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!