What can you say about the $k$-th cohomology group of a closed orientable $n$-manifold for $k = n$ and $k = n-1$?

Question

What can you say about the $k$-th cohomology group of a closed orientable $n$-manifold for:

(1) $k = n$, and

(2) $k = n - 1$.

Poincaré Duality tells us that for $M$ a closed $R$-orientable $n$-manifold, $H^k(M;R)\longrightarrow H_{n-k}(M;R)$ is an isomorphism for all $k$.

$k=n$, by Poincaré Duality we have that

$$H^k(M)\cong H_{n-k}(M)\cong H_{n-n}(M)\cong H_0(M).$$

$k=n-1$, by Poincaré Duality we have that

$$H^k(M)\cong H_{n-k}(M)\cong H_{n-(n-1)}(M)\cong H_{1}(M).$$

Does my logic follow here? And is there anything else I can say in regards to the answer of this question? The question is sort of vague to begin with, I don't know if there is something I'm missing that I'm not saying.

Answer 1

Your logic does follow, but you should include the coefficients - when (co)homology groups are written without coefficients, it is (often) implicit that $\mathbb{Z}$ coefficients are meant.

You've shown

\begin{align*} H^n(M; R) &\cong H_0(M; R)\\ H^{n-1}(M; R) &\cong H_1(M; R). \end{align*}

However, more can be said, but that may be beyond the scope of the question. The $R$-module $H_0(M; R)$ is free, with generators the connected components of $M$; in particular, if $M$ is connected, $H^n(M; R) \cong H_0(M; R) \cong R$.