Here’s my question:
Let M be a connected, compact, orientable, smooth n-manifold ($ n \in \mathbb{N}_{\geq 2} $). Let V be a neighborhood of p diffeomorphic to $\mathbb{R}^n$ and let U = M \ {p}. Show that: $$ \delta: H_{dR}^{n-1}(U \cap V) \rightarrow H_{dR}^{n}(M) $$ Is an isomorphism.
My thoughts:
Surjectivity:
I know that the integration map $Int:H_{dR}^{n}(M) \rightarrow \mathbb{R}$ is an isomorphism.
Let $\omega$ denote the oriented (n-1)-form of the sphere on $U \cap V \cong \mathbb{R}^n \backslash \{0\}$.
$\delta[\omega] \neq 0$ since $\omega$ is not exact on $\mathbb{R}^n \backslash \{0\}$. (I've proven $\omega$ is not exact on $\mathbb{R}^n \backslash \{0\}$ separately)
Since $Int$ is an isomorphism, $H_{dR}^{n}(M)$ is one-dimensional and therefore spanned by [$\delta(\omega)$].
I don't feel very comfortable with these ideas, so any help would be appreciated.
A direct solution is okay, if you feel like it would still be insightful.
This website is somewhat new to me (second post), so I would also greatly appreciate any help with improving my question.
The Mayer-Vietoris sequence goes $\cdots \rightarrow H^{n - 1}(U) \oplus H^{n - 1}(V) \rightarrow H^{n-1}(U \cap V) \rightarrow H^{n}(M) \rightarrow H^{n}(U) \oplus H^{n}(V) \rightarrow 0$. Note that since $V \cong \mathbb{R}^{n}$ we have $H^{n}(V) = 0$. We also already know that $H^{n-1}(U \cap V)$ and $H^{n}(M)$ are both $\mathbb{R}$, so in particular they have the same dimension. To prove that the map between them (which is $\delta$) is an isomorphism, it suffices to show that it's surjective. By the exactness of the sequence, it suffices to show that $H^{n}(U) = 0$. By Poincaré duality we have $H^{n}(U) \cong (H_{c}^{0}(U))^{*}$, but since $U$ is not compact we have $H_{c}^{0}(U) = 0$. Hence $H^{n}(U) = 0$ and $\delta$ is an isomorphism.