Cup product with arbitrary coefficients I apologize in advance if this question is inappropriate for this forum. 
I am reading Hatcher's book and have the following problem: Let $X$ be a space, $G$ an abelian group and $R$ a ring. In his proof of the universal coefficient theorem in $\S$ 3.1, he considers a natural homomorphism $h\colon H^k(X;G)\to \text{Hom}(H_k(X),G)$. However, when computing cup product structures in $\S$ 3.2, specifically in examples 3.8 and 3.9, he seems to be using a homomorphism $\tilde{h}\colon H^k(X;R)\to \text{Hom}(H_k(X;R),R)$. 
I think I can see that such a homomorphism always exists when $R$ is a ring. More precisely, given a cocyle $\phi\colon C_k(X)\to R$, we can define $\tilde{\phi}\colon Z_k(X)\otimes R\to R$ by $\tilde{\phi}(\alpha \otimes r)=\phi(\alpha)r$, and the fact that $\delta\phi=\phi\partial=0$ means that this gives rise to a homomorphism $H_k(X;R)\to R$. In other words, the construction yields a map $\tilde{h}\colon H^k(X;R)\to \text{Hom}(H_k(X;R),R)$.
What I would like to know is: 
(a) Is there a way to define something similar to $\tilde{h}$ with $G$ in place of $R$?
(b) Is this map $\tilde{h}$ surjective as $h$ is?
(c) What is its kernel?
Thank you.
 A: Claim 1: If $R$ is a principal ideal domain, then there is a short exact sequence 
$$0 \to Ext^1_R(H_{n-1}(X;R),R) \to H^n(X;R) \xrightarrow[]{\tilde{h}}  Hom_R(H_n(X;R),R) \to 0\hspace{10pt}(\ast)$$
This answers b) and c). 
Claim 2: If $R$ or $H_\ast(X)$ is torsion-free, then $\tilde{h} = h$ (for $G=R$). 
Proof: Denote the simplicial complex of $X$ by $S(X)$. Then $C := S(X) \otimes_{\mathbb Z} R$ is a complex of free $R$-modules. Hence, by the universal coefficient theorem (cf. Hatcher, after Corollary 3.4) we have the short exact sequence 
$$0 \to Ext^1_R(H_{n-1}(C),R) \to H^n Hom_R(C,R) \xrightarrow[]{\tilde{h}} Hom_R(H_n(C),R)\to 0.$$
But $Hom_R(C,R) = Hom_R(S(X) \otimes_{\mathbb Z} R,R) \cong Hom(S(X),R)$ (cf. III (3.3) in Brown: Cohomology of Groups) and $H_n(C) = H_n(X;R)$. Hence the short exact sequence above transforms into $(\ast)$. 
Moreover, by universal coefficients, $$H_n(X;R) = H_n(X) \otimes_{\mathbb Z}R  \oplus Tor_1^{\mathbb Z}(H_{n-1}(X),R) = H_n(X) \otimes_{\mathbb Z}R$$ if $R$ or $H_{n-1}(X)$ is torsion-free. Hence $$Hom_R(H_n(X;R),R) = Hom_R(H_n(X) \otimes_{\mathbb Z} R,R) \cong Hom(H_n(X),R).$$ 
This shows claim 2. 
Remark: a) Example 3.8 from Hatcher uses $R=\mathbb{Z}/2$. 
b) Example 3.9 uses $R=\mathbb{Z}/m$. Hence $(\ast)$ doesn't apply (for general $m$). In particular it can't be used to show that $\tilde{h}$ is surjective. This may be the reason for Hatcher's alternative argumentation in this example. 

Added: I want to add an application of the short exact sequence $(\ast)$ that is of particular importance: Let $R=k$ be a field. Thus $H_\ast(X;k)$ is a projective $k$-module and hence $Ext_k^1(H_{n-1}(X;k),k)=0$ for all $n$. Now $(\ast)$ yields an isomorphism
$$H^n(X;k) \xrightarrow[]{\sim}Hom_k(H_n(X;k),k),$$
i.e. $H^n(X;k)$ is just the dual space of $H_n(X;k)$. This property is frequently used in the literature. 
