Intuition about calculation in Mayer-Vietoris sequence

Question

I'm interested in calculating de Rham cohomology groups of torus $\mathbb{T}^{2}=\mathbb{S}^{1}\times \mathbb{S}^{1}$, my approach is Mayer-Vietoris Sequence, but i don't know find open sets $U$ and $V$ Of $\mathbb{T}^{2}$, such that $\mathbb{T}^{2}=U\cup V$, and $U$ and $V$ has "easy" de Rham cohomology groups for calculate. I would like hints !! Thks

Answer 1

One option is to choose a point in the torus $X$ and an open ball $V$ around it. Consider a smaller closed $B\subseteq V$ and let $U=B^c$. Then $U\cap V$ is homotopic to a circle, $U$ is homotopic to $S^1\vee S^1$, and $\{U, V\}$ covers the torus. Moreover, $V$ is contractible so it has zero reduced de Rham cohomology.

The Mayer-Vietoris sequence is the following, since $U,V$ and $U\cap V$ have trivial cohomology in degrees $>1$, and are all connected, so there is no $\widetilde{H}^0$:

$$0\to H^1(X) \to H^1(U)=\mathbb Z^2\to \mathbb Z = H^1(U\cap V) \to H^2(X) \to 0$$

Now the claim is that the middle map is zero. Recall that map is built by choosing a partition of unity for $U$ and $V$. But you can choose $\psi_V$ in such a way that the support of the forms generating $H^1(U)$ is disjoint from the support of $\psi_V$, giving the result.

This means that the restriction map $H^1(X) \to H^1(U)=\mathbb Z^2$ is an isomorphism, and any two "angle forms" in $X$ that restrict to those in $U$ generate $H^1(X)$, that $H^2(X)$ is $\mathbb Z$, and that all other cohomology groups vanish.

Note that the same method using Van Kampen gives that $\pi_1(X) = \langle a,b \mid aba^{-1}b^{-1}\rangle =\mathbb Z^2$, so that $H_1(X) = \mathbb Z^2$, which also allows you conclude that the middle map has to be trivial.