(A related question is this 
[On the fundamental group of hypersurfaces](https://mathoverflow.net/questions/51301/on-the-fundamental-group-of-hypersurfaces)).


Let $X$ be a simply connected projective complex manifold of dimension at least 3. 
Let $Y\subset X$ be a smooth hypersurface. What is $\pi\_1(Y)$ then?
If $Y$ is a hyperplane section, then $\pi\_2(X,Y)=0$ by the Lefschetz
theorem, hence $\pi\_1(Y)=0$. The question is, can we say anything about the fundamental 
group of a hypersurface without assuming ampleness?



What is the fundamental group of a hypersurface?
I think the answer is no. Take any smooth projective surface $S$ in $\mathbb{P}^N$, and choose a generic projection to $\mathbb{P}^3$. The image is a singular surface $S'$ (with a double curve, some triple points etc.). There exists a birational morphism $P\rightarrow \mathbb{P^3}$ such that the strict transform of $S'$ in $P$ is isomorphic to $S$. Then $P$ is simply-connected, while $S$ is arbitrary.


