What is the proof of "a second order anti-symmetric tensor remains anti-symmetric in any coordinate system"? We continuously use that a second order anti-symmetric tensor remains anti-symmetric in any coordinate system,  but could not find the proof anywhere.
 A: Such a tensor represents a skew bilinear form $B$ on $V$ or $V^*$. The property
$B(x,y)=-B(y,x)$, resp. $B(\phi,\psi)=-B(\psi,\phi)$, is expressed in the antisymmetry of the matrix of $B$ in any coordinate system. Therefore  the simple answer to your question is: This is obvious.
A: Let's say we have two local coordinates $x^\mu$ and $y^\mu$. Say we have an anti-symmetric contravariant tensor $t^{\mu\nu}$ in coordinates $x^\mu$. We are to show that in coordinates $y^\mu$, the same tensor $$t'^{\mu\nu}=\frac{\partial y^\mu}{\partial x^\rho}\frac{\partial y^\nu}{\partial x^\lambda}t^{\rho\lambda}$$ in $y^\mu$ coordinates is also anti-symmetric. This is easy: $$t'^{\mu\nu}=\frac{\partial y^\mu}{\partial x^\rho}\frac{\partial y^\nu}{\partial x^\lambda}t^{\rho\lambda}=\frac{\partial y^\mu}{\partial x^\rho}\frac{\partial y^\nu}{\partial x^\lambda}(-t^{\lambda\rho})=-\frac{\partial y^\nu}{\partial x^\lambda}\frac{\partial y^\mu}{\partial x^\rho}t^{\lambda\rho}=-t'^{\nu\mu}.$$
For covariant tensors, the proof is the same. Say we have an anti-symmetric covariant tensor $t_{\mu\nu}$ in coordinates $x^\mu$. The same tensor $$t'{}_{\mu\nu}=\frac{\partial x^\rho}{\partial y^\mu}\frac{\partial x^\lambda}{\partial y^\nu}t_{\rho\lambda}$$ in coordinates $y^\mu$ is also anti-symmetric: $$t'{}_{\mu\nu}=\frac{\partial x^\rho}{\partial y^\mu}\frac{\partial x^\lambda}{\partial y^\nu}t_{\rho\lambda}=\frac{\partial x^\rho}{\partial y^\mu}\frac{\partial x^\lambda}{\partial y^\nu}(-t_{\lambda\rho})=-\frac{\partial x^\lambda}{\partial y^\nu}\frac{\partial x^\rho}{\partial y^\mu}t_{\lambda\rho}=-t'_{\nu\mu}.$$
