Top Cohomology Group of Non-Compact, Non-Orientable Manifolds

Question

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 more than 8 years ago but the accepted answer 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.

Answer 1

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.