Let $M$ be a connected, non-compact, non-orientable topological manifold of dimension $n$.  

**Question: Is the top singular cohomology group $H^n(M,\mathbb Z)$ zero?**  

This naïve question does not seem to be answered in the standard algebraic topology treatises, like those by Bredon, Dold, Hatcher, Massey, Spanier, tom Dieck, Switzer,...  

**Some remarks.**  

a) Since $H\_n(M,\mathbb Z)=0$ (Bredon, 7.12 corollary) we deduce by the universal coefficient theorem: $$ H^n(M,\mathbb Z) =\operatorname {Ext}(H\_{n-1}(M,\mathbb Z), \mathbb Z)\oplus \operatorname {Hom} (H\_n(M,\mathbb Z),\mathbb Z)=\operatorname {Ext}(H\_{n-1}(M,\mathbb Z),\mathbb Z )$$
But since $H\_{n-1}(M,\mathbb Z)$ need not be finitely generated I see no reason why $\operatorname {Ext}(H\_{n-1}(M,\mathbb Z),\mathbb Z)$ should be zero.  

b) Of course the weaker statement $H^n(M,\mathbb R) =0$ **is** true by the universal coefficient theorem, or by De Rham theory if $M$ admits of a differentiable structure.  

c) This question was asked [on this site](https://mathoverflow.net/q/101554/450) more than 8 years ago but the [accepted answer](https://mathoverflow.net/a/101570/450) is unsubstanciated since it misquotes Bredon.  

Indeed, Bredon states in (7.14, page 347) that $H^n(M,\mathbb Z)\neq0$ for $M$ compact, orientable or not, but says **nothing** in the non-compact case, contrary to what the answerer claims.



What is the top cohomology group of a non-compact, non-orientable manifold?
I believe you can deduce this from the corresponding statement in the orientable case. Let $\tilde M$ be the oriented double cover. Make an exact sequence of cochain complexes
$$
0 \to C^\bullet(M;\mathbb Z^t)\to C^\bullet(\tilde M;\mathbb Z)\xrightarrow{p\_!} C^\bullet(M;\mathbb Z)\to 0,
$$
where $\mathbb Z^t$ is the local system on $M$ corresponding to the non-orientability of $M$. (The map $p\_!$ is dual to the transfer map taking each singular simplex of $M$ to the sum of its two lifts to $\tilde M$.) Then there is an exact sequence
$$
H^n(\tilde M;\mathbb Z)\to H^n(M;\mathbb Z)\to H^{n+1}(M;\mathbb Z^t)
$$
with the first and third groups trivial.


