Why is the matrix ($\frac{\partial \tilde x^j}{\partial x^i}$) invertible(Proposition 11.9 Lee's Introduction to smooth manifolds)?

Question

In the following proof

Why is the matrix ($\frac{\partial \tilde x^j}{\partial x^i}$) invertible? I don't see any reason for that

As a side thing there seem to be a typo in the equation (which is not in the official errata of the book), I think it should be

$\Phi \circ \tilde \Phi^{-1}(p,( \xi_1,...,\xi_n))=...$

instead of

$\Phi \circ \tilde \Phi^{-1}(p,(\tilde \xi_1,...,\tilde\xi_n))=...$

because we are using the following equation:

Answer 1

The coordinates $x = (x_1, \ldots, x_n)$ are a diffeomorphic function of the coordinates $\tilde{x} = \left( \tilde{x}_1, \ldots, {\tilde{x}_n} \right)$. This is an abuse of notation, using $x$ to denote both points in $U$ and to denote the transition diffemorphism $x : \tilde{U} \rightarrow U$, $x (\tilde{x}) = x$. The inverse of this transition diffeomorphism is denoted $\tilde{x}$. Differentiating the relation $\tilde{x}^i (x) = \tilde{x}^i$ with respect to $\tilde{x}^j$ yields \begin{eqnarray*} \frac{\partial \tilde{x}^i}{\partial x^k} \frac{\partial x^k}{\partial \tilde{x}^j} & = & \delta_{i j} . \end{eqnarray*} Hence $\left( \frac{\partial \tilde{x}^i}{\partial x^j} \right) = \left( \frac{\partial x^i}{\partial \tilde{x}^j} \right)^{- 1}$. Hence $\left( \frac{\partial \tilde{x}^i}{\partial x^j} \right)^T = \left( \frac{\partial \tilde{x}^j}{\partial x^i} \right)$ is invertible.