This is a follow up on my previous question. I would like to test whether I understand the first part of the answer given to me there by rewriting it in my own words.
Please could someone tell me if this is correct?
(Like before I'd like you to be meticulous/nitpicky.)
My goal was to find two non-equivalent closed non-exact smooth $1$-forms on $T=S^1 \times S^1$.
We know that the angle form $d\theta$ on $S^1$ is closed but not exact. If we let $U = S^1 - N$ and $V = S^1 - S$ where $N$ and $S$ are the north and southpole respectively then $(U\times U, (\theta_U, \varphi_U)$ and $(V \times V, (\theta_V,\varphi_V)) $ are two smooth charts for $T$ such that on $U\cap V$ we have $d \theta_U = d\theta_V$ and $d\varphi_U = d \varphi_V$.
We denote by $d\theta$ and $d\varphi$ the differential $1$-forms that agree with the corresponding $1$-forms $d \theta_U, d\theta_V$ and $d\varphi_U, d \varphi_V$ on $U$ and $V$ respectively.
We now claim that $d\theta, d\varphi$ are closed but not exact and that they do not differ by an exact $1$-form.
Since for this angular parametrisation one of the three Euclidean coordinates $x,y,z$ is constant both $d\theta$ and $d \varphi$ are an expression of the form ${x dy - y dx \over x^2 + y^2} $ so that it is clear that they are closed.
To see that they are not exact we use the curves $$ \gamma (t) = (t,0)$$ and $$ \gamma' (t) = (0,t)$$
We have
$$ \oint_\gamma d \theta = \int_0^{2\pi} (1,0)\cdot(1,0) dt = 2 \pi$$ and similarly for $d\varphi$. Hence by Stokes' theorem both forms are not exact.
Finally, assume that $d\theta - d\varphi$ was exact. Then by Stokes' theorem we'd have
$$ \oint_\gamma d\theta - d\varphi = 0$$
but we just saw that this is not the case hence $d\theta$ and $d \varphi$ do not differ by an exact $1$-form and hence are not equivalent.
That all looks pretty decent to me. As an alternative, you could just say that given a 1-form $\alpha$ on the first factor and a function $f$ on the second factor, you can define a 1-form on the product, $$ \beta_p(v) = f(p) \alpha(D\pi_1(p)[v]) $$ where $\pi_1$ is the projection on the first factor. A natural name for $\beta$ is $\alpha \times f$. So the two things you've got are $$ d\theta \times 1 \\ 1 \times d\phi $$ Your proof of non-exactness still applies. I just didn't have to define the 1-forms via charts, which is a little nicer.
(You still need to prove closed-ness, but it's probably not a bad idea to convince yourself that $d(\alpha \times \beta)$, where $\times$ is properly generalized, is something like $(d\alpha) \times \beta \pm \alpha \times (d\beta)$, and then the result follows.)