Q: Let $M$ be a smooth compact manifold, and suppose there is a smooth map $F:M \rightarrow S^{1}$ whose derivative is non-zero at every point. Prove that the de Rham cohomology space $H^{1}(M)$ is non-zero.
So here is my trouble with this question. I have the $1$-form $d\theta$ on $S^{1}$ and I would like to say that $F^{*}(d\theta)$ is a non-exact form. If $M$ was $1$-dimensional then I could conclude the map has a non-zero degree, hence the $\int_{M}F^{*}(d\theta)=2\pi k$ is non zero. However the problem does not say that $M$ is $1$-dimensional. However, by the way it says "derivative" instead of "Jacobian" I'm inclined to think $M$ is $1$-dimensional and the problem is just sloppy, but then $M$ would have to be either the circle or an interval.
So, my question is, is there a way to solve this replacing "derivative" with "Jacobian" in the problem for an arbitrary dimensional manifold $M$? Are there any general tools other than degree for dealing with pullbacks when little is known about one or either of the manifolds in question?
Many thanks!
Edit: So above instead of "Jacobian" I should have said $F$ is a smooth submersion, or something along those lines.
$\newcommand{\dd}{\mathrm{d}}$Step 1. Note that $\omega := F^{\ast}(\dd \theta)$ is closed.
Step 2. Note that $\omega$ is nonvanishing.
Explicitly, we have $$\omega(x)(X) = \dd \theta(F(x))(D_x F(X)).$$ Since $DF$ is nonzero at every point, it is onto. As $\dd \theta$ is nonvanishing, we get the result.
Step 3. Conclude that $\omega$ cannot be exact.
Indeed, if $\omega = \dd f$ for some $f \in C^{\infty}(M)$, then consider a point of global extremum (here is where we use compactness). At that point, we have $\dd f = 0$. (We are assuming that $M$ is boundaryless.)