$H_{n-1}(M;\mathbb{Z})$ is a free abelian group

Question

need help with this problem:

show that if $M$ is closed connected oriented n-manifold then $H_{n-1}(M;\mathbb{Z})$ is a free abelian group.

thanx.

Answer 1

$$O \to Ext(H_{n-1}(M),\mathbb{Z}) \to H^n(M) \to Hom(H_n(M),\mathbb{Z}) \to 0$$ As the latter arrow is an isomorphism when M is closed, connected and orientable, it follows that $Ext(H_{n-1}(M),\mathbb{Z})=0$. You just need to understand why it implies your $n-1$ torsion group $T_{n-1}$ is $0$...