$H_2(M)$ is free abelian for any simply connected $4$-manifold

Question

In Naber's book "Topology, Geometry and Gauge Fields. Foundations", it is stated that for each $4$-manifold $M$ which is smooth, closed, connected and simply connected we have $H_0(M) = H_4(M)= \mathbb Z$, $H_1(M)=H_3(M)=0$ and $H_2(M)$ is a finitely generated free abelian group. The statements about $H_k(M)$, $k\neq 2$ are clear for me. It is also clear that $H_2(M)$ is finitely generated (since there is a finite triangulation). But how to see that this group is also free?

Answer 1

By the Universal coefficient theorem (for cohomology) and the assumption that $M$ is simply connected, we have $$H^2(M)=\operatorname{Hom}(H_2(M),\mathbb{Z})\oplus \operatorname{Ext}(H_1(M),\mathbb{Z})= \operatorname{Hom}(H_2(M),\mathbb{Z}),$$ which is torsion free, hence free abelian (because it is finitely generated). By Poincaré duality, $H^2(M)\cong H_2(M)$ and we're done.