# (Local) simple connectedness of irreducible algebraic varieties

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?

• Please define "locally simply connected". For which topology?
– abx
Commented Oct 13, 2021 at 13:11
• Zariski topology. Commented Oct 13, 2021 at 15:23
• 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.
– abx
Commented Oct 13, 2021 at 17:12
• Yes, if you mean that $Y$ has codimension $\geq 2$. Removing a closed subvariety of codimension at least 2 doesn't change $\pi_1$.
– abx
Commented Oct 14, 2021 at 3:55
• 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. Commented Oct 14, 2021 at 10:03