DeRham Cohomology of $\mathbb{R} \setminus \{0\}$

Question

I've learned that the dimension of $H_{dR}^0(M)$ — the $0^{\text{th}}$De Rham Cohomology Group of a manifold $M$ — equals the number of connected components of $M$. But if $M = \mathbb{R} \setminus \{0\}$, then according to Wikipedia:

$$\dim H_{dR}^0(M) = 1.$$

But $\mathbb{R} \setminus \{0\}$ clearly has two connected components, so I'm confused as to how this can be possible. Any clarification would be much appreciated! Thanks.

Answer 1

You are correct. The statement in Wikipedia is not correct. They are trying to cover the de Rham cohomology groups $H^k_{\text{dR}}(\mathbb{R}^n\setminus\{0\})$ for all $k$ and $n$ simultaneously, but overlooked this exceptional case.