# A “dual” universal coefficient theorem

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

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
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