Show that \(H_*(X; \QQ) = H_*(X;\ZZ)\tensor \QQ\) \(H^*(X; \ZZ) = \hom(H_*(X; \ZZ), \QQ)\)
Solution
1. The homology statement is correct:\[ \boxed{H_n(X;\mathbb Q)\cong H_n(X;\mathbb Z)\otimes\mathbb Q.} \]
The universal coefficient theorem for homology gives a natural short exact sequence \[ 0\to H_n(X;\mathbb Z)\otimes\mathbb Q\to H_n(X;\mathbb Q)\to \operatorname{Tor}_1^{\mathbb Z}(H_{n-1}(X;\mathbb Z),\mathbb Q)\to0. \] The group \(\mathbb Q\) is flat over \(\mathbb Z\), so the Tor term vanishes.
2. The cohomology formula in the card has a coefficient typo. The correct statement is\[ \boxed{H^n(X;\mathbb Q)\cong \operatorname{Hom}_{\mathbb Z}(H_n(X;\mathbb Z),\mathbb Q).} \]
The cohomological universal coefficient theorem gives \[ 0\to \operatorname{Ext}^1_{\mathbb Z}(H_{n-1}(X;\mathbb Z),\mathbb Q) \to H^n(X;\mathbb Q) \to \operatorname{Hom}_{\mathbb Z}(H_n(X;\mathbb Z),\mathbb Q) \to0. \] The abelian group \(\mathbb Q\) is divisible, hence injective as a \(\mathbb Z\)-module, so the Ext term vanishes.
3. As literally written, \(H^*(X;\mathbb Z)=\operatorname{Hom}(H_*(X;\mathbb Z),\mathbb Q)\) is false.
Take \(X=S^1\). Then \(H^1(S^1;\mathbb Z)\cong\mathbb Z\), whereas \[ \operatorname{Hom}(H_1(S^1;\mathbb Z),\mathbb Q)\cong\operatorname{Hom}(\mathbb Z,\mathbb Q)\cong\mathbb Q. \]