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Universal coefficient theorem allows us to calculate $H^*(X,M)$ from $H_*(X,Z)$. Do we have a "dual" universal coefficient theorem that allows us to calculate $H_*(X,M)$ from $H^*(X,Z)$?

Here $Z$ is the set of integers.

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Thanks. It is nice to know. It appears that $H_\*(X,Z)$ contained most info. This question is motivated by another question mathoverflow.net/questions/111087/… which is unfortunately closed :-( Any light on that question will be greatly appreciated. –  Xiao-Gang Wen Nov 2 '12 at 1:42

2 Answers 2

up vote 6 down vote accepted

Yes, there is such a universal coefficient theorem.

$$0 \to Ext(H^{q+1}(X,R), G) \to H_q(X, G) \to Hom(H^q(X, R), G) \to 0$$

see Theorem 6.5.12 in Spanier's textbook "Algebraic Topology". It's on page 248.

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@Ryan Budney: Thanks. But it is a little confusing. Is the above $R$ the field of real numbers, or $R=Z$ the set of integers? –  Xiao-Gang Wen Nov 2 '12 at 2:59
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R is a principal ideal domain and G is an R-module. You also need $H_\ast(X;R)$ to be of finite type, meaning each $H_i(X;R)$ is a finitely generated $R$-module. –  Greg Friedman Nov 3 '12 at 2:36
    
Thanks. That helps a lot. We should really include this result in Wiki. –  Xiao-Gang Wen Nov 3 '12 at 4:43

The proof of Spanier's 6.5.12 starts from a free chain complex with homology of finite type and replaces it by a quasi-isomorphic free chain complex of finite type. Then the conclusion reduces directly to application of the usual universal coefficient theorem for computing cohomology of chain complexes from homology. The reference to EKMM should be to p. 82, which states "A reference to Adams [1] is mandatory''; [1] is "Lectures on generalized cohomology". Springer Lecture Notes 99, 1969, pp 1--138. As recalled on EKMM p. 82, Adams shows how to deduce further spectral sequences from those listed on that page, by duality. Adams' (UCT 4), page 5 op cit, gives the spectral sequence that generalizes the result cited from Spanier to generalized homology/cohomology theories.

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