TITLE: Is every complex rational algebraic variety simply connected for the Euclidean topology?
QUESTION [5 upvotes]: Is it true that every quasi-projective rational irreducible algebraic  complex variety is simply connected for the Euclidean topology?
Of course, this is false if we replace "complex" with "real" or if we forget "rational".

REPLY [12 votes]: I want to mention the positive direction.  Let $X$ be a smooth, projective variety over $\mathbb{C}$, resp. over an algebraically closed field of arbitrary characteristic.  Let $Z\subset X$ be a proper closed subset.  If $X$ is (separably) rationally connected and if $Z$ has codimension $\geq 2$ in $X$, then $X\setminus Z$ is simply connected for the Euclidean topology, resp. the algebraic fundamental group is trivial.  This was first proved by Campana over $\mathbb{C}$: there is an excellent explanation in Debarre's book, "Higher dimensional algebraic geometry".  There is another proof by Kollár that extends to positive characteristic: there is an excellent explanation in one of Debarre's Bourbaki seminars.  
One might hope to drop the hypothesis that $Z$ has codimension $2$ if we assume that every general pair of points of $X\setminus Z$ is connected by a rational curve completely contained in $X\setminus Z$.  Unfortunately there are many counterexamples, such as the smooth locus of the singular cubic surface with equation $xyz-w^3=0$.  What is true is that the fundamental group is finite.  However, it is quite open to understand the fundamental group of the smooth locus of $\mathbb{Q}$-log-Fano varieties, cf. work of Chenyang Xu, Zhiyu Tian, etc.