# De Rham cohomology of punctured manifold

I'm trying to solve the following problem (Lee's Intro to Smooth Manifolds, 17-6):

Let $M$ be a connected smooth manifold of dimension $n \geq 3$. For any $x \in M$ and $0 \leq p \leq n-2$, prove that the map $H^p_{dR}(M) \to H^p_{dR}(M \setminus \{x\})$ induced by inclusion $M\setminus\{x\} \hookrightarrow M$ is an isomorphism. Prove that the same is true for $p = n-1$ if $M$ is compact and orientable.

Let $U \approx \mathbb R^n$ be a coordinate chart for $x$, and let $V = M \setminus \{x\}$. Then $M = U \cup V$, and $U \cap V \simeq \mathbb S^{n-1}$, so $H_{dR}^p(U)\oplus H_{dR}^p(V) \cong H_{dR}^p(V)$, and $H_{dR}^p(U \cap V) \cong \mathbb R$ if $p = 0$ or $p=n-1$, and $H_{dR}^p(U \cap V) \cong 0$ otherwise. The Mayer-Vietoris sequence for $M$ is therefore $$\cdots \to H_{dR}^{p-1}(\mathbb S^{n-1}) \to H_{dR}^p(M) \xrightarrow{\ell^*} H_{dR}^p(M \setminus \{x\}) \to H_{dR}^p(\mathbb S^{n-1}) \to \cdots$$ where $\ell : M\setminus\{x\} \to M$ is inclusion. For $p \neq 0, 1, n-1$, this gives us the exact sequence $0 \to H_{dR}^p(M) \to H_{dR}^p(M \setminus \{x\}) \to 0$, from which it immediately follows that $\ell^*: H_{dR}^p(M) \to H_{dR}^p(M \setminus \{x\})$ is an isomorphism. For $p=0$, because $M$ is connected, so is $M \setminus \{x\},$ and so $H_{dR}^0(M) \cong H_{dR}^0(M \setminus \{x\}) \cong \mathbb R$, with basis the constant function $f \equiv 1$ in both cases, so clearly $\ell^*$ is an isomorphism.

For $p=1$ and $p=n-1$, I'm having trouble. The map $H_{dR}^{p-1}(\mathbb S^{n-1}) \to H_{dR}^p(M)$ is induced by the map $\delta : H_{dR}^{p-1}(U \cap V) \to H_{dR}^p(M)$ defined in the following way: for $[\omega] \in H_{dR}^{p-1}(U \cap V)$, there are $\eta \in \Omega^{p-1}(U)$ and $\eta' \in \Omega^{p-1}(V)$ so that $\omega = \eta|_{U \cap V} - \eta'|_{U \cap V}$, so we define $\delta [\omega] = [\sigma]$, where $\sigma = d\eta$ on $U$ and $\sigma = d\eta'$ on $V$.

My thought is to prove $\delta = 0$ for $p=1$ and $p=n-1$, but I don't know how. It looks to me as though $\delta$ defined in this way would always be $0$ on cohomology, since $[\sigma|_U] = 0$ and $[\sigma|_V]=0$ in both $H_{dR}^p(U)$ and $H_{dR}^p(V)$ respectively, but I'm sure I'm oversimplifying something. Is there a way to show why $\delta = 0$ for this Mayer-Vietoris sequence? Or is there a better way to approach this?

EDIT: See the comments for the solution to the $p=n-1$ case.

• This is probably related – Giuseppe Negro May 21 '18 at 22:48
• For $p=1$ observe that $H^0\mathbb{R}^n\rightarrow H^0S^{n-1}$ is onto, which suffices to show that $\delta:H^0S^{n-1}\rightarrow H^1M$ is trivial. Since $H^1\mathbb{R}^n=0$, we have $H^1M\cong H^1(M-x)$. For the case of $p=n-1$ you obviously need to use the fact that $M$ is orientable, and I'm not sure exactly how Lee has approached this. – Tyrone May 22 '18 at 10:19
• For $p=n-1$, since $H_{dR}^{n-2}(U \cap V)=0$, the map $H_{dR}^{n-1}(V)\cong H_{dR}^{n-1}(U)\oplus H_{dR}^{n-1}(V) \to H_{dR}^{n-1}(M)$ is injective. To show it is surjective, you need to show that $\delta : H_{dR}^{n-1}(U \cap V) \to H_{dR}^n(M)$ is injective. – Aweygan May 22 '18 at 14:40
• To do this, if we assume $M$ is boundaryless (which I think is legal here, since the exercise doesn't say $M$ is with or without boundary), it suffices to show that there is some form $\omega\in\delta[\sigma]$ (here $[\sigma]$ is a generator of $H_{dR}^{n-1}(U \cap V)$) such that $\int_M\omega\neq0$. – Aweygan May 22 '18 at 14:44
• Alternatively, one can show $\delta \neq 0$. For if it were, we'd have $H^n(M) \cong H^n(M \setminus \{x\})$, which is demonstrably false. – D Ford May 22 '18 at 18:10

For $p=1$, the Mayer-Vietoris sequence gives us the following exact sequence: $$\cdots \to H_{dR}^0(U) \oplus H_{dR}^0(M \setminus \{x\}) \to H_{dR}^0(\mathbb S^{n-1}) \to H_{dR}^1(M) \to H_{dR}^1(M \setminus \{x\}) \to H_{dR}^1(\mathbb S^{n-1}) \to \cdots$$ where $U \approx \mathbb R^n$ is a coordinate neighborhood of $x$. Since $H_{dR}^0(U) \cong H_{dR}^0(M \setminus \{x\}) \cong H_{dR}^0(\mathbb S^{n-1}) \cong 0$, this becomes: $$\cdots \to \mathbb R^2 \to \mathbb R \to H_{dR}^1(M) \to H_{dR}^1(M \setminus \{x\}) \to 0 \to \cdots$$ hence $H_{dR}^1(M) \to H_{dR}^1(M \setminus \{x\})$ is surjective. To prove injectivity, we need $H_{dR}^0(\mathbb S^{n-1}) \to H_{dR}^1(M)$ to be trivial. For this, it's enough to show $H_{dR}^0(U) \oplus H_{dR}^0(M \setminus \{x\}) \to H_{dR}^0(\mathbb S^{n-1})$ is surjective, which is obvious since all these $0$-degree cohomology groups are generated by constant functions. So $H_{dR}^0(\mathbb S^{n-1}) \to H_{dR}^1(M)$ is trivial, giving us that $H_{dR}^1(M) \to H_{dR}^1(M \setminus \{x\})$ is an isomorphism.
For $p=n-1$, if $M$ is connected, compact, and orientable, then $H_{dR}^n(M) \cong \mathbb R$ and $H_{dR}^n(M \setminus \{x\}) \cong 0$, since every orientable connected manifold has top cohomology $\mathbb R$ if and only if $M$ is compact, and $0$ otherwise. Hence the top of the Mayer-Vietoris sequence becomes: $$0 \to H_{dR}^{n-1}(M) \to H_{dR}^{n-1}(M \setminus \{x\}) \to H_{dR}^{n-1}(\mathbb S^{n-1}) \to H_{dR}^n(M) \to 0$$ Hence $H_{dR}^{n-1}(M) \to H_{dR}^{n-1}(M \setminus \{x\})$ is injective, and $H_{dR}^{n-1}(\mathbb S^{n-1}) \to H_{dR}^n(M)$ is surjective. Since $H_{dR}^{n-1}(\mathbb S^{n-1}) \cong H_{dR}^n(M)\cong \mathbb R$, we actually get $H_{dR}^{n-1}(\mathbb S^{n-1}) \to H_{dR}^n(M)$ is an isomorphism. Hence $H_{dR}^{n-1}(M \setminus \{x\}) \to H_{dR}^{n-1}(\mathbb S^{n-1})$ is trivial, so $H_{dR}^{n-1}(M) \to H_{dR}^{n-1}(M \setminus \{x\})$ is surjective, making $H_{dR}^{n-1}(M) \to H_{dR}^{n-1}(M \setminus \{x\})$ an isomorphism.