I'm studying for general relativity and as such I'm learning tensor algebra. I'm following Schutz's book where denotes covectors with a tilde such as $\tilde{w}$ and vectors with the familiar arrow $\vec{e}$.
He defines the gradient of $\phi$ as the one-form $\tilde{d}\phi$ such that is has elements $\tilde{d}\phi = (\partial_i\phi) = \left( \dfrac{\partial\phi}{\partial x^{0}}, \dfrac{\partial\phi}{\partial x^{1}}, \dfrac{\partial\phi}{\partial x^{2}}, \dfrac{\partial\phi}{\partial x^{3}} \right)$. They of course transform as covectors such that $$(\tilde{d}\phi)_{\bar{a}} = {\Lambda^{b}}_\bar{a} (\tilde{d}\phi)_b$$ then it can be deduced that ${\Lambda^{b}}_\bar{a} = \dfrac{\partial x^b}{\partial x^\bar{a}}$.
He then claims that you can use these gradients to form a basis for one-forms such that $\tilde{d}x^a = w^a$. I don't understand this derivation because he uses the following relationships: $$\dfrac{\partial x^b}{\partial x^a} = {\delta^a}_b \ \ \ \text{and}\ \ \ \tilde{w}^a(\vec{e}_b) = {\delta^a}_b$$
Why is the first relationship true? I understand that for Cartesian coordinates this is true but what about in the general case? Couldn't I always define a new coordinate system $(x', y')$ where $x' = x'(y')$ and $y' = y'(x')$? Clearly then their various partial derivatives do not necessarily vanish. Does this mean the claim that $\tilde{d}x^a$ forming a covector basis only holds for Cartesian coordinates?
