How to determine whether a differential $1$-form is globally welldefined? This is a question that occurred after working on finding a generator of the first de Rham cohomology group of the torus. It was pointed out to me that the differential $1$-form
$$ dx + dy$$
was well defined on all of $T$ and that this is generally not the case.
Hence my question(s):

(1) Given a differential form, how do I find out whether it is
  globally well defined or only locally?
(2) In this concrete example how can I prove that $dx + dy$ is well
  defined on $T$?

Since the two questions seem so closely related I posted them in one question but if this is inappropriate please leave a comment and I will split them into two separate quetions. 
 A: Take T to be the quotient space of $\pi:\ \mathbb{R}^2\rightarrow \mathbb{R}^2 / \mathbb{Z}^2$, where $\mathbb{Z}^2$ acts on $\mathbb{R}^2$ by the translations $t_{mn}: (x,y)\mapsto(x+m,y+n)$ for arbitrary integers $m$ and $n$. Those actions obviously leaves $dx$ and $dy$ unchanged:  $$t_{mn}^*(dx)=d(x+m)=dx,\ t_{mn}^*(dy)=d(y+n)=dy$$
And if $\tilde X$ is the lift up of a vector field $X$ on T, that is, for any $q=\pi(p)$, $\pi_*(\tilde X_p)=X_q$, and $z$ is another point in the pre-image of $q$ (that is $\pi(z)=q$ too), and $t$ is an action maps $p$ to $z$, we have $\pi_* (t_*(\tilde X_p))=\pi_*(\tilde X_p)$.
Thus $dx$ and $dy$ are well-defined on the quotient space T, since, for any form $\tilde\omega$ preserved by deck transformations $t$ (that is $t^*\tilde\omega=\tilde\omega$), we can always push down it to a form $\omega$ on the quotient space like:
$$\omega_q(X_q)=\tilde\omega_p(\tilde X_p)$$ This $\omega$ is well defined for if we  lift up $X$ to a different point $z$ (where $\pi(z)=q$ and $t(p)=z$) we have:
$$\tilde\omega_z(\tilde X_z)=t^*(\tilde\omega_z)(\tilde X_p)=\tilde\omega_p(\tilde X_p)$$
