I am trying to compute $H^{i}(\mathbb{R}^2,\mathbb{R}^2-S^1)$.
For that, I have used the long exact sequence of relative cohomology:
$0\to H^0(\mathbb{R}^2,\mathbb{R}^2-S^1)\to H^0(\mathbb{R}^2)\to H^0(\mathbb{R}^2-S^1)\to H^1(\mathbb{R}^2,\mathbb{R}^2-S^1)\to H^1(\mathbb{R}^2) \to H^1(\mathbb{R}^2-S^1)\to H^2(\mathbb{R}^2,\mathbb{R}^2-S^1)\to 0$.
Now, $\mathbb{R}^2-S^1$= $A\cup B$ where $B$ is contractible and $A$ is homotopy equivalent to $S^1$. Therefore, $$H^0(\mathbb{R}^2-S^1)=\mathbb{Z}^2, H^1(\mathbb{R}^2-S^1)=\mathbb{Z}, H^2(\mathbb{R}^2-S^1)=0.$$
So, $H^2(\mathbb{R}^2,\mathbb{R}^2-S^1)=\mathbb{Z}$ and I also have $$0\to H^0(\mathbb{R}^2,\mathbb{R}^2-S^1)\to \mathbb{Z} \to \mathbb{Z}^2\to H^1(\mathbb{R}^2,\mathbb{R}^2-S^1)\to 0.$$
Nevertheless, I do not know how to compute $H^0(\mathbb{R}^2,\mathbb{R}^2-S^1)$. I know that if $X$ is a space, $H^{0}(X)=\mathbb{Z}^n$ where $n$ is the number of connected components of $X$. But what does it mean in the case of relative cohomology? Thanks in advance!
$H^0(\mathbb{R}^2,\mathbb{R}^2-S^1)$ is, from the long exact sequence, the kernel of $H^0(\mathbb{R}^2)\to H^0(\mathbb{R}^2-S^1)$.
But this map sends the cocycle that sends every point to $1$ (which corresponds to $1\in \mathbb{Z}$) to the cocycle that does the same thing (and corresponds to $(1,1)\in\mathbb{Z}^2$), therefore it is injective.
Another way to see that it must be injective, is that it's a map $\mathbb{Z\to Z}^2$ and such a map is either injective or $0$ (can you prove that ? ). Now it can't be $0$ ( to see that, though, you must probably argue in a way similar to the one above, so I don't know if it's really that helpful)