So, last time we were talking about Brown representability. We were, in particular, trying to show that a contravariant functor 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 sends filtered colimits into filtered limits, just that for coproducts. In fact, generally 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 and , we have shown how to extend the pair to an -universal pair for any . We’d now like to do the same for an -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 6Let be inclusions of subcomplexes in , with the limit. Then the canonical map

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 is a family of elements with such that . We have to piece these together into one element of .

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 with the usual attaching (and with quotiented to a point, where is the basepoint). We then attach this to the mapping cylinder of , and so on.

Let us be precise about this. Namely, we consider the infinite disjoint union

where the elements and are identified, as well as anything involving the basepoint .

There are two subcomplexes that we will need. Namely, is the subcomplex consisting of the for even, and for odd. Then is homotopy equivalent to and is homotopy equivalent to . Further . Moreover, .

There is an element which restricts to each of the even , and an element which restricts to each of the odd . The intersections glue on by hypothesis on the . We get an element which restricts to each of the on .

I claim now that is actually homotopy equivalent to , in a way compatible with the and . Then the element will correspond to the element of 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 that extends the projections . We need to construct a map in the reverse direction.

To do this, first consider the inclusion of on the first part of the mapping cylinder. Inductively, given , let us suppose that we have a map which is homotopic to the inclusion . We will extend this to . Namely, we have the inclusion on . The restriction is homotopic to . The homotopy extension property implies that we can homotope to something which agrees with on . We call this homotoped map .

Putting together all the , we get a map in the other direction, . We saw by construction that is homotopic to the identity, and same for . We want to say the same for the direct limit of these maps. For this we prove:

Lemma 7Suppose is a direct system of spaces such that the maps are cofibrations. Suppose is a pair of compatible systems of maps, such that . Then is homotopic to .

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

*Proof:* We shall construct by an infinite composition (which will be well-defined).

First, there is a homotopy from . The homotopy extension property implies that we can extend this to a homotopy from to some . Here agrees with on but may otherwise be arbitrary.

Next, is homotopic to . Since agree on , we can take the homotopy stationary on by a basic property of cofibrations (which I will not prove here). We can extend this map to a homotopy of . Let us call this map . Then homotopes to some while being stationary on , where agrees with on .

Similarly, we can homotope into some while not touching , where agrees with on . And so on. We get an infinite sequence of homotopies . If we splice all these together such that the homotopy is carried out in the time , we get a homotopy from . The fact that has the weak topology and the become stationary on any fixed for large implies that is continuous, and it is a homotopy from to .

As a result, we find:

Proposition 8Let be a pair. Then there is a -universal pair such that is a subcomplex and .

*Proof:* Indeed, we know that we can find a tower of pairs of -universal elements (where the tower means that is a subcomplex of and ). If we consider the direct limit and an element that pulls back to all the (by the above lemma), we find that the map

is a bijection for all , because the homotopy groups commute with filtered colimits, by compactness of and .

**3. -universal pairs **

We are now going to show that an -universal pair is in fact a universal object for the functor. Namely, we need to show that any element can be obtained by pulling back by a unique (up to homotopy) . We are going to show existence and uniqueness in one fell swoop, with the following.

Lemma 9Let be an -universal pair. Let and . Suppose is a subcomplex of and restricts to .Then there is a extending which pulls back to . 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 is a relative CW complex):

Here is covered by the two subcomplexes , whose intersection is . Moreover, there are elements that pull back to the same thing on . The Mayer-Vietoris axiom implies that there is pulling back to each of .

Now, there is a bigger complex and a -universal element that restricts to on . Let us draw a diagram:

This means that pulls back to . I claim that this is enough to make a homotopy equivalence. Indeed, in the maps

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

homotopy commutative (where is the usual map). By the homotopy extension property, we can make it actually commutative by homotoping . The claim is that sends back to . This follows from the fact that pulls back to and pulls back to . This completes the lemma.

Finally, we can prove the Brown representability theorem.

*Proof:* In fact, we are going to prove that if is an -universal pair, then the map sending to is a bijection. This will imply that is a representing pair.

To do this, let us first see surjectivity. The above lemma with the one-point space shows that, given , there *exists* such that . We now want to show that is unique up to homotopy. So suppose pull back to the same thing, and consider the “reduced product” where is the basepoint.

We have a map out of the “boundary” into . applied to the boundary is (it’s a coproduct) and pulls back to something in which extends to something in (alternatively, is in the diagonal). The lemma again implies that extends to the whole reduced product, which states that are homotopic.

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