# Is every homology theory given by a spectrum?

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?

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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. –  Eric Peterson May 5 '11 at 6:42
@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. –  Mark Grant May 6 '11 at 13:43

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 $\pi_*Map(-,E)$. –  Tom Goodwillie May 5 '11 at 15:05
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 $X_{\alpha}$.