The following fact is used in a proof of Preissman's theorem.
Let $f:\mathbb{T}^2\to N$ be a smooth map from the 2-torus to a manifold $N$. Suppose that $d_xf:T_x\mathbb{T}^2\to T_{f(x)}N$ has rank 1 everywhere. Then we would like to show that $f$ factorizes through a map $g:\mathbb{S}^1\to N$.
My attempt:
Now clearly, being of rank 1, $d_xf$ induces a never vanishing vector field $X$(after endowing the torus with a metric). My intuition suggest to show that $X$ is gradient of a map $h:\mathbb{T}^2\to \mathbb{S}^1$ (notice that this would imply that since $\nabla h \neq 0$, $h$ is a submersion and provides a fibration $S\to \mathbb{T}^2\to \mathbb{S}^1$. The fiber $S$ must be $1$-dimensional and closed, thus is $\mathbb{S}^1$). At this point we would have a fibration where $f$ is constant along the fiber of $h$ and we could pass $f$ to the quotient. The problem is then: how to show that $h$ exists? note that we would not be happy with a map to $\mathbb{R}$ (also it would be impossible to have that because of the homotopy type of the torus).
