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?

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    $\begingroup$ Yes, but I I don't have time to write a long answer right now. If $U$ is a so called Artin neighbourhood (= "bon voisinage" in SGA4, exp XI), then the fundamental group is free by free by free etc. This should be residually finite. $\endgroup$ Jun 14, 2017 at 15:07
  • $\begingroup$ @DonuArapura Interesting! Many thanks. If you have the time, I would be very happy to see a bit of the details. $\endgroup$
    – Randy
    Jun 14, 2017 at 17:25

1 Answer 1


(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?

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    $\begingroup$ If a group has a subnormal series $1=G_0\le G_1\le\dots\le G_n=G$ such that each $G_i$ is normal in $G_{i+1}$ and $G_{i+1}/G_i$ is free and finitely generated, then $G$ is residually finite. (It's probably false if one only assumes that $G$ is finitely generated and that the $G_{i+1}/G_i$ are free.) It's certainly enough here. $\endgroup$
    – YCor
    Jun 14, 2017 at 21:55

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