What happens if one defines homology groups \(h_n(X; G)\) as the homology groups of the chain complex \(\cdots \to \operatorname{Hom}(G, C_n(X)) \to \operatorname{Hom}(G, C_{n-1}(X)) \to \cdots\)? More specifically, what are the groups \(h_n(X; G)\) when \(G = \mathbb{Z}\), \(\mathbb{Z}_m\), and \(\mathbb{Q}\)?
Solution
For a space \(X\), each singular chain group \(C_n(X)\) is a free abelian group. We compute \[ h_n(X;G)=H_n\bigl(\operatorname{Hom}(G,C_\bullet(X))\bigr) \] for the three specified groups \(G\).
1. If \(G=\mathbb Z\), then\[ \boxed{h_n(X;\mathbb Z)\cong H_n(X;\mathbb Z).} \]
Evaluation at \(1\) gives a natural isomorphism \[ \operatorname{Hom}(\mathbb Z,C_n(X))\cong C_n(X) \] for every \(n\), and under this identification the induced differential is the ordinary singular boundary. Thus the two chain complexes are naturally isomorphic.
2. If \(G=\mathbb Z_m\), then\[ \boxed{h_n(X;\mathbb Z_m)=0} \]for every \(n\).
A homomorphism \(\mathbb Z_m\to C_n(X)\) must send \(1\) to an element annihilated by \(m\). Since \(C_n(X)\) is free abelian and therefore torsion-free, the only such element is \(0\). Hence \[ \operatorname{Hom}(\mathbb Z_m,C_n(X))=0 \] in every degree.
3. If \(G=\mathbb Q\), then\[ \boxed{h_n(X;\mathbb Q)=0} \]for every \(n\).
Let \(\varphi:\mathbb Q\to C_n(X)\) be a homomorphism and put \(c=\varphi(1)\). For every positive integer \(r\), \[ c=r\,\varphi(1/r), \] so \(c\) is divisible by every \(r\). A free abelian group has no nonzero element divisible by every positive integer, hence \(c=0\). Then for \(a/b\in\mathbb Q\), \[ b\varphi(a/b)=a\varphi(1)=0, \] and torsion-freeness gives \(\varphi(a/b)=0\). Thus every such homomorphism is zero.
So this construction behaves very differently from ordinary homology: for torsion or divisible coefficient groups it can vanish identically.