# Computing the de Rham cohomology of $S^2$

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.

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 $$H^1(S^2)=0.$$
• Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism! – BOlivianoperuano84 Dec 12 '18 at 20:46