Part of inverse Function Theorem tells us that for
$f$: M $\rightarrow$ N, $J(f^{-1})( \textbf{y}) = [(Jf)(f^{-1}( \textbf{y}))]^{-1}$.
That is the Jacobian of the inverse function at $ \textbf{y}\in N$ is equal to the inverse of Jacobian at $x = f( \textbf{y}) \in M$. Because the Jacobian is defined as $J_{ij} = \frac{\partial f_{i}}{\partial x_{j}}$ it could be thought of as a matrix representation of the push forwards. My question is:
- Is the inverse Jacobian same thing as the pull back? To take the inverse of Jacobian, we have to take the adjoint and divide by the determinant. On the other hand, the pull back is simply a transpose of the push forward in matrix form.
- is the Jacobian a (1,1) tensor or is it (0,2) or (2,0) tensor? We know that $J^{-1}J = I$ but if it is a (1,1) tensor how is it different from pushforward/pull back?
Could someone elaborate on the differences between the Jacobian, inverse Jacobian vs Push forward, pull back?