Is the matrix $dF_p$ invertible when $F=(y^1, \dots, y^n)$ and $(dy^1\big|_p,\dots, dy^n\big|_p)$ is a basis for $T_p^*M$?

Question

Suppose $y^i:M\to \mathbb{R}$ is smooth (for $i=1,\dots, n$) and we define $F:M\to \mathbb{R}^n$ as $F(q)=(y^1(q), \dots, y^n(q))$. If we know that $(dy^1\big|_p,\dots, dy^n\big|_p)$ is a basis for $T_p^*M$, then can we say that the matrix $dF_p$ is invertible? Here I mean that $dF_p$ is just the Jacobian matrix $\frac{\partial y^i}{\partial x^j}$ for some local coordinates $x^j$. I am not sure how to use the fact that $(dy^1\big|_p,\dots, dy^n\big|_p)$ is a basis for $T_p^*M$ to show that the jacobian is invertible. Edit: $M$ is a smooth manifold.

Answer 1

This is my attempt.

As you say, for some local coordinates $x^i$, you can write the matrix $dF$ as $\frac{\partial y^i}{\partial x^j}$.

We also have the two bases $\{dx^i\},\{dy^i\}$ of $T^*_pM$, so we can write $dy^i = \sum_j a^i_j dx^j$, for a nonsingular matrix $(a^i_j)$ (you can show it maps a basis to a basis).

Write $$a^i_k = \sum_j a^i_j dx^j(\partial_{x^k}) = dy^i(\partial_{x^k}) = \partial_{x^k}y^i = \frac{\partial y^i}{\partial x^k}$$ where I used the fact $df(X) = Xf$.

Answer 2

$\operatorname dy^1|_p,\dots,\operatorname dy^n|_p$ are the columns of $\operatorname dF_p$. Since they are linearly independent and there are $n$ of them, where $n=\operatorname {dim}M$, $\operatorname dF_p$ has full rank, so is invertible.