$M$ orientable implies $H_{n-1}(M, \mathbb{Z})$ is free Abelian group. Let $M$ be a compact connected $n$-manifold without boundary, where $n \ge 2$. How do I see that if $M$ is orientable, then $H_{n-1}(M, \mathbb{Z})$ is a free Abelian group?
 A: An orientable manifold is $R$-orientable for any $R$. Now if $M$ is closed connected $n$ dimensional $R$-orientable manifold then $H_n(M;R)=R$.
Again universal co-efficient theorem for homology says that , If $C$ is a chain complex of free abelian groups, then there are natural short exact sequence $0\to H_m(C)\otimes G\to H_m(C;G)\to Tor(H_{m-1}(C);G)\to 0$ is a split exact sequence for all $m,G$.
And $Tor$ has following properties,
$Tor(A,B)= Tor(T(A),B)$ where $T(A)$ is the torsion subgroup of $A$, and $Tor(Z_n,B)= ker(B\to_n B)$, in particular $Tor(Z_n,Z_n)=Z_n$.
Since homology groups of $M$ are finitely generated, now if $H_{n-1}(M;\mathbb{Z})$ contained torsion, then there exists a prime $p$ s.t if we use universal coefficient theorem for homology then $H_n(M,Z_p)$ would be larger that $Z_p$. (contradiction).
A: As the closed manifold is orientable, we have the powerful tool of Poincaré Duality. This means we have an isomorphism by capping with the fundamental class
$$H_{n-1}M \stackrel \sim \to H^1M.$$
Now it is easy to see that by definition (from here on we don't use any orientability by the way) or universal coefficients, that $H^1M$ is just $Hom(H_1M,\mathbb Z)$ (which is itself equal to $Hom(\pi_1M,\mathbb Z)$ as $\mathbb Z$ is abelian. So we get an isomorphism $H_{n-1}M \cong \mathbb Z^{b_1M}$, which means freeness.

By the way an interesting way to see the Poincaré Duality Map under said identifications is the following. For $x\in H_{n-1}M$ choose an $(n-1)$-submanifold $N\subset M$ representing $x$. Then
$$
H_{n-1}M \stackrel \sim \to Hom(\pi_1M,\mathbb Z),
$$
$$
[N] \mapsto (\gamma \mapsto N \pitchfork \gamma),
$$
is given by mapping $[N]$ to the homomorphism on $\pi_1M$ which evaluates a (transverse) loop to the signed intersection with $N$.
