I am trying to show: if when $f^\prime = 0$, then $f^{\prime\prime} \neq 0 \Leftrightarrow (f \circ \phi^{-1})^\prime = 0$, $(f \circ \phi^{-1})^{\prime\prime} \neq 0$.
But the problem is, because $(f \circ \phi^{-1})^\prime = f^\prime(\phi(x))\phi^\prime(x)$, when $(f \circ \phi^{-1})^\prime = 0$, it could be $f^\prime = 0$ or $\phi^\prime = 0$. Can I simply regard $\phi$ as a change of coordinate function hence its Jacobian is nonsingular, so $\phi^\prime \neq 0$?
$(f \circ \phi^{-1})^{\prime\prime} = (f^\prime(\phi(x))\phi^\prime(x))^\prime = f^{\prime\prime}(\phi(x))\phi^\prime(x)\phi^\prime(x) + f^\prime(\phi(x))\phi^{\prime\prime}(x)$
Assuming $f^\prime(\phi(x))=0, \phi^\prime \neq 0$: $(f \circ \phi^{-1})^{\prime\prime} = f^{\prime\prime}(\phi(x))\phi^\prime(x)\phi^\prime(x) \neq 0$
Assuming $\phi^\prime = 0: (f \circ \phi^{-1})^{\prime\prime} = 0$ Failed.
Thank you very much!
This is just a calculus question. The given condition "coordinate chart" is strong. I found this in my calculus book: Advanced Calculus of Several Variables, Edwards
Definition: Let $\phi: U \rightarrow \mathbb{R}^n$ be a $\mathscr{C}^1$ mapping defined on an open subset $U$ of $\mathbb{R}^k (k \leq n)$. Then we call $\phi$ regular if the devrivatire matrix $\phi^\prime(u)$ has maximal rank $k$, for each $u \in U$.
Definition: Given $p \in M$, there exists an open set $U \subset \mathbb{R}^k (k<n)$ and a regular $\mathscr{C}^1$ mapping $\phi: U \rightarrow \mathbb{R}^n$ such that $p \in \phi(U)$, with $\phi(U^\prime)$ being an open subset of $M$ for each open set $U^\prime \subset U$. Then the mapping $\phi$ is called a coordinate patch for $M$ provided it is injective.
Certainly, a full rank matrix can't be an empty matrix.