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I'm reading Brown's theorem in Hatcher and Brown's original paper. I am confused by the assumptions concerning the generalized cohomology theory, which is defined axiomatically by the Eilenberg-Steenrod axioms except the dimension axiom. Brown replaced the dimension axiom by the condition that the nth cohomology group of a point is countable for nonzero n, and Brown said this condition couldn't be weakened. The wedge axiom used by Hatcher to define reduced cohomology theory, however, implies that the nth cohomology group of a point is trivial. This seems contradictory. Can somebody please help me?

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Reduced cohomology of a point is by definition trivial, whereas I guess Brown was defining non-reduced cohomology. –  David Roberts Jun 25 '13 at 3:56
    
Thanks! So Hatcher's version is more powerful than Brown's original one? –  Emmy Jun 25 '13 at 4:08
    
I wouldn't have thought so - one can recover ordinary cohomology from reduced cohomology by adding a disjoint basepoint to a space, so the two should be interconvertible to some extent. However, I don't see that assuming reducedness would give you the cohomology of a point to be countable, so you may be right. –  David Roberts Jun 25 '13 at 5:17
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I haven't checked what versions are given in the sources that you mention. However, one important issue is as follows. If you assume that your original cohomology theory $H$ is defined only on finite CW complexes, then you need the coefficients to be countable. If you assume that $H$ is defined on all CW complexes, then countability is not required. –  Neil Strickland Jun 25 '13 at 8:26

2 Answers 2

Hatcher's Theorem 4E.1 in his 'Algebraic Topology' book is equivalent to Brown's Theorem II version (4.6) in his 'Cohomology theories' paper. Countability of coefficients is only required for Theorem II version (4.5). The first theorem is for cohomology theories defined on all CW-complexes, and the second one for those defined on finite CW-complexes. It looks like if the countability hypothesis were necessary if we wanted to represent cohomology theories defined only on finite CW-complexes, but this hypothesis was removed by Adams in 'A variant of E. H. Brown's representability theorem'.

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The most general reference I know for Brown Representability for cohomology is Neeman's book on triangulated categories. There even fewer of the Eilenberg-Steenrod axioms are assumed (additivity is not). It's so general that it should cover both cases you list. In this sense, Brown was wrong and his ideas could be made to work more generally, without a countability hypothesis. For homology theories the story is more subtle and countability is often needed. In Neeman's book this theorem is one of the big goals, so it's stated in the introduction and proved close to the end (chapter 12 I believe). Check it out on Google books for a preview

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the countability hypothesis is for the representability theorem of cohomology theories defined on compact objects. In his book, Neeman doesn't consider this problem, which he actually deals with in a paper entitled 'On a theorem of Brown and Adams'. Concerning the representability theorem in the boo, Neeman does require additivity. In fact, he always works with additive functors, so additivity is implicit. Actually, it is a necessary hypothesis since representable functors are additive. –  Fernando Muro Jun 25 '13 at 21:46
    
"Actually, it is a necessary hypothesis since representable functors are additive." I don't understand the logic behind that, @FernandoMuro: for example, representable functors are always, well, representable, but being representable is not necessary as a hypothesis in theorems about representability. –  Omar Antolín-Camarena Jun 26 '13 at 4:17
    
@Omar: you mean that you may start with a non-additive cohomology theory and end up showing it is representable and hence additive? Well, it may be the case, your right, but Neeman (and enveryone else considering Brown representability) assumes additivity. –  Fernando Muro Jun 26 '13 at 7:43
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I know that these Brown representability type theorems do have additivity as a hypothesis, I just thought you meant that they are forced to have it as a hypothesis because they are characterizing representable functors and those are always representable. If that is what you meant, I don't understand the logic behind it, @FernandoMuro. –  Omar Antolín-Camarena Jun 26 '13 at 13:07
    
@Omar, I hoped I had explained myself enough in my second comment. Actually, if you know enough about Brown representability you must have understood me from the beginning. I think you're insisting in a very formal point, like if I now said that I don't see what you mean when you say that "representable functors [...] are always representable". Of course, I understand you mean "cohomological functors [...] are always representable". –  Fernando Muro Jun 26 '13 at 14:08

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