# If all problems in NP are polynomially reducible to SAT (Cook & Levin) can one also assume that all to SAT polynomially reducible problems are in NP?

From Cook & Levin's theorem we know that all Problems in NP are polynomially reducible to SAT: $$\forall_{L\in NP}: L \leq_{p}SAT$$.

Is the opposite also true? That is if we know that a language L is reducible to SAT in polynomial time, can it be also stated that it must be in NP?

$$L \leq_{p}SAT \Rightarrow L \in NP$$

I am struggling to understand if this holds true.

Yes, this is true. More generally, if $$A \le_p B$$ and $$B \in \mathbf{NP}$$, then $$A \in \mathbf{NP}$$. (Most intro theory textbooks include a proof of this result.)
In fact, one way of defining what the class $$\mathbf{NP}$$ is is as “the class of all decision problems that polynomial-time reduce to $$SAT$$.”
So, given a string $$x$$, one way to check if $$x \in L$$ is to run the reduction to SAT (in polynomial time) on $$x$$ and check if the resulting formula is satisfiable (in non-deterministic polynomial time). This is clearly a non-deterministic poly-time algorithm to check membership in $$L$$; in other words, $$L$$ is in $$NP$$.