13
$\begingroup$

Let $E$ be a spectrum. For any CW complex $X$, define $h_*=\pi_i(E\wedge X)$. Then we know that $h_*$ form a homology theory. In other words, there functors satisfy the homotopy invariance, maps a cofiber sequence of spaces to a long exact sequence of abelian groups, also satisfy the wedge axiom in the definition of a homology theory. I want to know the converse case. Is every homology theory given by a spectrum in such way?

Thanks for all your comments. This is not really a problem. Anybody knows how to close it?

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ A reference is probably best: see Switzer 14.35-36 for homology representation, which relies on various cohomology representation results from chapter 9 --- say, 9.21 and onward. $\endgroup$
    – Eric Peterson
    May 5, 2011 at 6:42
  • $\begingroup$ @yeshengkui: There is no need to close questions that have been satisfactorily answered, as these are unlikely to attract new answers and hence keep bubbling up to the top of the front page. $\endgroup$
    – Mark Grant
    May 6, 2011 at 13:43

2 Answers 2

Reset to default
19
$\begingroup$

For homology theories on CW-complexes or homology theories that map weak equivalences to isomorphisms, that's Brown's representability theorem, which you can find in any textbook on stable homotopy theory. You forgot the important axiom of excision, by the way. The short answer is yes.

Improve this answer
$\endgroup$
3
  • 10
    $\begingroup$ To fill in a step: Brown's theorem applies to cohomology theories. To pass from cohomology to homology you use duality, observing that $S$-duality sets up a one to one correspondence between homology theories and cohomology theories (for finite complexes) in which $\pi_*(-\wedge E)$ corresponds to $\pi_*Map(-,E)$. $\endgroup$
    – Tom Goodwillie
    May 5, 2011 at 15:05
  • $\begingroup$ I see. I thought the homological one also went by the name of Brown rep. thm. $\endgroup$
    – Tilman
    May 5, 2011 at 18:49
  • $\begingroup$ Well, I think some people call it that, but I believe that what Brown proved was a theorem about which contravariant functors from the homotopy category of spaces to sets are representable, which is used to show that cohomology theories are representable. $\endgroup$
    – Tom Goodwillie
    May 6, 2011 at 0:40
16
$\begingroup$

The answer is yes, if you replace the wedge axiom with the stronger direct limit axiom $h_{i}(X) = \mathrm{lim}\ h_{i}(X_{\alpha})$, where $X$ is the direct limit of subcomplexes $X_{\alpha}$.

As well as Switzer, this is discussed in Chapter 4.F of Hatcher's "Algebraic Topology", Adams' little blue book "Stable homotopy and generalised homology", and Adams' paper "A variant of E. H. Brown's representability theorem".

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ So what is an example of a homology theory that satisfies Milnor's additivity axiom but not the direct limit axiom? $\endgroup$
    – Bruno Stonek
    Jan 29, 2016 at 11:11
  • $\begingroup$ I don't know, this might be a good MO question! An Exercise in Hatcher's 4.F is to show that the wedge axiom implies the direct limit axiom on the category of countable CW complexes, so to find such a thing we'd have to go to uncountable CW complexes. $\endgroup$
    – Mark Grant
    Jan 29, 2016 at 11:30
  • 1
    $\begingroup$ Ok, I asked it here: mathoverflow.net/questions/229612/… $\endgroup$
    – Bruno Stonek
    Jan 29, 2016 at 12:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.