I'm really trying to understand differential forms so any critiques at all of the argument below would be greatly appreciated.
I want to show that $dx, dy\in \Omega^1(\mathbb{R}^2)$ induce well-defined forms $\omega_1,\omega_2 \in \Omega^1(T^2), T^2 = \mathbb{R}^2/\mathbb{Z}^2$. Then compute $d\omega_i$ and $\int_{C_j}\omega_i$, where $C_j$ is the image of the $j$-axis under the projection $\mathbb{R}^2\rightarrow\mathbb{R}^2/\mathbb{Z}^2$.
What I have:
Identify $T^2$ with $S^1\times S^1$ via some orientation preserving diffeomorphism. Let $\phi_{1}:(0,1)^2\rightarrow T^2$ be given by $(t,s)\mapsto(\cos2\pi t,\sin2\pi t,\cos2\pi s, \sin2\pi s)$ and $\phi_{2}:(-\frac{1}{2},\frac{1}{2})^2\rightarrow T^2$ defined the same. These are diffeomorphisms whose inverses are identical in overlap so we have an atlas. Then the pullback $(\phi^{-1}_i)^*(dx)$ is defined for $p\in T^2$ as $$(\phi^{-1}_i)^*(dx)(p) = dx(\phi_i^{-1}(p))\circ d(\phi_i^{-1})_p,$$ so since the $\phi_i^{-1}$ agree on the overlap, this form is well defined on all of $T^2$. Similarly we get a pulled back $(\phi_i^{-1})^*(dy)$. Then the exterior derivative of both is zero since $d(\phi_i^{-1})^*(dx) = (\phi_i^{-1})^*(d^2x) = 0$. Lastly, to take the integral, we can take restrictions of $\phi_2$ to parametrize the $C_j$ so that, for example, $$\int_{C_1}\omega_1 = \int_{(0,1)\times\{0\}}(\phi_2)^*(\phi_2^{-1})^*(dx) = 1. $$ and $$\int_{C_2}\omega_1 = \int_{\{0\}\times(0,1)}(\phi_2)^*(\phi_2^{-1})^*(dx) = 0. $$