Inspired by this answer given by Noam, which (I think) implies that a set $A \in NP$ if and only if there is polynomial-time computable function $f$ from random string to an element in $A$ such that $A$= $\{ a: f(x)=a, x \in S \}$ where $S$ is the set of all witnesses $x$ for $NP$ set $A$. If P=NP then every coNP set is efficiently samplable in Noam's sense.

The following characterization of NP is taken from Theory of computational complexity;

"A binary relation $R \subseteq\Sigma^* \times \Sigma^*$ is called polynomial honest if there exists a polynomial function $p$ such that $<x,y> \in R$ only if $|x|\le p(|y|)$ and $|y|\le p(|x|)$. A function $f:\Sigma^* \to \Sigma^*$ is polynomially honest if the relation $\{ <x, f(x)>: x\in \Sigma^*\}$ is polynomial honest.

Therefore, $A \subseteq \Sigma^*$ is in $NP$ if and only if $A=Range(f)$ for some polynomial honest and polynomial time computable function $f$."

Now, assume that there exists a universal witness set $W \in P$ such that for every $NP$-complete set $D_i \subseteq \Sigma^*$ there is some polynomially honest and polynomial-time computable function $g_i:W\to \Sigma^*$ such that $D_i=Range(g_i)$.

Does the existence of a universal witness set for $NP$ implies the impossibility of efficient sampling of coNP sets?