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I am trying to compute the de Rham cohomology of $S^2$ using Mayer Vietoris sequence. I considered the open cover $U$, where $U$ is the whole $S^2$ minus the north pole, and $V$, where $V$ is the whole $S^2$ minus the South Pole. Then, these two spaces are homotopic to a point. Even more, the intersection $U\cap V \cong S^1 \times [0,1] \cong S^1$. Thus, with this information we can start considering the following long exact sequence.

$$ 0 \xrightarrow{p} H^0(S^2) \xrightarrow{\alpha_0} H^0(U) \oplus H^0(V)\cong \mathbb{R} \oplus \mathbb{R} \xrightarrow{\beta_0} H^0(U\cap V)\cong \mathbb{R} \xrightarrow{d_0} H^1(S^2) \xrightarrow{\alpha_1} H^1(U) \oplus H^1(V) \cong 0 \xrightarrow{\beta_1} H^1(U \cap V) \cong \mathbb{R} \xrightarrow{d_1} H^2(S^2) \xrightarrow{\alpha_2} 0 \xrightarrow{\beta_2}0$$

I am struggling to compute $H^k(S^2)$. For instance, to compute $H^0(S^2)$, I was thinking of using the exactness of the sequence. We know that $0 = \text{im } p = \ker \alpha_0$ and so by first isomorphism theorem, we know that $H^0(S^2) = H^0(S^2)/\ker(\alpha_0) \cong \text{im } \alpha_0 = _{\text{(by exactness)}} \ker \beta_0 $. But then, I know how to figure out the kernel of $\beta_0$. I know the answer for the $0$-th de Rham cohomology is $H^0(S^2) \cong \mathbb{R}$, but I really don't know if this approach is correct and what I am missing to reach my desire conclusion.

Thanks for your help!

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Since $$ 0 \to H^1(U\cap V)\cong \mathbb{R} \xrightarrow{d_1}H^2(S)\to 0$$ is exact then $d_1$ is an isomorphism, so $H^2(S)\cong \mathbb{R}.$

Now note that if $M$ is any manifold, then $$ H^0(M)=\{f\in C^\infty(M) \text{ | } df=0\}=\{f\in C^{\infty}(M) \text{ | }f \text{ is locally constant}\},$$ so if $M$ has $n$ connected components, then $H^0(M)\cong \mathbb{R}^n.$ Since $S^2$ is connected, this shows that $H^0(S^2)=\mathbb{R}.$

Finally, look at the exact sequence $$ 0\to H^0(S^2)\cong \mathbb{R} \to H^0(U)\oplus H^0(V)\cong \mathbb{R}\oplus \mathbb{R}\to H^0(U\cap V) \cong \mathbb{R}\to H^1(S^2) \to 0$$ and use the following:

Proposition. If $0\to A_1\to A_2\to \cdots \to A_n \to 0$ is an exact sequence of vector spaces then $$ \sum_{i=1}^n(-1)^i\dim A_i=0.$$

Then you find that $1-2+1-\dim H^1(S^2)=0,$ so $H^1(S^2)=0.$

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  • $\begingroup$ Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism! $\endgroup$ – BOlivianoperuano84 Dec 12 '18 at 20:46

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