[Math] Normal Varieties

Question

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

Answer 1

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!