Let X be a complex normal variety and U a subvariety that is open in the analytic topology. Then the map $\pi_1(U) \to \pi_1(X)$ coming from the map $U \subset V$ is surjective - why is this?
edited to include complex
This isn't strictly what you're looking for, but I don't have the rep to leave this as a comment. In the algebraic situation, this follows from Grothendieck's general yoga, which says that the map of (etale) fundamental groups induced by $U\to X$ is surjective precisely when every connected (etale) cover of $X$ is still connected when pulled back to $U$. When $X$ is normal (or more generally, geometrically unibranch, if I'm not mistaken), then one checks easily that every connected etale cover of $X$ is still connected over the generic point of $X$ (I'm assuming $X$ itself is connected).
I'm looking forward to Donu Arapura's geometric answer!
I would like to risk an answer that does not use the language of algebraic geometry. For a pair (complex analytic variety $X$; closed analytic subvariety $Y$), $U=X\setminus Y$,
In order to show that the fundamental group of $X\setminus Y$ surjects onto the fundamental group of $X$, it is sufficient to show that every loop in $X$ can be homotoped into $X\setminus Y$. Since $Y$ it is contained in the simplicial subcomplex of codimension $2$ it is enough to show that any loop in $X$ can be homotoped so it does not touch any simplex of codim $2$, but this is true for every $PL$ space that is a manifold in codim $2$.