I am working towards an undertanding of de Rham cohomology. For this reason I am trying to find generator(s) of $H^n_{dR}(\mathbb R^2 -\{(0,0)\})$ and currently I am working on the case $n=1$. I believe I should use the Poincaré lemma for this. Let me give its statement here:
If $X$ is a contractible open subset of $R^n$, any smooth closed $p$-form $\alpha$ defined on $X$ is exact, for any integer $p > 0$.
What's not clear to me is how I can apply it even though $\mathbb R^2 -\{(0,0)\}$ is not contractible. Please could somebody help me with this?
I calculated the (co-)homology groups using cellular homology.
For a closed 1-form $\omega$, define $f: Z^1(M)\rightarrow R,\ f(\omega)=\oint_C \omega$ along the unit circle $C$ in $M=\mathbb{R}^2-\{0\}$. Exact forms are obviously in the kernel of the map since $\oint_C dg=g(1,0)-g(1,0)=0$. On the other side if $f(\omega)=0$ then the integral $g(p)=\int_{(1,0)}^p\omega$ is path independent thus it defines smooth function $g$ and $\omega=dg$ is exact.
So $H_{DR}^1(M)=Z^1(M)/d\Omega^0(M)=Z^1(M)/\ker(f)=\operatorname{img}(f)=\mathbb{R}$