# Normal Varieties

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

• If someone else doesn't give an answer, I'll try to write one later. In outline: reduce it to a local statement (normality enters because the links are connected) then use van Kampen. – Donu Arapura Aug 29 '10 at 15:39
• I guess you are working over $\mathbb{C}$? E.g., over $\mathbb{R}$ the inclusion $\mathbb{A}^1 \subset \mathbb{P}^1$ gives a counterexample. – Pete L. Clark Aug 29 '10 at 15:47
• Is normality really needed? A subvariety that is open in the analytic topology is also open in the Zariski topology, and consequently has complement with "topological" codimension at least 2. But loops are 1-dimensional, so in $X$ they can be deformed to be entirely in $U$. That's a sketch of a proof, ignoring care needed to deal with singularities (in $X$ and $X-U$), so does a problem really arise when the singularities are worse than normal? – BCnrd Aug 29 '10 at 15:49
• I'm assuming this is over $\mathbb{C}$, but it should be stated. Also a more descriptive title would help. – Donu Arapura Aug 29 '10 at 15:49
• BCnrd, if $X$ is a nodal rational curve, and $U$ is the smooth part, $\pi_1(U)=\mathbb{Z}$ maps to $0$, but $\pi_1(X)=\mathbb{Z}$. – Donu Arapura Aug 29 '10 at 15:54

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 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 $Y$), $U=X\setminus Y$,
there exists a triangulation such that $Y$ is a subcomplex (see, for example Triangulations of algebraic sets - Hironaka 1974, can be found with google books). In other words $X$ is a simplicial complex, and $Y$ is a subcomplex. Now, if $X$ is normal its singularities are in real codimension at least $4$. I.e. $X$ is a $PL$ manifold in codimension $4$.
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 $2$.