I want to compute the De Rham Cohomology of $\mathbb{R}^2\backslash\{k\text{ points}\}$ using Mayer Vietoris Argument.
My challenge is how to pick a good cover.
My Attempt for $k=2$
Let $$Z=\mathbb{R}^2\backslash\{\text{$k$ points}\},X=\{(x,y)\in\mathbb{R}^2\mid x>0\}\backslash\{(2,1)\} \text{ and }$$ $$Y=\{(x,y)\in\mathbb{R}^2\mid x<1\}\backslash\{(-1,1)\}.$$ Then $X\cup Y=Z$ (my choice of open cover). Observe that $X\cap Y$ is homotopy equivalent to $\mathbb{R}$ and $X,Y$ is homotopy equivalent to $S^1$.
The sequence of inclusion $X\cap Y \xrightarrow{} X\dot\cup Y \xrightarrow{} Z$ induce the long exact Mayer Vietoris Sequence
$$0\rightarrow H^0(Z)\rightarrow H^0(X)\bigoplus H^0(Y)\rightarrow H^0(X\cap Y)\rightarrow H^1(Z)\rightarrow H^1(X)\bigoplus H^1(Y)\rightarrow H^1(X\cap Y)\rightarrow H^2(Z)\rightarrow 0 $$
which is the same as
$$0\rightarrow \mathbb{R}\rightarrow \mathbb{R}^2\rightarrow\mathbb{R}\rightarrow H^1(Z)\rightarrow \mathbb{R}^2\rightarrow 0\rightarrow H^2(Z)\rightarrow 0.$$
By dimension argument $\dim H^1(Z)=2$ which implies $H^1(Z)=\mathbb{R}^2$ and $H^2(Z)=0$.
I would like to know if this is correct. How do I choose cover when $k>2$ points are removed. Thanks
Best Answer
Your argument is right for the case $k=2$. For $k>2$ the argument works similarly by induction.
If you're allowed the choose where your punctures are, you can suppose $k-1$ of the punctures are to the left of the $y$-axis, that the last puncture is to the right, and that there is an open strip around the $y$-axis containing none of the punctures. Something like $$X = \{(x, y)\in \mathbb{R}^2\ |\ x<1\}\setminus \{p_1,\dots,p_{k-1}\}$$ and $$Y = \{(x, y)\in \mathbb{R}^2\ |\ x> -1\}\setminus \{p_k\}$$ where the $x$-coordinates of $p_1,\dots,p_{k-1}$ are $<-1$ and the $x$-coordinate of $p_k$ is $>1$. Then $X\cup Y = Z$ and their intersection is contractible.
Now by induction $H^1(X)\cong \mathbb{R}^{k-1}$ and $H^2(X) = 0$. The Meyer-Vietoris sequence is then
$$0\to \mathbb{R} \to \mathbb{R}\oplus\mathbb{R} \to\mathbb{R} \to H^1(Z)\to \mathbb{R}^{k-1}\oplus\mathbb{R} \to 0 \to H^2(Z) \to 0$$