integral over rectangle to compute de Rham cohomology

Question

I am trying to reconstruct Seub's proof to calculate the first de Rham cohomology of the punctered plane : how to compute the de Rham cohomology of the punctured plane just by Calculus? In particular, when $M = \mathbb{R}^2 \setminus \{0\}$: "Fix permanently some point $m_0 \in M$, whichever you like best. For any $m\in M$, consider a rectangle $R$ in $\mathbb{R}^2$ whose boundary contains $m_0$ and $m$ but avoids the origin." As I understand it, 0 must be in the interior of the rectangle $R$. But I have a hard time to understand this, for example say that $m_0=(1,1)$ and $m=(2,2)$, then there is no such rectangle. Or am I misunderstanding something here?

Answer 1

You are misunderstanding. It is not necessary that $0$ is in the interior of the rectangle. The point is that we want to find an antiderivative of $\alpha$ by taking a path integral (this is basically a two-dimensional form of the fundamental theorem of calculus, which says that for nice functions on $\mathbb{R}$ you can find an antiderivative by integrating). However, in order to get a well-defined function by integrating and verify that its derivative really is $\alpha$, we need the integral to be independent of the path we choose. Now if you take two paths along a rectangle, those paths will automatically give the same integral if $0$ is not in the interior of the rectangle, so actually we would prefer if $0$ were not in the interior. But if $0$ is in the interior, the integrals will differ by $\lambda(\alpha)$, and so since we are assuming $\lambda(\alpha)=0$, we're actually still OK. So in this context we don't care whether $0$ is in the interior of our rectangles.