TITLE: n-th integral cohomology of a non-compact manifold of dimension n
QUESTION [11 upvotes]: Let $M$ be a smooth connected oriented without boundary non-compact manifold
of dimension n.
Let $k$ be a principal ideal, e. g. the integers $Z$
Let $H_n(M)$ and $H^n(M)$ be the homology and cohomology in degree n of $M$
with coefficients in $k$.
It is well-known that $H_n(M)=0$.
Is $H^n(M)$ also trivial ?
By the universal coefficient theorem for cohomology,
$H^n(M)=Ext(H_{n-1}(M),k)$.
Therefore if $k$ is a field,  $H^n(M)=0$. I would like to know this answer over
a principal ideal: $Z$.
It is also known (Bredon's book) that   $H_{n-1}(M)$ is without torsion.
But I believe that $H_{n-1}(M)$ in the non-compact case, is a not a finitely 
generated $k$-module.
Therefore I don't know if $H_{n-1}(M)$ is free, i. e. projective.
Since there exists abelians groups without torsion, non-free, e. g. the rationals $Q$
I suppose that this must be well-known.
But I could not find a reference.

REPLY [2 votes]: Maybe too late, and close to the Fernando's argument. Let $c$ be an infinite path going through each center of the $n$-simplex of a triangulation. Let $U$ be a neighbourhood of $c$, we can assume $U$ has the homotopy type of $[0,+\infty[$ and $\partial U$ is homeomorphic to $\mathbb R^{n-1}$. Then $V= \overline{X\setminus U}$ retracts on the $n-1$-skeleton, so $H^{n-1}(V)=0$ and we conclude by Mayer Vietoris
$$H^{n-1}(U\cap V) \to H^n(X) \to H^n(U)\oplus H^n(V)$$