Am I right that de Rham cohomology $H^k(S^2\setminus \{k~\text{points}\})$ of $2-$dimensional sphere without $k$ points are $$H^0 = \mathbb{R}$$ $$H^2 = \mathbb{R}^{N}$$ $$H^1 = \mathbb{R}^{N+k-1}?$$
I received this using Mayer–Vietoris sequence. And I want only to verify my result.
If you know some elementery methods to compute cohomology of this manifold, I am grateful to you.
Calculation:
Let's $M = S^2$, $U_1$ - set consists of $k$ $2-$dimensional disks without boundary and $U_2 = S^2\setminus \{k~\text{points}\}$. $$M = U_1 \cup U_2$$ each punctured point of $U_2$ covered by disk (which contain in $U_1$). And $$U_1\cap U_2$$ is a set consists of $k$ punctured disks (which homotopic to $S^1$). Than collection of dimensions in Mayer–Vietoris sequence $$0\to H^0(M)\to\ldots\to H^2(U_1 \cap U_2)\to 0$$ is $$0~~~~~1~~~~~k+\alpha~~~~~k~~~~~0~~~~~\beta~~~~~k~~~~~1~~~~~\gamma~~~~~0~~~~~0$$ whrer $\alpha, \beta, \gamma$ are dimensions of $0-$th, $1-$th and $2-$th cohomolody respectively. $$1 - (k+\alpha) + k = 0,$$ so $$\alpha = 1.$$ $$\beta - k + 1 - \gamma = 0,$$ so $$\beta = \gamma + (k-1).$$ So $$H^0 = \mathbb{R}$$ $$H^2 = \mathbb{R}^{N}$$ $$H^1 = \mathbb{R}^{N+k-1}$$
Thanks a lot!