I am studying cohomology using Chapter 3 of Hatcher's Algebraic Topology, and he defines (singular) cohomology with coefficients as follows:
Given a space $X$ and an abelian group $G$, we first take the singular chain complex with $\mathbb{Z}$ coefficients, dualize by applying $\text{Hom}(-, G)$ to obtain the cochain complex, and then we let the $n$th cohomology group with $G$ coefficients $H^{n}(X; G)$ to be the $n$th homology group of this cochain complex.
But what if we instead want to take $G$ coefficients everywhere, and start out with the singular chain complex with $G$ coefficients? In other words, what if we instead took $$ \cdots \xrightarrow[]{} C_{n+1}(X; G) \xrightarrow[]{\partial_{n+1}} C_{n}(X; G) \xrightarrow[]{\partial_{n}} C_{n-1}(X; G) \xrightarrow[]{} \cdots$$ to begin with (where $C_{n}(X; G)$ consists of formal sums of singular $n$-simplices with $G$ coefficents), dualized by applying $\text{Hom}(-, G)$ to obtain the cochain complex $$\cdots \xleftarrow[]{} \text{Hom}(C_{n+1}(X; G), G) \xleftarrow[]{\delta_{n+1}} \text{Hom}(C_{n}(X; G), G) \xleftarrow[]{\delta_{n}} \text{Hom}(C_{n-1}(X; G), G) \xleftarrow[]{} \cdots,$$ and defined the alternate $n$th cohomology group with $G$ coefficients $h^{n}(X;G)$ to be the $n$th homology group of this cochain complex?
I know that if you put the universal coefficient theorem for homology and the universal coefficient theorem for cohomology, you can express $h^{n}(X;G)$ in terms of $\mathbb{Z}$-homology $H_{*}(X)$. To be more precise, from the universal coefficient theorems we have the following split short exact sequences for a chain complex $C$ of free abelian groups: $$0 \to H_{n}(C) \otimes G \to H_{n}(C; G) \to \text{Tor}(H_{n-1}(C), G) \to 0$$ and
$$0 \to \text{Ext}(H_{n-1}(C), G) \to H^{n}(C; G) \to \text{Hom}(H_{n}(C), G) \to 0.$$
Taking $C$ to be $$ \cdots \xrightarrow[]{} C_{n+1}(X; G) \xrightarrow[]{\partial_{n+1}} C_{n}(X; G) \xrightarrow[]{\partial_{n}} C_{n-1}(X; G) \xrightarrow[]{} \cdots,$$ it follows from the second split short exact sequence that $$h^{n}(X; G) = H^{n}(C; G) \cong \text{Ext}(H_{n-1}(C), G) \oplus \text{Hom}(H_{n}(C), G) = \text{Ext}(H_{n-1}(X; G), G) \oplus \text{Hom}(H_{n}(X; G), G).$$
Then, taking $C$ to be the singular chain complex for $X$ with $\mathbb{Z}$ coefficients, it follows from the first split short exact sequence that $$H_{n}(X; G) \cong (H_{n}(X) \otimes G) \oplus \text{Tor}(H_{n-1}(X), G).$$
Putting these two together, we have \begin{align*} h^{n}(X; G) &\cong \text{Ext}\left((H_{n-1}(X) \otimes G) \oplus \text{Tor}(H_{n-2}(X), G), G\right) \\ & \hphantom{1} \hphantom{1} \hphantom{1} \oplus \text{Hom}((H_{n}(X) \otimes G) \oplus \text{Tor}(H_{n-1}(X), G), G)\\ &\cong \text{Ext}(H_{n-1}(X) \otimes G, G) \oplus \text{Ext}(\text{Tor}(H_{n-2}(X), G), G) \\ & \hphantom{1} \hphantom{1} \hphantom{1} \oplus \text{Hom}(H_{n}(X) \otimes G, G) \oplus \text{Hom}(\text{Tor}(H_{n-1}(X), G), G). \end{align*}
My question is: is there any way to simplify the ugly expression for $h^{n}(X; G)$ above to better see what is going on?
One observation:
- I know that by the tensor-hom adjunction we can write $$\text{Hom}(H_{n}(X) \otimes G, G) \cong \text{Hom}(H_{n}(X), \text{Hom}(G, G)).$$ So, $h^{n}(X; G)$ contains information about what happens when we dualize $H_{n}(X)$ with respect to $\text{Hom}(G, G)$, which is sort of what we expect (because we dualized with respect to $G$ twice).
I don't really know what to do with $\text{Ext}$ and $\text{Tor}$ when the group $G$ is not $\mathbb{Z}$, so I'm a bit stuck on the other terms. I tried making the assumption that $H_{*}(X)$ was finitely generated, but it didn't help much.