The increased rate of posting the past few days will probably taper off; we’ve just had a long weekend.

Next, I shall start talking specifically about the homotopy properties of CW complexes. CW complexes are nice because they are built up in a recursive manner using very simple objects (namely, ${n}$-cells). They are also very flexible since every space is, up to something called weak homotopy equivalence which we will define someday, a CW complex.

So let ${X, Y}$ be CW complexes. This means that they are built up from the ${n}$-skeletons ${X^n, Y^n}$; indeed, ${X, Y}$ have the inductive limit (i.e., weak) topology on the ${X^n, Y^n}$. Moreover, each ${X^{n+1}, Y^{n+1} }$ is built out of ${X^n, Y^n}$ by attaching ${n+1}$-cells ${D^{n+1}}$ via attaching maps ${S^{n} \rightarrow X^n, Y^n}$. Nonetheless, a general (continuous) map ${f: X \rightarrow Y}$ will not respect the cellular structure. It will not necessarily be the case that ${f(X^n) \subset Y^n}$. Nevertheless, the point of this post is that we can get that, up to homotopy. First, we give this a name:

Definition 1 A map ${f: X \rightarrow Y}$ of CW complexes is called cellular if ${f(X^n) \subset Y^n}$ for all ${n}$.

The principal result for today is:

Theorem 2 Let ${f: X \rightarrow Y}$ be a continuous map of CW complexes. Then ${f}$ is homotopic to a cellular map; if there is a subcomplex ${K \subset X}$ on which ${f|_K}$ is cellular, then the homotopy can be taken stationary on ${K}$.

So this is the CW-complex version of the simplicial approximation theorem. There are some differences. One, you don’t have to do any of that subdivision business! Two, a cellular map, unlike a simplicial map, can still be extremely complex. Nonetheless, if ${f}$ is both a cellular map and an inclusion, then ${f}$ can be viewed as the inclusion of a subcomplex.

Despite these differences, we can actually prove the cellular approximation theorem by essentially the simplicial theorem. The basic idea is that if a map is not cellular, then it sends an ${n}$-cell into an ${n+1}$-cell. But an ${n}$-cell is, intuitively, too small to fill up an ${n+1}$-cell. This is not true if we allow space-filling curves and things like that, but it is if we allow only smooth maps (by Sard’s theorem, actually). So we can push it off to the boundary.

Honestly, I don’t feel like going into the proof, because it is kind of technical. Perhaps I’m just being lazy, but this being a blog, I don’t feel the need to emphasize nitty-gritty details when those are done nicely in textbooks. As for myself, I’m not sure how well my time would be spent writing it up, anyway, as the main ideas are not that complicated. So let me tell you about those ideas.

The point is that given ${X \rightarrow Y}$, you look at cells of ${X}$ which aren’t mapped into cells of the appropriate dimension in ${Y}$. So consider a cell ${e_n^\alpha}$ which is mapped into some union ${e_{m_1}^{\beta_1} \cup \dots \cup e_{m_k}^{\beta_k}}$ where some of the ${m_i > n}$. Then we homotope ${f}$ to make it, as a collection of maps between subsets of euclidean space, piecewise-linear on a substantial portion of the domain. The piecewise linearity shows that the image of ${f}$ has to miss a point. Then we can deformation retract the homotoped version of ${f}$ onto the boundaries of the things out of ${e_n^\alpha}$‘s league. You have to use the fact that the inclusion of a subcomplex is a cofibration to extend these homotopies to the whole complex.

I think it will be more fun, and enjoyable, for readers of this blog if I move on and not spend too much time on this technical result. We will come back to similar ideas, anyway, when I talk about the excision theorem for homotopy groups.