Given $\xi \in (T_p M)^{*}$, is it always possible to choose a real function $\phi$ such that $\mathrm{d}\phi_p = \xi$?

Question

Of course, here $M$ is a smooth manifold. Some authors seem to implicitly assume the answer to this question is yes when defining the principal symbol of a differential operator in a coordinate free manner, but I can't see why it's true. I tried to somehow use the fact that given any $v \in T_pM$ there exists a smooth curve $\alpha$ such that $\alpha(0) = p$ with $\alpha'(0) = v$ but I didn't get anywhere. I'd appreciate any help.

Answer 1

Let $(\varphi,U)$ be a chart containing $p$, and let $\xi=\sum_{i=1}^n\xi_i \text{d}\phi^i$ be its coordinate decomposition. The function $$f:U\to \mathbb{R}\\ f(x)=\xi_1\phi_1(x)+\dots+\xi_n\phi_n(x)$$ satisfies $\text{d}f_p=\xi$

If you know what a differential form is, a natural generalization of this question is whether, given a $n$-form $\omega$, there's an $n-1$-form $f$ such that $\text{d}f=\omega$. A necessary condition, as one easily finds, is $\text{d}\omega=0$. The study of "how much" this condition fails to be sufficient leads to the The Rham cohomology.