So, today I am going to talk about the Brown representability theorem. This is a central fact in algebraic topology, proved in the 1950s by Edgar Brown in a paper in the Annals. It states that under mild conditions a contravariant functor on the homotopy category ${CW_*}$ of pointed CW complexes is representable. As we saw yesterday, this guarantees the existence of Eilenberg-Maclane spaces. More importantly, it guarantees—at least for CW complexes—things like universal bundles. I will have more to say about these applications in the future. First, let us try to understand the result itself.

So let ${F: CW_* \rightarrow \mathbf{Sets}_*}$ be a contravariant functor to the category of pointed sets. We require that ${F}$ satisfy two axioms below. First, like any representable functor, it should send coproducts to products. Since the coproduct in ${CW_*}$ is the wedge sum, we require that (for ${\alpha}$ ranging over some index set)

$\displaystyle F(\bigvee X_\alpha) = \prod F(X_\alpha)$

under the canonical map ${F(\bigvee X_\alpha) \rightarrow \prod F(X_\alpha)}$ that arises by taking the product (over ${\beta}$) of the maps ${F(\bigvee X_\alpha) \rightarrow F(X_\beta)}$.

Second, we require that ${F}$ satisfy the following Mayer-Vietoris axiom. If ${X = X_1 \cup X_2}$ for subcomplexes ${X_1, X_2}$, then if ${q_1 \in F(X_1)}$ and ${q_2 \in F(X_2)}$ “glue together,” i.e. become the same element of ${F(X_1 \cap X_2)}$, they both come from a fixed element of ${F(X)}$. This is a sheaf-theoretic condition.

The first observation is that any representable functor ${F}$ must satisfy the coproduct axiom and the Mayer-Vietoris (i.e. sheafish) axiom. The coproduct axiom is automatic for any category.

The sheaf axiom is less trivial. Let ${X = X_1 \cup X_2}$ as before. Suppose that ${F}$ is represented by a pointed complex ${Y}$. Given “gluable” elements ${f_1: X_1 \rightarrow Y}$ and ${f_2: X_2 \rightarrow Y}$, by assumption the restrictions ${f_1|_{X_1 \cap X_2}, f_2|_{X_1 \cap X_2}}$ are equivalent (i.e. homotopic). This does not immediately mean we can glue the maps. However, by the homotopy extension property for CW pairs, we can homotope ${f_2}$ to some ${f_2'}$ such that ${f_1, f_2'}$ become equal (not merely homotopic) on ${X_1 \cap X_2}$. Together these define a map ${X \rightarrow Y}$ that becomes equivalent to ${f_1, f_2}$.

These axioms are not too difficult to check in many interesting cases. For instance, they are true for singular cohomology. This is why the following is highly important:

Theorem 1 (Brown) If ${F: CW_* \rightarrow \mathbf{Sets}_*}$ is a contravariant functor satisfying the coproduct and Mayer-Vietoris axioms, then ${F}$ is representable.

This is the result that I would like to begin to prove today.

${n}$-universal elements

The idea behind the proof of Brown representability is that one is working with CW complexes. CW complexes are built up in a piece-by-piece manner from the comparatively simple objects of ${n}$-cells. And in a sense, the way to construct the representing object is to construct it for the spheres.

To prove Brown representability, we will need to find a space ${Y}$ and an element ${u \in F(Y)}$ such that if ${X}$ is any space, then the elements of ${F(X)}$ are in bijection with the ${f^*(u)}$ for ${f}$ ranging over the homotopy classes of maps ${X \rightarrow Y}$.

As I stated above, the strategy will be to do this for the spheres. Namely, we shall find a space ${Y}$ such that the above claim is true for the spheres of any dimension. We shall then show that the space will work as a representing object in general.

Definition 2 Let ${(u, Y)}$ be a pair with ${Y \in CW_*}$ and ${u \in F(Y)}$. The pair is said to be ${n}$-universal if the map ${\hom_{CW_*}(S^k, Y) \rightarrow F(S^k)}$ sending ${f: S^k \rightarrow Y}$ to ${f^*(u)}$ is bijective for ${k and epimorphic for ${k=n}$. A pair is ${\infty}$-universal if it is ${n}$-universal for all ${n}$.

The point is that we will find a universal pair, and show that this is indeed a representing object (a “universal object” in the categorical sense).

The process of finding universal elements is contained in the following lemma. It is an approach that comes up over and over: to add enough spheres to make the map surjective, and then to attach various cells to reduce the number of homotopy classes of maps.

Now, to make notation clearer, let us note that ${F}$ is a presheaf on the category ${CW_*}$, that is a contravariant functor to ${\mathbf{Sets}_*}$. Motivated by standard notation, let us fix the following convention: if ${A \subset A'}$ is a subcomplex of a CW complex, and ${t \in F(A')}$, we write ${t|_A}$ for the pull-back of ${t}$ to ${F(A)}$.

Lemma 3 Let ${(Y, u)}$ be a ${n}$-universal pair. Then there is a space ${Y' \in CW_*}$ obtained by attaching cells and an ${n+1}$-universal element ${u' \in F(Y')}$ such that ${u'|_Y = u}$.

Proof: The first thing to note that is that ${F(S^k)}$ is a group in a natural way. This actually formally follows from general nonsense. In fact, ${F}$ sends coproducts in ${CW_*}$ to products, so ${F}$ sends cogroup objects to group objects in ${\mathbf{Sets}_*}$ (i.e. groups). This group structure is deduced from the cogroup law in ${S^k}$. This is the same way that ${\pi_k(Y) = \hom_{CW*}(S^k, Y)}$ has a group structure.

In particular, we find that the map

$\displaystyle \pi_k(Y) \rightarrow F(S^k), \quad f \rightarrow f^*(u)$

is a group homomorphism. To say that it is injective is thus to say that it has trivial kernel.

We already know that this injectivity is true for ${k. We want to attach various ${n+1}$-cells to make this true for ${k=n}$.

Namely, we let ${\left\{f_\alpha\right\}}$ be a collection of maps ${S^n \rightarrow Y}$ that forms a system of representatives for the kernel of ${\pi_n(Y) \rightarrow F(Y), f \rightarrow f^*(u)}$. We can assume that the ${f_\alpha}$ are cellular. We attach ${n+1}$-cells to ${Y}$ by these maps ${f_\alpha}$ to get a new space ${Y''}$. Doing so makes all the maps ${f_\alpha: S^n \rightarrow Y}$ nullhomotopic in ${Y'}$.

The claim is that ${u \in F(Y)}$ extends to the bigger space ${Y''}$. The way to see this is that ${Y''}$ is the (reduced) mapping cone of ${\bigvee f_\alpha: \bigvee S^n \rightarrow Y}$. Now we will use a general fact:

Lemma 4 Suppose ${F}$ satisfies the Mayer-Vietoris axiom Let ${f: A \rightarrow B}$ be a cellular map in ${CW_*}$. Suppose ${u \in F(B)}$ pulls back to the trivial (basepoint) of ${F(A)}$. Then ${u}$ extends to the mapping cone ${B \cup_f CA}$.

This lemma follows by the cover of ${X \cup_f CA}$ by ${X}$ and ${CA}$.

We see that in our case, there is ${u'' \in F(Y'')}$ such that ${u''}$ pulls back to ${u}$. The claim is that ${u''}$ is ${n}$-universal, and moreover

$\displaystyle \pi_n(Y'') \rightarrow F(S^n), \quad f \rightarrow f^*(u'')$

is injective (thus an isomorphism). Indeed, if ${f}$ pulls ${u''}$ back to the trivial element, we can first assume by homotoping ${f}$ that ${f}$ lands inside ${Y}$, thanks to the cellular approximation theorem. In that case, it follows that ${f}$ pulls ${u = u''|_Y}$ to the trivial element, so it is one of the ${f_\alpha}$ (at least up to homotopy). But these ${f_\alpha}$ are trivial in ${Y''}$.

On the other hand, we still have to get an ${n+1}$-universal element. We haven’t done that yet.

Namely, we have to get surjectivity on the ${n+1}$-level. For this, we just augment the space by wedging along a whole bunch of spheres.

Let ${\left\{t_\beta\right\}_{\beta \in B}}$ be a listing of elements in ${F(S^{n+1})}$. We can form

$\displaystyle Y' = Y'' \vee \bigvee_B S^{n+1}.$

Let us choose an element ${u' \in F(Y')}$ extending ${u''}$ such that ${u'|_{Y''} = u''}$ (this is what extension means) and ${u''}$ restricted to the ${\beta}$th factor of ${S^{n+1}}$ is ${t_{\beta}}$. This is possible by the coproduct condition.

This last condition means that the inclusion of ${S^{n+1}}$ on the ${\beta}$th factor pulls ${u'}$ back to ${t_{\beta}}$. In particular,

$\displaystyle \pi_{n+1}(Y'') \rightarrow F(S^{n+1})$

is surjective. Since we have adjoined only cells of dimension ${\geq n+1}$, the cellular approximation lemma implies that any map ${S^k \rightarrow Y'}$ comes from a map ${S^k \rightarrow Y}$. So ${u'}$ pulls back to the trivial element if only if the associated map ${S^k \rightarrow Y}$ does. As before, we conclude that this implies that ${S^k \rightarrow Y}$ is homotopically trivial in ${Y'}$.

The point is that, using this, we will be able to construct $\infty$-universal elements by a filtered colimit argument, and we will be able to show that these are actually representing pairs by the Whitehead theorem. This, however, needs to be finished next time.