# at.algebraic topology – 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.