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Brown representability states that any contravariant functor from the homotopy category $CW_*$ of pointed CW complexes to the category of pointed sets is representable if it turns coproducts into products and satisfies a type of Mayer-Vietoris gluability axiom, which I like to think of as a weak version of "the functor sends push-outs into pull-backs" as any representable functor must. The proof very much relies on the fact that CW complexes can be built up in a steady and predictable manner, as it uses Whitehead's theorem that a weak homotopy equivalence is automatically a homotopy equivalence. Namely, one shows that an element which is "universal" for the spheres is actually a "universal element" for this functor (in the sense of Yoneda's lemma).

Brown representability has many interesting consequences, e.g. that there is a "universal" principal $G$-bundle for pointed CW complexes (where $G$ is a topological group) or that the Eilenberg-Maclane spaces represent the cohomology functors. However, in the former case, it's actually true that the universal bundle exists for any topological space, not just CW complexes. I don't know whether the cohomology functors are representable on the category of all pointed topological spaces (even if one restricts to non-pathological ones: say Hausdorff, with nondegenerate basepoint), though I would imagine that a CW complex couldn't do it. This leads me to ask:

Is there a version of Brown representability for arbitrary pointed topological spaces?

There is a version of it on the nLab in more generality, but I don't know enough about categorical homotopy theory to understand anything. Could someone perhaps translate some of that into the special case of topological spaces?

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What do you mean when you say that there are universal bundles for all spaces? As far as I know, bundles are only representable if you assume something like paracompactness of your spaces. –  Eric Wofsey Nov 27 '10 at 16:46
    
OK. I had not realized that -- I only know of them from Wikipedia and Brown's paper. –  Akhil Mathew Nov 27 '10 at 17:37
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@Eric - not so. Bundles are all representable if you restrict to those that are trivialised on a numerable cover. Of course, for paracompact spaces this implies bundles trivialisable over all open covers. –  David Roberts Nov 28 '10 at 0:53

4 Answers 4

up vote 13 down vote accepted

Is there a version of Brown representability for arbitrary pointed topological spaces?

The answer is: No and Yes.

If you take a particular construction of a generalised cohomology theory that makes sense for all topological spaces then there is no guarantee that it will be representable in the homotopy category of topological spaces. For example, the various flavours of ordinary cohomology are only guaranteed to coincide for CW complexes (okay, and stuff with the homotopy type of such). That they disagree elsewhere shows that at least one of them can't be representable. Another example is K-theory. It's great that for compact Hausdorff spaces (and some others) that K-theory is exactly what you get when you take vector bundles and group complete, but it's not true for other spaces.

That's the "No". Now for the "Yes". The point about the "Yes" is that Brown representability is so good to have that when we move outside the realm of CW-complexes, we often define our cohomology theory to be (homotopy) homs into the representing object (found by restricting to CW-complexes). That is, we take our "natural construction" of whatever cohomology theory it is, use Brown representability to find the representing object for CW-complexes, and then define the extension of the theory to all topological spaces to be $[X,\underline{E}]$. Then it is representable, but by construction rather than by any fancy theory.

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Thanks! . –  Akhil Mathew Nov 28 '10 at 1:18

As Andrew Stacey pointed out, certainly there is no representability theorem for arbitrary cohomology theories (even ordinary). For instance, singular cohomology is not representable on compact metric spaces. But with additional axioms, there might be such a theorem.

On locally compact separable metric spaces, Milnor's three additional axioms guarantee uniqueness of ordinary cohomology, and therefore representability (because Cech cohomology is representable). It is conceivable EDIT: and is the case, indeed, that generalized theories satisfying these axioms could also be representable on those spaces. The axioms are 1) cohomology of an (infinite) disjoint union is the product, 2) cohomology of the one-point compactification of an infinite disjoint union of compact spaces is the sum (Milnor calls this the Cluster Axiom), and 3) Wallace's strong excision (Milnor calls it Map Excision). The uniqueness theorem for ordinary cohomology was proved by Petkova, and with slightly different axioms earlier by her adviser Sklyarenko. The parallel story for ordinary homology is in Petkova's another paper, which seems to exist only in Russian. Milnor's papers, which together contain the three axioms, are this and the 1961 preprint published in this volume.

EDIT: Here is the proof of the claim on generalized theories. On compact metric spaces, the additivity (i.e. the cluster axiom) and map excision imply that the theory is naturally equivalent to the Cech extension of its restriction to polyhedra. This is proved similarly to the Milnor short exact sequence for homology (see his 1961 preprint linked above). Then on locally compact separable metric spaces, the multiplicativity axiom implies that the theory is again naturally equivalent to the Cech extension of its restriction to polyhedra. This is similar to the proof of Petkova's Theorem 9: a natural transformation between the two theories comes from their coincidence on polyhedra (and the categorical definition of direct limit), and it is an isomorphism by the proof of Milnor 's short exact sequence for cohomology (see his 1962 paper linked above) plus the five lemma. Therefore by the Lee-Raymond theorem (Example 3.12 in the appendix of Dold's "Lectures in Algbraic Topology") the theory is representable.

Beyond locally compact separable metric spaces I'm not aware of any reasonable sets of axioms guarateeing uniqueness. But if you accept strong shape invariance as an axiom (thought of as a little upgrade of homotopy invariance), along with some kinds of additivity/multiplicativity, perhaps representability would follow? Of course one has to agree on what is "strong shape", because beyond compact metric spaces there are some variations; but for instance this one should fit. For compact metric spaces, strong shape invariance is equivalent to map excision.

Certainly there exist constructions of representable extraordinary theories defined on all or almost all topological spaces, for instance here. Further results are discussed on the last four pages (pp. 461-464 in Chapter 22 "Generalized strong homology") in Mardesic's Strong Shape and Homology (available on Google books).

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Thanks! Do you have a reference or explicit example of how homotopy classes from $X \to K(G,n)$ could fail to coincide with $H^n(X, G)$ for a pointed space $X$? –  Akhil Mathew Nov 28 '10 at 1:17
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There are some examples here: mathoverflow.net/questions/1750. The set of homotopy classes $[X,K(G,n)]$ coincides with Cech cohomology, which they compare with singular cohomology. Also one can see directly that $[N^+,S^0]$ is countable, where $N^+$ is the one-point compactification of a countable discrete set, and so differs from the singular 0-cohomology; and that $[W,S^1]$ is nontrivial, where $W$ is the Warsaw circle, and so differs from the singular 1-cohomology. Note also the Barratt-Milnor example. –  Sergey Melikhov Nov 28 '10 at 1:57

Brown representability is true for arbitrary pointed (CGWH) spaces, as long as you take into account cofibrant replacement (use the Quillen model structure instead of the Hurewicz model structure). This means that $[A,B]$ does not necessarily represent homotopy classes of maps, but represents equivalence classes of spans $A\leftarrow Q(A)\to B$. That is, $[A,B]=[QA,B]$ for a fixed cofibrant replacement (iirc all spaces are Serre fibrant).

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To clarify, the important point to understand is that pointed CW complexes are the fibrant-cofibrant objects in the model category of pointed spaces, so any "generalization" you're going to get for general spaces is really an illusion. –  Harry Gindi Nov 27 '10 at 16:40
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Thanks! OK, so is the model structure on pointed spaces that you are using the one where fibrations are Serre fibrations, weak equivalences are weak homotopy equivalences, and cofibrations deduced from this? –  Akhil Mathew Nov 28 '10 at 1:04
    
@Akhil: Yes, this is the Quillen model structure (Serre model structure). –  Harry Gindi Nov 28 '10 at 9:10

Let me try again. The original question

Is there a version of Brown representability for arbitrary pointed topological spaces?

is interesting, and I don't see how it has been fully answered. That Brown's theorem, as stated, does not hold beyond CW complexes doesn't necessarily mean that there is no reasonable version of this theorem for arbitrary (or almost arbitrary) spaces, generalizing an equivalent formulation of Brown's original theorem. On the other hand, versions coming with disclaimers like

Then it is representable, but by construction rather than by any fancy theory.

or

any "generalization" you're going to get for general spaces is really an illusion

are truly admirable for honesty of the disclaimers, but leave some doubt as to whether this is the end of the story.

More to the point. Given a homotopy invariant contravariant functor $h$ on the category of polyhedra, Watanabe shows that $h$ admits a unique (up to natural equivalence) extension $\check h$ to all topological spaces that is homotopy invariant and also continuous in the sense of satisfying two rather simple axioms (Theorems 5 and 6 here). This extension is also known as the Cech extension, and Lee and Raymond show that if $h$ is representable, then so is $\check h$ (See also Dold's Lectures in Algebraic Topology, Appendix, Example 3.12.) I understand that if $h$ is a (generalized) cohomology theory, then so is $\check h$.

Let me summarize: if we restrict our attention to functors that are cohomology theories and in addition are "continuous", then yes, we do have Brown's theorem for arbitrary topological spaces, by a certain fancy theory. (Assuming that I'm not missing any fine points concerning e.g. non-Hausdorff spaces. Note that in Dold's Appendix, some things work better for paracompact spaces.)

In case Watanabe's axioms of a "continuous" functor cut no ice with you, then there are the axioms mentioned in my previous reply in this thread (strong excision, strong homotopy invariance, additivity and multiplicativity) EDIT: which do suffice for locally compact separable metric spaces. Doing something similar for more general spaces could involve a fancier theory, which is probably not developed as yet. Certainly, doing something similar with homology (and other covariant functors) would involve a fancier theory. Going in the direction of such fancier theory are the following uniqueness theorems for generalized homology theories on metric compacta: Theorem 5.3 in Bauer and Corollary 2.17 in Kaminker-Schochet.

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Thanks for the (second) detailed answer! –  Akhil Mathew Nov 29 '10 at 13:14

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