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I already know that a coordinate system must have a non-zero wedge product of the components, but does it go the other way, that is, does $df_1\wedge\cdots\wedge df_n(p)\ne0$ mean that $f$ is a coordinate system?

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up vote 5 down vote accepted

If you work in a local coordinate chart $(U,(x_1,\ldots,x_n))$ then you have that $$df_1\wedge\ldots\wedge df_n=\det\left(\frac{\partial f}{\partial x}\right)dx_1\wedge\ldots\wedge dx_n$$ where $\frac{\partial f}{\partial x}$ is the Jacobian matrix for $f(x)=(f_1(x),\ldots, f_n(x))$.
So $df_1\wedge\ldots \wedge df_n(p)\neq 0$ if and only if the Jacobian $\frac{\partial f}{\partial x}(p)$ is invertible; and by the inverse functon theorem this last condition means that $x\to f(x)$ is a diffeomorphism around $p.$

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Yes, because with respect to any other choice of coordinate system $t_i$, this is the same as saying that the Jacobian determinant of the functions $f_i(t_i)$ from $\mathbb{R}^n$ to $\mathbb{R}^n$ doesn't vanish at $P$, which says that the $f_i$ are a coordinate system at $P$ (this is the implicit function theorem).

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