From my theory of computation lecture I recall:
If $A \le_m B$ and $B$ is decidable then $A$ is decidable (uses a computable function as a reduction).
If $A \le_p B$ and $B$ is in P then $A$ is in P (uses a polynomial function reduction).
I wanted to know if there is an analogous notion of reduction for the class NP.
Judging from your replies to nir shahar's comments above, it seems that you're really looking for a class of reduction functions that gives a necessary and sufficient condition, rather than merely a sufficient one as your question arguably implies.
If so, then the answer is rather boring: for any problem B in NP, a problem A is in NP if and only if it is nondeterministic-polynomial-time Turing-reducible to B, that is, if and only if there exists a Turing reduction from A to B that runs in polynomial time on a nondeterministic Turing machine, or (equivalently) that can be verified in polynomial time on a deterministic Turing machine.
I've specified Turing-reducible because a many-one reduction is not always available: if B is trivial (always returns "yes" or always returns "no"), and A is not, then there is no many-one reduction from A to B, regardless of whether A is in NP. However, if we require B to be non-trivial, then a nondeterministic-polynomial-time many-one reduction does have to exist. (Of course, there are other kinds of reductions besides Turing reductions and many-one reductions; but those are the most commonly discussed.)
Likewise, a polynomial-time reduction is not always available unless P = NP, because if B runs in polynomial time and A does not, then there won't be a polynomial-time reduction from A to B, regardless of whether A is in NP. However, if we require B to be NP-complete, then a polynomial-time reduction does have to exist, simply because that's the definition of NP-completeness.
