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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.

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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 $b_n(N) = b_n(M)-1$.

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  • $\begingroup$ 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? :) $\endgroup$ – Riccardo Ceccon May 26 at 6:58
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    $\begingroup$ @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. $\endgroup$ – user10354138 May 26 at 8:22
  • $\begingroup$ Right, that's true! Thanks a lot! $\endgroup$ – Riccardo Ceccon May 26 at 9:37
  • $\begingroup$ 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. $\endgroup$ – z.z Jun 13 at 22:11
  • $\begingroup$ @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. $\endgroup$ – jgon Jun 13 at 22:18
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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 $i<n-2$ 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

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