Nov 29 '10 at 2:12
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I've wondered about this too -- the terminology and the fact that they were introduced by Quillen make it seem evident that Top is perhaps the model category (i.e., the model category on which the very definition of model category is modeled). –  Aaron Mazel-Gee Nov 29 '10 at 3:35
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Apparently it took a paper to prove this, Strom's "The Homotopy Category is a Homotopy Category"; cf. mathoverflow.net/questions/47756/… –  Aaron Mazel-Gee Dec 1 '10 at 7:01
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This is in many places. I think that the whole point of model categories is that people saw you could get a lot of mileage out of the adjunction between Top and sSet (geometric realization and singular simplices). Just how good is the combinatorial model for topological spaces? The answer is that it is a Quillen equivalence (whatever that means). But for that sentence to exist we need to talk about model categories. I would say that Top is one of the most basic examples, before simplicial sets and chain complexes. For one thing the model structure on sSet requires use of the model structure on Top (the standard definition of a weak equivalence of simplicial sets is that its geometric realization is a weak equivalence in Top). The model structure on chain complexes sort of came out of homological algebra, and i don't know how early it was realized that this model category stuff shed light on the homological algebra that was going on.

When thinking about this stuff I find the model theoretic approach helps me understand what is going on in top. Also, the weak equivalences are the maps that induce isos in homotopy for all choices of basepoints.

ps Peter May has a write up of a verification of the model category axioms for Top under Misc notes on his website. (There is tons of great stuff on there)

pps Dwyer and Spalinski is a great resource as well http://www.nd.edu/~wgd/ number 75

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Thanks for the answer! –  Akhil Mathew Dec 3 '10 at 13:33
    
your welcome. Peter really has some great notes on his website, you should wander around in it for a bit! –  Sean Tilson Dec 3 '10 at 17:07
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One remark. Logically you do not need the category of topological spaces to get the model structure on simplicial sets. See ncatlab.org/nlab/show/… –  Baby Dragon Mar 11 '13 at 0:13
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As Aaron observes in the comments, this is in Strom's "The homotopy category is a homotopy category."

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We should be careful to note that there are two Stroms who could have conceivably written such a paper. –  Sean Tilson Mar 19 '13 at 0:06
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