= Solution
An <orientation> of a topological n-manifold without boundary is a choice of generator $\mu_x$ of each local <homology> group $H_n(M,M\setminus\{x\};\mathbb Z)\cong\mathbb Z$, compatible in coordinate neighbourhoods. Equivalently, local <orientations> of coordinate discs are chosen so their overlap changes have degree $+1$. This trivializes the <orientation local system>.
Here the printed $H^n_{\mathrm{ct}}$ denotes <compactly supported cohomology>, which we write $H_c^n$. We prove the cyclicity assertion without assuming <Poincare duality>. By the <compactly supported cohomology of Euclidean space>, an open n-disc has one copy of $\mathbb Z$ in degree $n$ and zero in all other degrees. The compact-support <Mayer–Vietoris sequence> for open sets contains
$$
H_c^n(U\cap V)\longrightarrow H_c^n(U)\oplus H_c^n(V)
\longrightarrow H_c^n(U\cup V)\longrightarrow H_c^{n+1}(U\cap V).
$$
Use the finite good cover allowed in the paper. Induction on the number of discs shows that any union of its discs has zero compact-support <cohomology> above degree $n$: at each step the intersection with the new disc is itself a union of fewer discs with a good cover, so the displayed sequence and its higher-degree terms apply.
The same induction presents $H_c^n(M)$ by a generator $u_i$ for each cover disc, with overlap relations $u_i=\epsilon_{ij}u_j$, where $\epsilon_{ij}=\pm1$ is its <orientation> transition sign. Indeed $H_c^n$ of the intersection is generated by its overlap discs; extension by zero sends each generator to the two disc generators with these signs. Passing through the compact-support <Mayer–Vietoris sequence> imposes exactly those relations. This is <top compactly supported cohomology from orientation signs>.
Since $M$ is connected, the overlap graph is connected: a disconnection would split the union into two disjoint nonempty open subsets. A spanning tree expresses every generator as $u$ or $-u$, so $H_c^n(M)$ is cyclic. Additional overlap relations can only impose $2u=0$. If $M$ is oriented, use its compatible local <orientations>; every relation becomes $u_i=u_j$, and the presentation is exactly one free generator. Thus
$$
\boxed{H_c^n(M)\text{ is cyclic};\qquad H_c^n(M)\cong\mathbb Z\text{ if }M\text{ is oriented}.}
$$
In fact a nonorientable connected manifold has an orientation-reversing loop in this transition graph and gives $\mathbb Z/2$. In dimension zero the connected manifold is a point and the assertion is immediate.
For a closed oriented n-manifold, <Poincare duality> says that its <fundamental class> $[M]\in H_n(M;\mathbb Z)$ gives isomorphisms by <cap product>
$$
H^q(M;\mathbb Z)\xrightarrow{\ \cap[M]\ }H_{n-q}(M;\mathbb Z)\qquad\text{for all }q.
$$
For an oriented manifold that is not necessarily compact, the corresponding ordinary-homology statement is $H_c^q(M;\mathbb Z)\cong H_{n-q}(M;\mathbb Z)$, using the locally finite <fundamental class>. The version without compact supports on <cohomology> pairs with locally finite <homology> instead. Orientability and the absence of boundary are essential to these untwisted formulas; for a manifold with boundary one needs <Poincare-Lefschetz duality>, and for a nonorientable manifold one uses the <orientation local system>.
For the following cases, $H_0=\mathbb Z$ forces connectedness. The <universal coefficient theorem for cohomology> gives
$$
H^1(M;\mathbb Z)\cong\operatorname{Hom}(H_1(M;\mathbb Z),\mathbb Z),
$$
since the Ext term from $H_0=\mathbb Z$ vanishes. Combining with duality yields the <third homology of a closed oriented four-manifold> constraint $H_3\cong\operatorname{Hom}(H_1,\mathbb Z)$.
Back to article page
