For a fiber bundle $M\longrightarrow N$ where $\dim N=n$, a non-degenerate 1-form $\theta$ on $M$ generates the differential ideal $\mathcal{I}$, and the Lagrangian $\mathcal{L}$ is an $n$-form on $M$. All these data form a non-degenerate variational problem.
In Griffiths' book Exterior Differential Systems and the Calculus of Variations page 84, he writes that
For non-degenerate variational problems the rank of $\theta$ is everywhere the maximum possible value $n$.
Then the question comes: Suppose there's another non-degenerate variational problem whose fiber bundle $W\longrightarrow N$ fibers on the same base $N$. Let $\varphi$ be the non-degenerate 1-form on $W$, then it's rank is restricted to be $n$.
If there is a smooth map $f:M\longrightarrow W$, we can induce a pullback $f^*:\varphi\longrightarrow\theta$. We know 1-form is a section of cotangent space (which is a vector space). A constant rank linear transformation between two vector space is invertible. Does the same rank $n$ of $\varphi$ and $\theta$ mean that the $f^*$ is invertible?
Or let $f^*\varphi$ be a 1-form on $M$, then is it non-degenerate？ Is the map $f^*|_{\varphi}$ an isomorphism?
