For a smooth $n$-manifold $M$ and smooth, real-valued functions $y_1, \cdots , y_n$ defined on a neighborhood $U$ of some $p \in M$, such that $\{dy^1|_p, \cdots , dy^n|_p\}$ is a basis of $T^∗_pM$, show that $\phi = (y_1,\cdots, y_n): U \to \mathbb{R}^n$ is a chart map on a sub-neighborhood of $U$ containing $p$.
So I tried taking the differential of each $y^i$ and writing it as a linear combination of $\{dy^1|_p, \cdots , dy^n|_p\}$ with a change of basis. Then I fixed a dual basis to $\{dy^1|_p, \cdots , dy^n|_p\}$ (basis in the double dual, which is identified with the tangent space). I wanted to show that the kernel of each $dy^i_p$ is trivial, which would show that $d\phi_p$ is invertible. Then I use the implicit function theorem ( a hint I was given).
My issue is that, I don't know how to show that $dy^i_p$ have a trivial kernel. I'm afraid that I get lost in all the notation, and with the reason for why I need to perform a change of basis. If anyone could clarify this problem for me, that would be great.
If $\{dy^1|_p,\ldots, dy^n|_p\}$ is a basis for $T_p^*M$, then this means that the differential of $\phi = (y^1,\ldots, y^n)\colon U \to \Bbb R^n$ at $p$, i.e., the linear map $d\phi_p\colon T_pM \to \Bbb R^n$ is an isomorphism. (It is not true in general that the kernel of each $dy^i|_p$ is trivial, this is impossible by the rank-nullity theorem.)
This is because $d\phi_p = (dy^1|_p,\ldots, dy^n|_p)$. Namely, if we write $(e_1,\ldots, e_n)$ for the dual basis of $(dy^1|_p,\ldots,dy^n|_p)$, any $v \in T_pM$ can be written as $v = \sum_{i=1}^n dy^i|_p(v) e_i$, but $v \in \ker d\phi_p$ means that $dy^i|_p(v) = 0$ for all $i$, so that $v=0$. Then $\ker d\phi_p = \{0\}$ says that $d\phi_p$ is injective, but $\dim T_pM = \dim \Bbb R^n$, so $d\phi_p$ is an isomorphism, as claimed.
Now the inverse function theorem kicks in and we conclude that there is an open neighborhood $U'$ of $p$, contained in $U$, for which $\phi|_{U'} \colon U' \to \phi[U']\subseteq \Bbb R^n$ is a diffeomorphism onto the open set $\phi[U']$.