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Let $M$ be an $R$-module where $R$ is a P.I.D. we have the exact sequence

$$0\rightarrow \operatorname{Ext}_R(H_{q−1}(X;R),M) \rightarrow H^q(X;M) \rightarrow \operatorname{Hom}(H_q(X;R),M)\rightarrow 0$$

  1. Is $H_q(X;R)$ an $R$-module? and then $\operatorname{Hom}(H_q(X;R),M)$ means homomorphisms of $R$-modules? I know this is true when $R=\mathbb Z$ i.e. $H_q(X;\mathbb Z)$ is an abelian group so a $\mathbb Z$-module and in this case $\operatorname{Hom}(H_q(X;\mathbb Z),M)$ means group homomorphism. but what about general $R$?
  2. If $M$ is an $R_1$-module and an $R_2$-module at the same time which $R_i$ we choose to calculate $H^q(X;M)$ from the exact sequence above? and what is $\operatorname{Hom}(H_q(X;R),M)$ in each case depending on $R_i$ chosen?. For example if $M=\mathbb Q$ , then $Q$ is an abelian group and then $R_1=\mathbb Z$ and also $\mathbb Q$ is a field so it is a $\mathbb Q$-module so $H^q(X;\mathbb Q) \cong \operatorname{Hom}(H_q(X;R),\mathbb Q)$ but which $R$ we use in $H_q(X;R)$, the $\mathbb Z$ or $\mathbb Q$?

  3. If $M$ is a field we have that $H^q(X; M) \cong \operatorname{Hom}(H_q(X;R),M)$ what is the dual expression for homology in terms of cohomology?

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Please start accepting answers... – Rasmus Jun 7 '11 at 12:46
look in spanier's textbook. – Ryan Budney Jun 8 '11 at 15:40
@ Ryan Budney : i looked everywhere before posting including spanier but i'm still confused.. i will be thankful if you answer me especially for 2) even concisely!! thanks – palio Jun 9 '11 at 15:48

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