Let $T$ be the torus defined as the square $0\leq \theta_1, \theta_2\leq 2\pi$ with opposite sides identified. Let $F(\theta_1,\theta_2)=\cos\theta_1+\cos\theta_2$. Sketch the phase portrait for the system $-\text{grad}\, F$ in $T$. Sketch a three-dimensional representation of this phase portrait with $T$ represented as the surface of a doughnut.
I understand that the $-\text{grad}(F) = (\sin(\theta_1), \sin(\theta_2))$, but I'm very confused on how the torus is defined here. Why is a doughnut defined as a square?
I'm guessing in essence that we can find $\theta'_1$ and $\theta'_2$ and apply the domain to it, but the confusion on $T$ leaves me stuck?