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Let $\mathbb k$ be an algebraically closed field of characteristic zero.

I have two questions:

(1) Is an irreducible algebraic variety $X/\mathbb k$ of dimension at least 2 locally simply connected?

(2) Let $X/\mathbb k$ be an irreducible algebraic variety and $Y$ a closed subvariety of dimension at least 2. If $X$ is (locally) simply connected, is $X\setminus Y$ (locally) simply connected?

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    $\begingroup$ Please define "locally simply connected". For which topology? $\endgroup$
    – abx
    Commented Oct 13, 2021 at 13:11
  • $\begingroup$ Zariski topology. $\endgroup$
    – Alberto Saracco
    Commented Oct 13, 2021 at 15:23
  • $\begingroup$ For any Zariski open subset $U\subset X$, the homomorphism $\pi _1(U)\rightarrow \pi _1(X)$ is surjective. So if $X$ is not simply connected, $U$ is not either. $\endgroup$
    – abx
    Commented Oct 13, 2021 at 17:12
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    $\begingroup$ Yes, if you mean that $Y$ has codimension $\geq 2$. Removing a closed subvariety of codimension at least 2 doesn't change $\pi_1$. $\endgroup$
    – abx
    Commented Oct 14, 2021 at 3:55
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    $\begingroup$ For question (1) the answer is no (I mean never), as soon as $\dim(X)>0$. For each point $x\in X$ you can find $f\in \mathscr{O}_{X,x}^\times$ which is not a square. Then (assuming $\mathrm{char}(\mathbb{k})\neq2$) $\mathrm{Spec}(\mathscr{O}_{X,x}[\sqrt{f}])$ extends to a nontrivial étale cover of some neighborhood $U$ of $x$, so $U$ is not simply connected. $\endgroup$
    – Laurent Moret-Bailly
    Commented Oct 14, 2021 at 10:03

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