Kronecker Delta as a product of partial derivatives So, I'm currently teaching myself the basics of tensors, and one of the definitions I continually run into for Contravariant and Covariant tensors is that they transform according to $\bar A^i = \frac{\partial \bar x^i}{\partial x^j} A^j$ and $\bar A^i = \frac{\partial x^j}{\partial \bar x^i} A^j$ respectively, where the bars indicate the new coordinate system. Specifically, one of the texts I'm using is A.J. McConnell's Applications of Tensor Analysis, in which he describes that an arbitrary tensor, $a_{np}^m$, has the transformation $\bar a_{st}^r = c_m^r \gamma_s^n \gamma_t^p a_{np}^m$, where he has only defined $c$ as a linear transformation, and $\gamma$ as the cofactor of $c$ divided by the determinant, $|c|$. He then goes on to say that upon contraction of $r$ and $t$, that $c_m^r \gamma_r^p = \delta_m^p$.
Firstly, I assume that $c_m^r$ and $\gamma_r^p$ can, in at least some instances (pls correct me if I'm wrong), be viewed as $c_m^r = \frac{\partial \bar x^r}{\partial x^m}$, and $\gamma_r^p = \frac{\partial x^p}{\partial \bar x^r}$, based on my previously mentioned definitions.
My question, given that my assumption is correct, is why does $\frac{\partial \bar x^r}{\partial x^m} \frac{\partial x^p}{\partial \bar x^r} = \delta_m^p$ ? Or, at least, why does the case $m \neq p$ lead to $\frac{\partial \bar x^r}{\partial x^m} \frac{\partial x^p}{\partial \bar x^r} = 0$. The other case makes sense to me.
Moreover, in what circumstances does this apply? For all coordinate transformations? Or just a subset (i.e. orthonormal, orthogonal, non-curvilinear)? At the very least, non-linear coordinate transformations wouldn't necessarily have this condition, right? 
Any help would be greatly appreciated!
 A: The expression you are interested in obeys the chain rule. 
$$ \sum_{r}\frac{\partial \bar{x}^r }{\partial x^m } \frac{\partial x^p }{\partial \bar{x}^r } = \frac{\partial x^p }{\partial x^m} $$
Now consider $\frac{\partial x^p}{\partial x^m}$. This expression will equal $1$ if $p=m$ because $x^p$ and $x^m$ would refer to the same variable (the partial derivative of a variable with respect to itself is $1$). Then if $p\neq m$ the expression will equal zero because $x^p$ and $x^m$ would refer to different variables (the partial derivative of a variable with respect to another variable is $0$).
These observations mean that $\frac{\partial x^p}{\partial x^m} = \delta^p_m$. 
We conclude that, 
$$ \boxed{\sum_{r}\frac{\partial \bar{x}^r }{\partial x^m } \frac{\partial x^p }{\partial \bar{x}^r } = \delta^p_m }$$

$c$ and $\gamma$ are equal to the partial derivative expressions that you identified. Not that by definition $\gamma$ is the inverse of $c$ (the cofactor matrix divided by the determinant is the inverse matrix). 

These rules would apply to all smooth coordinate transformations from a manifold onto itself. If the coordinate transformation is very non-smooth to the point where it doesn't have a well defined derivative then you will definitely run into problems for sure.  
