De Rham cohomology of $S^2\setminus \{k~\text{points}\}$ 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!

It helps to use the fact that DeRahm cohomology is a homotopy invariant, meaning we can reduce the problem to a simpler space with the same homotopy type.  I think the method you are trying will work if you can straighten out the details, but if you're still having trouble then try this:  
$S^2$ with $1$ point removed is homeomorphic to the disk $D^2$.  If we let $S_k$ denote $S^2$ with $k$ points removed, then $S_k$ is homeomorphic to $D_{k-1}$ (where $D_{k-1}$ denotes $D^2$ with $k-1$ interior points removed).  
Hint: Find a nicer space which is homotopy equivalent to $D_{k-1}$.  Can you, for instance, make it $1$-dimensional?  If you could, that would immediately tell you something about $H^2(S_k)$.  If you get that far and know Mayer-Vietoris, you should be able to work out the calculation. 
