TITLE: Finding the de Rham cohomology of an open subset of $ \Bbb{R}^{n} $ minus a point.
QUESTION [7 upvotes]: Here’s my question:

Let $ n \in \mathbb{N}_{\geq 2} $. Suppose that $ U \subseteq \Bbb{R}^{n} $ is an open set and that $ x \in U $. Then show that
  $$
{H_{\text{dR}}^{n - 1}}(U \setminus \{ x \}) \neq 0.
$$

My thoughts:

I was trying to use that
  $$
                S
\hookrightarrow U \setminus \{ x \}
\hookrightarrow \Bbb{R}^{n} \setminus \{ x \}
$$
  (where $ S $ is a small sphere in $ U $ centered at $ x $) should induce a homotopy equivalence, so
  $$
      {H_{\text{dR}}^{n - 1}}(U \setminus \{ x \})
\cong {H_{\text{dR}}^{n - 1}}(S)
\cong \Bbb{R}.
$$

I get the feeling that there’s something horribly wrong with this line of thought, so I was wanting some help. I’m okay with a direct answer, as long as you think the solution alone would be illuminating.
P.S.: This is my first post. Any advice on how I could make my question better would be appreciated as well.

REPLY [2 votes]: You have the right idea, but you're line of thought is not entirely correct. The map
$$S^{n-1}
\hookrightarrow U \setminus \{ x \}
\hookrightarrow \Bbb{R}^{n} \setminus \{ x \}$$ is indeed a homotopy equivalence, so the composition
$$H^{n-1}_{dR}(S^{n-1})
\rightarrow H^{n-1}_{dR}(U \setminus \{ x \})
\rightarrow H^{n-1}_{dR}(\Bbb{R}^{n} \setminus \{ x \})$$ is an isomorphism. Since $H^{n-1}_{dR}(S^{n-1})\neq0$ we get $H^{n-1}_{dR}(\Bbb{R}^{n})\neq0.$ (Otherwise the composition would not be injective.)
However, your claim $H^{n-1}_{dR}(U\backslash\{x\})\cong \mathbb{R}$ is not correct. An open annulus $U\subseteq\mathbb{R}^2$ serves as a counterexample, since it holds $H^1_{dR}(U\backslash\{x\})\cong\mathbb{R}\oplus\mathbb{R}$.