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".



Is every complex rational algebraic variety simply connected for the Euclidean topology?
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. 


