I have been trying to find a proof that the de Rham Cohomology group for a single point is equal to:
$H^{k}_{dR}=\mathbb{R}$ when $k=0$ and $H^{k}_{dR}=0$ when $k>0$.
I am sure this result is correct, but it seems to be quoted in many places almost religiously without proof; as though the results is obvious. If it is obvious, I cannot see it. Can anyone on here provide a proof of this result. I have searched extensively for one but cannot find one. It is quite fundamental in that a single point is a very simple example of a manifold. Thus I hope this question is useful in the community.
Since $M = \{ \text{pt} \}$ is a zero-dimensional manifold, there are no $k$ forms on $M$ for $k > 0$. The $0$-forms on $M$ are by definition smooth real valued functions on $M$ and, in our case they are identified naturally with the real numbers (each function is determined uniquely by where it sends the point to) so the de-Rham complex is
$$ 0 \rightarrow \underbrace{\mathbb{R}}_{\Omega^0(M)} \rightarrow \underbrace{0}_{\Omega^1(M)} \rightarrow \dots $$
Hence, $H_{dR}^k(M) = 0$ for $k > 0$ and $H_{dR}^0(M) = \mathbb{R}$.