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This question is motivated by Exercise 1.7 from Differential Forms in Algebraic Topology by Bott & Tu, book I'm working over on my own. The original question in the text concerns the de Rham cohomology of $\mathbb{R}^2$ with points $P$ and $Q$ deleted. I have tried to simplify it a bit caring only about one point. So I'm trying to:

Compute in a rigorous way de Rham cohomology of $\mathbb{R}^2$ with one point $P$ deleted and find the closed forms that represent the cohomology classes.

There are two related questions:

first

second

I have already solved the exercise in several ways:

  • Using singular cohomology and the isomorphism between singular and the de Rham cohomologies.
  • Using Stokes and the ideas of Example 24.4 of Loring's book Introduction to Smooth Manifolds.

However, I want to solve the exercise rigorously using only what is previously covered in the book: the definition of the de Rham cohomology.

Since I have already solve it by other means, I already know the solution, so I am only interested in the ideas and the heuristics of another approach which uses only what I stated above.

Any help would be appreciated.

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    Maybe you should add more detail about the definitions given prior to the exercise in the book. People may more easily understand the situation. Anyway I am throwing in some glimpses of ideas to start with. What you will need to do is to prove that $xdy-ydx$ is (probably the only?) 1-form that is exact, but not closed. To do that assume it was d(.), then you will need to integrate it over a circle around the origin to get a contradiction with the stokes theorem. – 2017-02-23
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    You are right @Behnam. I was being stupid thinking that I couldn't use Stokes but of course I can since I am not assuming Stokes on Manifolds, only in the Euclidean space. I will give it other attempt! And regarding the definitions....they were only defined the concepts of differential forms, differential complexes and de Rham cohomology (also with compact supports). – 2017-02-23
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    I think this covers what you need (1 point only removed): https://math.stackexchange.com/questions/612837/how-to-compute-the-de-rham-cohomology-of-the-punctured-plane-just-by-calculus – 2017-07-25
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    @aytio thanks for the comment! I will have a look at it. – 2017-07-25

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Fact Homotopy equivalent spaces have isomorphic De Rham Cohomology.

As $\mathbb{R}^2\backslash${point} is homotopy equivalent to $S^1$ (define retraction as in image below), enter image description here

we have that

$H_{Dr}^n(\mathbb{R}^2\backslash\text{\{point\}})=H_{Dr}^n(S^1)=\begin{cases} \mathbb{R}, n=0,1\\ 0, \text{else} \end{cases}$