If A is polynomial time reducible to B and B is in NP, then A is in NP
I think both hold too.

Co-NP version:some Problem A reduce to B, 

if B is co-NP,$\overline{B}$ is in NP,and no-instance of B has some verify way and certificate bit. and A's no-instance can use it too.

So this A is complement of NP problem $\overline{A}$. this means complexity class of A is co-NP.