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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.

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.

Remember that a ring is called hereditary, if submodules of projective modules are projective. If this holds only for finitely generated submodules, the ring is called semi-hereditary.

Claim 1: If the ring $R$ is hereditary, 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.

Remarks: a) Note that the universal coefficient theorem from Hatcher requires $R$ to be a PID. But it actually holds for hereditary rings (I think this can be found in Spanier, otherwise in Cartan-Eilenberg).

b) Example 3.8 from Hatcher uses $R=\mathbb{Z}/2$ what's hereditary as a field.

c) Example 3.9 uses $R=\mathbb{Z}/m$. Since a finite product of hereditary rings is hereditary, $R$ is hereditary if $m$ is square free. Moreover we have $Hom_R(-,R) = Hom(-,R)$. I don't know if $R$ is hereditary or semi-hereditary for general $m$ (see Mariano's comment). In order to determine if $\tilde{h}$ is surjective, these cases require a case-by-case analysis as done in 3.9 by Hatcher.

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.

Remember that a ring is called hereditary, if submodules of projective modules are projective. If this holds only for finitely generated submodules, the ring is called semi-hereditary.

Claim 1: If the ring $R$ is hereditary, 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.

Remarks: a) Note that the universal coefficient theorem from Hatcher requires $R$ to be a PID. But it actually holds for hereditary rings (I think this can be found in Spanier, otherwise in Cartan-Eilenberg).

b) Example 3.8 from Hatcher uses $R=\mathbb{Z}/2$ what's hereditary as a field. Example 3.9 uses $R=\mathbb{Z}/m$. Since a finite product of hereditary rings is hereditary, $R$ is hereditary if $m$ is square free. Moreover we have $Hom_R(-,R) = Hom(-,R)$. I don't know if $R$ is hereditary or semi-hereditary for general $m$.

Remember that a ring is called hereditary, if submodules of projective modules are projective. If this holds only for finitely generated submodules, the ring is called semi-hereditary.

Claim 1: If the ring $R$ is hereditary, 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.

Remarks: a) Note that the universal coefficient theorem from Hatcher requires $R$ to be a PID. But it actually holds for hereditary rings (I think this can be found in Spanier, otherwise in Cartan-Eilenberg).

b) Example 3.8 from Hatcher uses $R=\mathbb{Z}/2$ what's hereditary as a field.

c) Example 3.9 uses $R=\mathbb{Z}/m$. Since a finite product of hereditary rings is hereditary, $R$ is hereditary if $m$ is square free. Moreover we have $Hom_R(-,R) = Hom(-,R)$. I don't know if $R$ is hereditary or semi-hereditary for general $m$ (see Mariano's comment). In order to determine if $\tilde{h}$ is surjective, these cases require a case-by-case analysis as done in 3.9 by Hatcher.