De Rham cohomology of compact manifold minus one point

Let $$M$$ be a compact, connected, oriented and without boundary $$n$$-manifold, and $$N$$ the manifold obtained by removing an arbitrary point $$p\in M$$ from $$M.$$ Show that the Betti numbers (i.e. the dimentions of the De Rham cohomology groups) are $$b^i(N)=b^i(M) \quad i=0,\dots,n-1;$$ $$b^n(N)=b^n(M)-1.$$

My idea was to use the Mayer-Vietoris theorem by considering the open sets $$U=N$$ and $$V,$$ where $$V$$ is an arbitrary neighbourhood that is also the domain of a chart $$\varphi:V\to\mathbb{R}^n.$$

The resulting Mayer-Vietoris sequence is: $$\begin{multline} 0\to H^0(M)\to H^0(N)\oplus\mathbb{R}\to\mathbb{R}\to H^1(M)\to H^1(N)\to 0\to \\ \to \cdots\to 0 \to H^{n-1}(M)\to H^{n-1}(N)\to\mathbb{R}\to H^n(M)\to H^n(N)\to 0;\end{multline}$$ due to the fact that the cohomology of $$V\sim\mathbb{R}^n$$ is $$H^0(\mathbb{R}^n)=\mathbb{R}$$ and $$H^i(\mathbb{R}^n)=0$$ for all $$i\geq 1,$$ and the cohomology of $$V\cap U$$ is the cohomology of $$\mathbb{S}^{n-1},$$ that is: $$H^0(\mathbb{S}^{n-1})=H^{n-1}(\mathbb{S}^{n-1})=\mathbb{R}$$ and $$H^i(\mathbb{S}^{n-1})=0$$ for all $$i\ne0,n-1.$$

The short sequences in the middle $$0 \to H^i(M)\to H^i(N)\to 0, \ i=2,\dots,n-2$$ let me say that $$b^i(N)=b^i(M)$$ for these $$i$$'s, but for the others I just have the sequences: $$0\to H^0(M)\to H^0(N)\oplus\mathbb{R}\to\mathbb{R}\to H^1(M)\to H^1(N)\to 0,$$ $$0 \to H^{n-1}(M)\to H^{n-1}(N)\to\mathbb{R}\to H^n(M)\to H^n(N)\to 0;$$ that allow me to write the relations: $$b^0(M)-b^0(N)-b^1(M)+b^1(N)=0,$$ and $$b^{n-1}(M)-b^{n-1}(N)+1-b^n(M)+b^n(N)=0,$$ which are coherent with the thesis that I want to obtain but are not sufficient to conclude, and I can't say anything more (I'm not an expert in using arrows as mathematical objects). Any help is really much appreciated. Thanks to everybody.

The trick is to look at the actual arrows involved. Also for convenience sake, I will assume $$N$$ is actually $$M$$ delete an open ball, since they're homotopy equivalent.

The $$0$$, $$1$$ exact sequence

Consider the exact sequence $$0 \to H^0(M)\to H^0(N)\oplus \Bbb{R}\to \Bbb{R}\to H^1(M)\to H^1(N)$$ The trick is to observe that $$H^0(M)=\Bbb{R}$$, since $$M$$ is connected. Thus we have $$0 \to \Bbb{R} \to H^0(N)\oplus \Bbb{R}\to \Bbb{R},$$ and since $$N\ne \varnothing$$, we see that the dimension of $$H^0(N) \oplus \Bbb{R}$$ is greater than or equal to 2, but it's also at most 2, since it fits into this exact sequence. Thus $$H^0(N)=\Bbb{R}$$ and the map $$H^0(N)\oplus \Bbb{R}\to \Bbb{R}$$ is surjective.

We could also show this geometrically ($$N$$ must be connected, so $$H^0(N)=\Bbb{R}$$ from that consideration alone, and surjectivity of the map is clear geometrically).

Regardless, surjectivity of this map allows us to split the exact sequence as $$0\to H^0(M)\to H^0(N)\oplus \Bbb{R} \to \Bbb{R} \to 0$$ and $$0\to H^1(M)\to H^1(N).$$ If $$n> 2$$, then this fits in as $$0\to H^1(M)\to H^1(N)\to 0$$, giving $$H^1(M)\simeq H^1(N)$$. Otherwise this is part of the other exact sequence we don't understand, so let's look at that one next.

The $$n$$, $$n-1$$ exact sequence

We have $$0 \to H^{n-1}(M) \to H^{n-1}(N) \to \Bbb{R} \to H^n(M)\to H^n(N) \to 0$$

Consider the map $$H^{n-1}(N)\to \Bbb{R}$$. This is induced by restricting a differential form $$\omega$$ representing some cohomology class $$[\omega]$$, defined on $$N$$ to the boundary sphere of $$N$$.

Then observe that $$\int_{\partial N} \omega = \int_N d\omega = \int_N 0 = 0,$$ by Stokes' theorem and that $$\omega$$ is closed.

Therefore the image of $$[\omega]$$ in $$\Bbb{R} = H^{n-1}(\partial N)$$ is $$0$$. Thus the map $$H^{n-1}(N)\to \Bbb{R}$$ is $$0$$.

Therefore we can split the exact sequence again into the pieces $$0\to H^{n-1}(M)\to H^{n-1}(N)\to 0$$ and $$0\to \Bbb{R}\to H^n(M)\to H^n(N)\to 0,$$ and using that $$H^n(M)=\Bbb{R}$$, since $$M$$ is connected, compact, and orientable, we see that $$H^n(N)=0$$, as desired.

Alternatively without using that $$H^n(M)=\Bbb{R}$$, we already have $$b_n(N) = b_n(M)-1$$.

• Thanks a lot! This is exactly what I was looking for! Just one last doubt: why does the fact that the integral is zero over $\partial N$ imply that the form is (equivalent to) zero? :) – Riccardo Ceccon May 26 at 6:58
• @RiccardoCeccon $\partial N=S^{n-1}$, so $H^{n-1}(\partial N)\cong\mathbb{R}$ and $\int_{\partial N}\colon H^{n-1}(\partial N)\to\mathbb{R}$ is an isomorphism. – user10354138 May 26 at 8:22
• Right, that's true! Thanks a lot! – Riccardo Ceccon May 26 at 9:37
• If you are going to integrate on $N$, then you must assume that $N$is orientable. Given that information you would have already known that the cohomology of $N$ is $\mathbb R$ which contradicts the final answer. But I guess you can argue by contradiction instead. – z.z Jun 13 at 22:11
• @z.z I'm not following you. We know $N$ is orientable, indeed oriented, because $M$ is. I've assumed that it's $M$ minus an open ball, so it's compact. Thus we can integrate on it. What's more, this doesn't imply that the top cohomology is $\Bbb{R}$, because it's a manifold with boundary. – jgon Jun 13 at 22:18

Let $$U=N$$, $$V$$ the domain of a chart which contains the point removed.

We obtain from Mayer Vietoris the following sequence is exact:

$$H^{i-1}(U\cap V)\rightarrow H^i(X)\rightarrow H^i(U)\oplus H^i(V)\rightarrow H^i(U\cap V)$$

Remark that $$U\cap V$$ is a deformation retract of $$S^{n-1}$$, thus $$H^i(U\cap V)=0$$ if $$i\neq 0, n-1$$ and $$H^i(U)=0$$ if $$i\neq 0$$ since $$U$$ is diffeomorphic to $$\mathbb{R}^n$$, this implies that if $$i we have

$$0=H^{i-1}(U\cap V)\rightarrow H^i(X)\rightarrow H^i(U)\oplus (H^i(V)=0)\rightarrow H^i(U\cap V)=0$$

implies that $$H^i(X)=H^i(N)=H^i(N)$$.

For $$i=n-1$$,

$$0=H^{n-2}(U\cap V)\rightarrow H^{n-1}(X)\rightarrow H^{n-1}(U)\oplus H^{n-1}(V)\rightarrow (H^{n-1}(U\cap V)=\mathbb{Z})$$

$$\rightarrow( H^n(X)=\mathbb{Z})\rightarrow H^n(U)\oplus (H^n(V)=0)\oplus H^n(U\cap V)=0$$

implies that $$H^{n-1}(X)=H^{n-1}(V)=H^{n-1}(N)$$.

Here I use the fact that the top cohomology group of a manifold without a point vanishes which is shown here:

Top homology of an oriented, compact, connected smooth manifold with boundary