Show that if a closed orientable manifold \(M\) of dimension \(2k\) has \(H_{k-1}(M; \mathbb{Z})\) torsionfree, then \(H_k(M; \mathbb{Z})\) is also torsionfree.
Solution
Since \(M\) is a closed manifold, all its homology groups are finitely generated. Thus the hypothesis that \(H_{k-1}(M;\mathbb Z)\) is torsionfree implies that it is free abelian, and therefore \[ \operatorname{Ext}(H_{k-1}(M;\mathbb Z),\mathbb Z)=0. \] The universal coefficient theorem for cohomology gives \[ 0\to \operatorname{Ext}(H_{k-1}(M),\mathbb Z) \to H^k(M;\mathbb Z) \to \operatorname{Hom}(H_k(M),\mathbb Z)\to0, \] so \[ H^k(M;\mathbb Z)\cong\operatorname{Hom}(H_k(M),\mathbb Z). \] The group on the right is free abelian.
Poincaré duality for the closed orientable \(2k\)-manifold \(M\) gives \[ H^k(M;\mathbb Z)\cong H_{2k-k}(M;\mathbb Z)=H_k(M;\mathbb Z). \] Hence \(H_k(M;\mathbb Z)\) is free abelian, in particular torsionfree.