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Probably you can "google" this question, but I can't find anything relevant. The classical Brown Representability Theorem states: Denote $hCW_*$ the homotopy category of pointed CW-complexes. Let $F : hCW_* \to Set_*$ be a contravariant functor. Then $F$ is representable if and only if

  • $F$ respects coproducts, i.e. $F(\vee_{i \in I} X_i) = \prod_{i \in I} F(X_i)$ for all families $X_i$ of pointed CW-complexes.
  • $F$ satisfies a sort of mayer-vietoris-axiom: If $X$ is a pointed CW-complex which is the union of two pointed subcomplexes $A,B$, then the canonical map $F(X) \to F(A) \times_{F(A \cap B)} F(B)$ is surjective.

I just know two applications: classifying spaces for $G$-principal bundles (in particular, vector bundles) for a locally compact topological group $G$ and for generalized cohomology theories on $CW_*$; the yoneda-lemma also yields functorial relations (cf. Switzer, Algebraic Topology). I'm interested in other explicit applications. I've read that there are categorical generalizations, but in this question I'm just asking whether there are explicit functors defined on CW-complexes, whose representabilty is of interest and can be shown with the theorem above. Also, these examples should really differ from the two ones mentioned above. :-)

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    $\begingroup$ again, my question does not seem to be clear :-(. I repeat it: "these examples should really differ from the ones mentioned above" (cohomology/bundles)! $\endgroup$ Jan 11, 2010 at 22:54
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    $\begingroup$ I have difficulty understanding the question. It seems to be saying "I want applications of the theorem 'Every cohomology theory is representable' that don't use the words 'cohomology theory'.". It feels a bit like "Applications of the Fundamental Theorem of Algebra without mentioning polynomials or complex numbers.". $\endgroup$ Jan 12, 2010 at 12:11
  • $\begingroup$ again I don't understand what's unclear in my question. the theorem is: every functor satisfying these two properties is representable. cohomology consists of a sequence of functors, which also are valuend in abelian groups. $\endgroup$ Jan 12, 2010 at 12:28
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    $\begingroup$ Cohomology isn't necessarily a sequence of functors. Cohomology theories can be ungraded (or graded by some other set than the integers). So that leads me to the "abelian group" part and makes me think that your question is: "Are there any interesting cohomology-like theories that are Set-valued rather than abelian group-valued.". Is that the correct interpretation? $\endgroup$ Jan 12, 2010 at 12:47
  • $\begingroup$ yes, this seems to be my question.. $\endgroup$ Jan 12, 2010 at 15:03

4 Answers 4

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The super-classical example would be the use of the (?Serre's?) theorem that $H^n(X;G) = [X,K(G,n)]$ to deduce that co-dimension two knots have Seifert surfaces. This is written up in Kervaire and Weber's article in LNM 685 "A survey of multi-dimensional knots".

The basic idea goes like this: let $C$ be the complement of a co-dimension two knot in $S^n$. Apply Poincare/Alexander duality to deduce that $C$ is a homology $S^1 \times D^{n-1}$. So $H^1 C \simeq \mathbb Z$, and $H^1 \partial C$ is either $\mathbb Z^2$ or $\mathbb Z$ according to whether or not $n=3$ or $n>3$. In either case the restriction map is an injection. Serre's theorem gives you a map $C \to S^1$ which you make transverse to a point (and a standard projection map on the boundary), this makes the preimage of this point a Seifert surface for the knot. By Seifert-surface I mean an orientable co-dimension one submanifold of $S^n$ whose boundary is the knot.

Serre and Thom used these ideas repeatedly in their early attacks on the Steenrod realization problem. Well, the version where you're trying to realize the homology classes by embedded submanifolds. This is a relative version of their arguments.

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    $\begingroup$ That's a lovely proof! $\endgroup$ Jan 12, 2010 at 15:24
  • $\begingroup$ This argument went unnoticed until 1963 when Kervaire pointed it out. It's a good example of how some pretty significant ideas can take time to filter across fields, even though knot theory and cobordism are so close. Thom and Serre had set up the machinery by 1954. $\endgroup$ Jan 14, 2010 at 19:01
  • $\begingroup$ Is this really an example of Brown's representability theorem in action? Maybe I'm misunderstanding, but this argument seems to rely only on $H^1(X,\mathbb{Z}) = [X, S^1]$, and thus $S^1 = K(\mathbb{Z},1)$. We don't really need Brown representability to see that. $\endgroup$
    – John Baez
    Oct 9, 2021 at 19:18
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Brown Representability combined with the Landweber Exact Functor theorem allows one to construct homotopy types out of purely algebro-geometric data, and is in particular the starting point for theories such as that of topological modular forms. Thus, this old theorem underlies one of the key themes which modern algebraic topology is fleshing out.

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    $\begingroup$ A lot of the chromatic picture comes from what you describe (as far as I understand it, which isnt much). The infusing of Lubin-Tate Theory for example! Also, this way of constructing spectra does not seem to be the most desired to anyone other than a homotopy theorist, I mean that while we get an honest spectrum out of it we don't really know anything at all about the spectrum except for it's coefficients, which is enough to do some things but not everything. for example I am not sure you can get adams operations out of Hopkins-Hovey's construction of K-theory using the Landweber exact... $\endgroup$ Mar 30, 2010 at 19:28
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    $\begingroup$ ...Functor theorem and you certainly don't learn anything about vector bundles. this might be a direct quote from the audio file on online.kitp.ucsb.edu/online/mp03/hopkins4 $\endgroup$ Mar 30, 2010 at 19:29
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If you look in Brayton Gray's book Homotopy Theory, I think that you will find that this theorem gives any generalized homology theory is representable by homotopy classes into a spectrum. I may have misread something here, and it has been a loooong first day of classes.

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    $\begingroup$ cohomology theory i think is what you mean, and also there is a theorem for homology that says homology is the homotopy of a the space smashed with an appropriate spectrum, the hypothesis are similar. $\endgroup$ Mar 30, 2010 at 2:07
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I think it might be pretty hard to get an answer that says nothing about cohomology. I interpret/think of Brown Representability as saying that if you want to think about these types of invariants you should really look at spectra. That somehow cohomology is really about spectra. Once you see this you then interpret the axioms for a cohomology theory or rather what you know about singular cohomology and try to think of the structure you have on the representing object that comes from this and you get ring spectra. I dont know if non-topologists really think about these things or if i just think that they should. They bring up interesting questions about operads... that may have come up independently.

You also get cohomology operations for any cohomology theory from this, but only as a theoretical tool, that is i am not completely sure how to use Brown Representability to compute Adams operations, in fact i dont think that we know all of $K^*K$.

This is a pretty biased answer though, but it is a unifying idea of algebraic topology that we should look at ring spectra, or rather the representing objects of all "nice" cohomology theories.

anyone should feel free to appropriately fix this, especially if they have better historical or computational information

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