Let $S$ be a smooth connected variety over the complex numbers. The fundamental group might not be residually finite (i.e., the homomorphism $\pi\_1(S(\mathbb C)) \to \pi\_1^{\mathrm{et}}(S)$ might not be injective). 


Is there a dense Zariski open subset $U\subset S$ such that $\pi\_1(U(\mathbb C)) $ is residually finite?



Do complex varieties have a dense open subset with residually finite fundamental group?
(I'm converting my comment to an answer.)


In SGA4 exp XI, Artin constructs a nonempty Zariski open $U\subset S$ which admits a sequence $U=U\_n \to U\_{n-1}\to\ldots $ which are topological fibrations with curves as fibres. Without loss of generality, these fibrations, and the base of the fibrations, can be taken to be non proper. It follows that $\pi\_1(U)$ admits a finite filtration by normal subgroups such that the quotients are free and *finitely generated* (cf. YCor's comment). An easy induction shows that $\pi\_1(U)$ is residually finite. Note that the base case of the induction, where $\pi\_1(U)$ is free, is discussed here: [Why are free groups residually finite?](https://mathoverflow.net/questions/20471/why-are-free-groups-residually-finite) 


