What is the top cohomology group of a non-compact, non-orientable manifold? 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.
 A: 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.
A: I think it's worth pointing out that at least for smooth or PL $n$-manifolds $M$ that are connected but not compact, something much stronger holds than the $n$th homology vanishing -- the manifold is actually homotopy equivalent to an $(n-1)$-dimensional simplicial complex!  This is a theorem of Whitehead from
J. H. C. Whitehead, The immersion of an open 3-manifold in euclidean 3-space, Proc. London Math. Soc. (3) 11 (1961), 81–90.
I gave a modern treatment of it in my note here.  In that note, I say that the manifold is smooth, but really all the proof uses is PL (I should fix this sometime).
I don't know if this holds if the manifold is not smooth or PL, but I suspect it does.
