Inverse of covariant tensor of rank two is contravariant. I'm studying tensors on my own, using "Tensor Calculus" from David C. Kay, and there is this theorem in page $29$:

Suppose that $(T_{ij})$ is a covariant tensor of order two. If the matrix $[T_{ij}]_{nn}$ is invertible on $\scr U$, with inverse matrix $[T^{ij}]_{nn},$ then $(T^{ij})$ is a contravariant tensor or order two.

Here, $\scr U$ is only the domain. He gives no proof at the moment (and I don't know if he will later), but I want to take a shot at this. Basically:
Have: $${\overline{T}}_{ij} = T_{rs}\frac{\partial x^r}{\partial {\overline{x}}^i}\frac{\partial x^s}{\partial {\overline{x}}^j}, \qquad T_{ik}T^{kj} = \delta_i^j, \quad T^{ik}T_{kj} = \delta^i_j.$$  
Want: $${\overline{T}}^{ij} = T^{rs}\frac{\partial {\overline{x}}^i}{\partial x^r}\frac{\partial {\overline{x}}^j}{\partial x^s}$$
Can I also say that ${\overline{T}}_{ik}{\overline{T}}^{kj} = \delta_i^j$? Why? 
My only ideas were: $${\overline{T}}_{ij} = T_{rs}\frac{\partial x^r}{\partial {\overline{x}}^i}\frac{\partial x^s}{\partial {\overline{x}}^j} \implies {\overline{T}}^{ki}{\overline{T}}_{ij} = {\overline{T}}^{ki}T_{rs}\frac{\partial x^r}{\partial {\overline{x}}^i}\frac{\partial x^s}{\partial {\overline{x}}^j} \implies {\overline{T}}^{ki}T_{rs}\frac{\partial x^r}{\partial {\overline{x}}^i}\frac{\partial x^s}{\partial {\overline{x}}^j}  = \delta^k_j,$$ but I don't know what to do with this. At least the indices are ok (on a side note, I never know if the correct is "indices" or "indexes", English is not my first language), so this is not entirely nonsense. 
And: $$T^{ik}T_{kj}={\overline{T}}^{ik}{\overline{T}}_{kj} \implies T^{ik}T_{kj}={\overline{T}}^{ik}T_{rs}\frac{\partial x^r}{\partial {\overline{x}}^k}\frac{\partial x^s}{\partial {\overline{x}}^j},$$ and maybe cancelling something here, somehow?
 A: You can indeed say that ${\overline T}_{ik}{\overline T}^{kj}=\delta_i^j$. This is because the definition of $T^{ij}$ is that in each basis its components are such that it is the inverse of $T_{jk}$.
Your line of working
$${\overline T}_{ij} = T_{rs}\frac{\partial x^r}{\partial {\overline x}^i}\frac{\partial x^s}{\partial {\overline x}^j} \implies {\overline T}^{ki}{\overline T}_{ij} = {\overline T}^{ki}T_{rs}\frac{\partial x^r}{\partial {\overline x}^i}\frac{\partial x^s}{\partial {\overline x}^j} \implies {\overline T}^{ki}T_{rs}\frac{\partial x^r}{\partial {\overline x}^i}\frac{\partial x^s}{\partial {\overline x}^j}  = \delta^k_j$$
is correct, and in fact you are very nearly done! You just need to use the fact that $\frac{\partial x^i}{\partial {\overline x}^j}$ has the inverse $\frac{\partial {\overline x}^j}{\partial x^k}$. We also already know the inverse of $T_{rs}$ is $T^{st}$, so starting with your equation
$${\overline T}^{ki}T_{rs}\frac{\partial x^r}{\partial {\overline x}^i}\frac{\partial x^s}{\partial {\overline x}^j}  = \delta^k_j$$
try to move everything apart from ${\overline T}^{ki}$ to the right hand side. You should end up with what you want.
