Last time, we explained the idea of a cofibration in terms of a useful homotopy extension property. We showed that, for Hausdorff spaces, cofibrations turn out always to be closed immersions. Moreover, we showed that a pair {(X, A)} with {A \subset X} closed is a cofibration precisely when it satisfies a technical condition of {A} being a neighborhood deformation retract. However, we have yet to give useful examples. The main result in this post is that a relative CW complex (for instance, a CW pair) leads to a cofibration.


1. The mapping cylinder

It turns out that, up to homotopy equivalence, every map is a cofibration. The method of showing this is to use the mapping cylinder. So let {f: X \rightarrow Y} be a map. Recall that the mapping cylinder {M_f} is the quotient space {(X \times I) \cup Y} where {(x,1)} is identified with {f(x) \in Y}. We have an inclusion map {X \rightarrow M_f} sending {x \rightarrow (x,0)} and a projection map {M_f \rightarrow Y} sending {(x,t) \rightarrow f(x)}. (more…)

The idea of a cofibration will be fundamental as we talk further about homotopy theory, as will its dual idea of a fibration. The point of the cofibration condition is the following. Oftentimes, we have a subspace {A \subset X} and a map {A \stackrel{f}{\rightarrow} Y}. We’d like to know when we can extend this map over all of {X}. One useful criterion can be given by algebraic functors. For instance, singular homology can be used to show that the map {1:S^{n} \rightarrow S^n} does not extend over {D^{n+1}}.

Many of the things we care about in algebraic topology are homotopy invariant, though. As a result, it would be nice to know when the question of how {A \rightarrow Y} extends depends only on the homotopy class of {A}. This is precisely the definition of a cofibration. Dualizing gives the definition of a fibration, which I will talk about some other time.

One can, not surprisingly, approach the business of fibrations and cofibrations in an axiomatic manner. This is the theory of model categories, due to Quillen. I don’t know anything about that though.