Integral of form pulled back from torus to sphere is zero I'm trying to show that when we pull back (with any map $f: S^2 \to T^2$  any 2-form on the 2-torus to the 2-sphere it's integral is zero.
I understand we can choose coordinates $(\theta_1,\theta_2)$ for $T^2$ and then any 2-form can be written as a multiple of $\omega = d\theta_1 \wedge d\theta_2$. I'm trying to show:
$$\int_{S^2} f^*(d\theta_1 \wedge d\theta_2) = 0$$
Now I'm trying to use the fact that every closed 1-form on $S^2$ is exact, however that seems to be obvious here anyway as $f^*(d\theta_1) = d(\theta_1 f)$ so I'm now trying to show that the integral of $d(\theta_1 f) \wedge d(\theta_1 f)$ is zero but I am struggling to proceed.

Denote by $\pi \colon \mathbb{R}^2 \rightarrow \mathbb{T}^2$ the universal cover map. Since $S^2$ is simply connected, the map $f \colon S^2 \rightarrow \mathbb{T}^2$ factors through $\pi$ and we can find a smooth $g \colon S^2 \rightarrow \mathbb{R}^2$ such that $f = \pi \circ g$. If $\omega$ is a two-form on $\mathbb{T}^2$, then we have
$$ \int_{S^2} f^{*}(\omega) = \int_{S^2} (\pi \circ g)^{*}(\omega) = \int_{S^2} g^{*}(\pi^{*}(\omega)).$$
The form $\pi^{*}(\omega)$ is a two-form on $\mathbb{R}^2$, hence exact and so $g^{*}(\pi^{*}(\omega))$ is also exact and the integral is zero by Stoke's theorem.
