So, last time we were talking about Brown representability. We were, in particular, trying to show that a contravariant functor $F$ on the homotopy category of pointed CW complexes satisfying two natural axioms (a coproduct axiom and a Mayer-Vietoris axiom) was actually representable. The approach thus far was to construct pairs which were “partially universal,” that is universal for a finite set of spheres, by a messy attaching procedure.

There is much work left to do. The first is to show that we can get a pair which is universal for all the spheres. This will use a filtered colimit argument. However, we don’t know that $F$ sends filtered colimits into filtered limits, just that for coproducts. In fact, generally $F$ will not do this, but it will do something close. So we will have to appeal to a mapping telescope argument which will, incidentally, use the Mayer-Vietoris property.

Next, we will have to show that a pair which is universal for the spheres is universal for all spaces. This will use a bit of diagram-chasing and the fact that, to some extent, CW complexes are determined by the ways you can map spheres into them. This is the Whitehead theorem.

Let’s get to work.

2. The mapping telescope argument

Given a space ${Y \in CW_*}$ and ${u \in F(Y)}$, we have shown how to extend the pair ${(Y,u)}$ to an ${n}$-universal pair for any ${n}$. We’d now like to do the same for an ${\infty}$-universal element. This is going to require some kind of filtered colimit condition. A priori, we never assumed anything about filtered colimits in the axioms. But from the wedge axiom, we actually get that.

The following is the form that condition takes:

Lemma 6 Let ${Y_1 \hookrightarrow Y_2 \hookrightarrow Y_3 \hookrightarrow \dots}$ be inclusions of subcomplexes in ${CW_*}$, with ${Y = \varinjlim Y_n}$ the limit. Then the canonical map

$\displaystyle F(Y) \rightarrow \varprojlim F(Y_n)$

is a surjection of pointed sets.

In general it is not a bijection, even for singular cohomology (the kernel has to do with the derived functors of the inverse limit).

Proof: Suppose ${\left\{u_n\right\}}$ is a family of elements with ${u_n \in F(Y_n)}$ such that ${u_n|_{Y_{n-1}} = u_{n-1}}$. We have to piece these together into one element of ${F(Y)}$.

We shall use a trick of considering a “mapping telescope.” I believe this trick is due to Milnor; it will make the direct limit homotopy equivalent to something which looks more like a coproduct. Namely, we start with the reduced mapping cylinder ${Y_1 \times I \cup Y_2}$ with the usual attaching (and with ${\ast \times I}$ quotiented to a point, where ${\ast}$ is the basepoint). We then attach this to the mapping cylinder of ${Y_2 \rightarrow Y_3}$, and so on.

$\displaystyle Z = \bigsqcup Y_i \times [i, i+1] / \sim$

where the elements ${(y_i, i+1) \in Y_i \times [i, i+1]}$ and ${(y_i, i+1) \in Y_{i+1} \times [i+1, i+2]}$ are identified, as well as anything involving the basepoint ${\ast}$.

There are two subcomplexes ${A , B \subset Z}$ that we will need. Namely, ${A}$ is the subcomplex consisting of the ${Y_i \times [i, i+1]}$ for ${i }$ even, and ${B}$ for ${i}$ odd. Then ${A}$ is homotopy equivalent to ${\bigvee_{\mathrm{even}} Y_i}$ and ${B}$ is homotopy equivalent to ${\bigvee_{\mathrm{odd}} Y_i}$. Further ${A \cup B = Z}$. Moreover, ${A \cap B = \bigvee Y_i}$.

There is an element ${u'_1 \in F(A)}$ which restricts to each of the even ${u_i}$, and an element ${u'_2 \in F(B)}$ which restricts to each of the odd ${u_i}$. The intersections glue on ${A \cap B = \bigvee Y_i}$ by hypothesis on the ${u_i}$. We get an element ${u' \in F(Z)}$ which restricts to each of the ${u_i}$ on ${Y_i \times [i, i+1] \simeq Y_i}$.

I claim now that ${Z}$ is actually homotopy equivalent to ${Y}$, in a way compatible with the ${Y_i \rightarrow Y}$ and ${Y_i \rightarrow Z}$. Then the element ${u' \in F(Z)}$ will correspond to the element of ${F(Y)}$ that we want. In particular, we now will show that the mapping telescope is the direct limit, up to homotopy equivalence. This is going to imply the lemma.

Namely, we have a natural projection ${p: Z \rightarrow Y}$ that extends the projections ${Y_i \rightarrow [i, i+1] \rightarrow Y_i \hookrightarrow Y}$. We need to construct a map in the reverse direction.

To do this, first consider the inclusion of ${Y_0 \hookrightarrow Z}$ on the first part of the mapping cylinder. Inductively, given ${n}$, let us suppose that we have a map ${g_n: Y_n \rightarrow Z}$ which is homotopic to the inclusion ${Y_n \rightarrow Y_n \times [n, n+1]}$. We will extend this to ${Y_{n+1}}$. Namely, we have the inclusion ${i_{n+1}:Y_{n+1} \hookrightarrow Z}$ on ${Y_{n+1} \times [n+1, n+2]}$. The restriction ${i_{n+1}|_{Y_{n}}}$ is homotopic to ${g_n}$. The homotopy extension property implies that we can homotope ${i_{n+1}}$ to something which agrees with ${g_n}$ on ${Y_n}$. We call this homotoped map ${g_{n+1}}$.

Putting together all the ${g_n}$, we get a map in the other direction, ${g: Y \rightarrow Z}$. We saw by construction that ${p \circ g_n: Y_n \rightarrow Y_n}$ is homotopic to the identity, and same for ${g_n \circ p: \bigsqcup_{i \leq n} Y_i \times [i, i+1] / \sim \rightarrow \bigsqcup_{i \leq n} Y_i \times [i, i+1] / \sim}$. We want to say the same for the direct limit of these maps. For this we prove:

Lemma 7 Suppose ${\left\{Y_n\right\}}$ is a direct system of spaces such that the maps ${Y_n \rightarrow Y_{n+1}}$ are cofibrations. Suppose ${f_n,g_n: Y_n \rightarrow Y_n}$ is a pair of compatible systems of maps, such that ${f_n \simeq g_n}$. Then ${f=\varinjlim f_n : Y \rightarrow Y}$ is homotopic to ${g = \varinjlim g_n: Y \rightarrow Y}$.

After we prove this lemma, we will be done with the initial lemma.

Proof: We shall construct ${H: Y \times I \rightarrow Y}$ by an infinite composition (which will be well-defined).

First, there is a homotopy ${Y_1 \times I \rightarrow Y_1}$ from ${f_1 \rightarrow g_1}$. The homotopy extension property implies that we can extend this to a homotopy ${H_1}$ from ${f}$ to some ${g^{(1)}: Y \rightarrow Y}$. Here ${g^{(1)}}$ agrees with ${g}$ on ${Y_1}$ but may otherwise be arbitrary.

Next, ${g^{(1)}|_{Y_2}}$ is homotopic to ${g|_{Y_2}}$. Since ${g^{(1)}, g}$ agree on ${Y_1}$, we can take the homotopy stationary on ${Y_1}$ by a basic property of cofibrations (which I will not prove here). We can extend this map to a homotopy of ${Y}$. Let us call this map ${H_2: Y \times I \rightarrow Y}$. Then ${H_2}$ homotopes ${g^{(1)}}$ to some ${g^{(2)}}$ while being stationary on ${Y_1}$, where ${g^{(2)}}$ agrees with ${g}$ on ${Y_2}$.

Similarly, we can homotope ${g^{(2)}}$ into some ${g^{(3)}: Y \rightarrow Y}$ while not touching ${Y_2}$, where ${g^{(3)}}$ agrees with ${g}$ on ${Y_3}$. And so on. We get an infinite sequence of homotopies ${H_n}$. If we splice all these together such that the homotopy ${H_n}$ is carried out in the time ${(1-2^{-n}, 1-2^{-n+1})}$, we get a homotopy ${H: Y \times I \rightarrow Y}$ from ${f,g}$. The fact that ${Y}$ has the weak topology and the ${H_n}$ become stationary on any fixed ${Y_m}$ for ${n}$ large implies that ${H}$ is continuous, and it is a homotopy from ${f}$ to ${g}$.

As a result, we find:

Proposition 8 Let ${(Y,u)}$ be a pair. Then there is a ${\infty}$-universal pair ${(Y', u')}$ such that ${Y \subset Y'}$ is a subcomplex and ${u'|_Y = u}$.

Proof: Indeed, we know that we can find a tower of pairs ${(Y,u) \subset (Y_1, u_1) \subset \dots \subset (Y_n , u_n) \subset \dots}$ of ${n}$-universal elements (where the tower means that ${Y_n}$ is a subcomplex of ${Y_{n+1}}$ and ${u_{n+1}|_{Y_n} = u_n}$). If we consider the direct limit ${Y=\varinjlim Y_n}$ and an element ${u \in F(Y)}$ that pulls back to all the ${u_n}$ (by the above lemma), we find that the map

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

is a bijection for all ${k}$, because the homotopy groups commute with filtered colimits, by compactness of ${S^n}$ and ${S^n \times I}$.

3. ${\infty}$-universal pairs

We are now going to show that an ${\infty}$-universal pair ${(Y, u)}$ is in fact a universal object for the functor. Namely, we need to show that any element ${t \in F(X)}$ can be obtained by pulling back ${u}$ by a unique (up to homotopy) ${f: X \rightarrow Y}$. We are going to show existence and uniqueness in one fell swoop, with the following.

Lemma 9 Let ${(Y, u)}$ be an ${\infty}$-universal pair. Let ${g: A \rightarrow Y}$ and ${t = g^*(u) \in F(A)}$. Suppose ${A}$ is a subcomplex of ${X}$ and ${t' \in F(X)}$ restricts to ${t \in F(A)}$. Then there is a ${g': X \rightarrow Y}$ extending ${g}$ which pulls ${u}$ back to ${t'}$. That is, we can draw a diagram

Proof: Right now, we still have a diagram

in which we can form the push-out (which is a CW complex since ${A \rightarrow X}$ is a relative CW complex):

Here ${Z}$ is covered by the two subcomplexes ${X, Y}$, whose intersection is ${A}$. Moreover, there are elements ${t', u}$ that pull back to the same thing on ${A}$. The Mayer-Vietoris axiom implies that there is ${t'' \in F(Z)}$ pulling back to each of ${t', u}$.

Now, there is a bigger complex ${Z' \supset Z}$ and a ${\infty}$-universal element ${u' \in F(Z')}$ that restricts to ${t''}$ on ${Z}$. Let us draw a diagram:

This means that ${u' \in F(Z')}$ pulls back to ${u \in F(Y)}$. I claim that this is enough to make ${q: Y \rightarrow Z'}$ a homotopy equivalence. Indeed, in the maps

$\displaystyle \pi_k(Y) \stackrel{q_*}{\rightarrow} \pi_k(Z) \stackrel{f \rightarrow f^*u}{\rightarrow} F(S^{k})$

the second arrow and the composite are isomorphisms, so the first one is too. Thus ${q}$ is an isomorphism on the homotopy groups, thus by the Whitehead theorem a homotopy equivalence. Composing a homotopy inverse of this with ${X \rightarrow Z'}$, we find a way to make the diagram

homotopy commutative (where ${A \rightarrow Y}$ is the usual map). By the homotopy extension property, we can make it actually commutative by homotoping ${X \rightarrow Y}$. The claim is that ${X \rightarrow Y}$ sends ${u}$ back to ${t'}$. This follows from the fact that ${u' \in F(Z')}$ pulls back to ${u}$ and ${X \rightarrow Z'}$ pulls ${u'}$ back to ${t'}$. This completes the lemma.

Finally, we can prove the Brown representability theorem.

Proof: In fact, we are going to prove that if ${(Y, u)}$ is an ${\infty}$-universal pair, then the map ${\hom_{CW_*}(X,Y) \rightarrow F(X)}$ sending ${f: X \rightarrow Y}$ to ${f^*(u)}$ is a bijection. This will imply that ${(Y,u)}$ is a representing pair.

To do this, let us first see surjectivity. The above lemma with ${A}$ the one-point space ${\ast}$ shows that, given ${t \in F(X)}$, there exists ${f: X \rightarrow Y}$ such that ${t = f^*(u)}$. We now want to show that ${f}$ is unique up to homotopy. So suppose ${f,g: X \rightarrow Y}$ pull ${u}$ back to the same thing, and consider the “reduced product” ${X \times I/\ast \times I}$ where ${\ast \in X}$ is the basepoint.

We have a map ${f \sqcup g}$ out of the “boundary” ${X \times \left\{0\right\} \cup_{\ast} X \times \left\{1\right\}}$ into ${Y}$. ${F}$ applied to the boundary is ${F(X) \times F(X)}$ (it’s a coproduct) and ${(f \sqcup g)^*(u)}$ pulls back to something in ${F(X) \times F(X)}$ which extends to something in ${F(X)}$ (alternatively, is in the diagonal). The lemma again implies that ${f \sqcup g}$ extends to the whole reduced product, which states that ${f,g}$ are homotopic.