I don't understand the notion of relativization. I expose with an example. Consider a class $A$ that contains $P$, e.g. $NP$. Why $P^A$ is not necessarily equal to $A$? I can naively think that if one can decide any problem in $A$, one can also decide any problem that uses polynomially many steps in a Turing machine plus calls to $A$. One can have at most polynomially many calls to $A$, thus $P^A$ seems there is only a polynomial overhead compare to $A$. If $A$ is $NP$, I could give a certification for all the (polinomially many) oracle calls, and also easily give a certification of the polynomial rest of the computation.I understand that if it is true, then $NP$ is the same as $coNP$, both being contained in $P^{NP}$ from the polynomial hierarchy. So, where is the error in my naive argument?
The meaning of relativization
complexity theory;oracle machines;relativization