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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

edited Aug 29 2010 at 15:58

asked Aug 29 2010 at 15:25

Robert Garbary
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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 2010 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 2010 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 2010 at 15:49

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I'm assuming this is over $\mathbb{C}$, but it should be stated. Also a more descriptive title would help. – Donu Arapura Aug 29 2010 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 2010 at 15:54

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Keerthi Madapusi Pera · Answer 1 · 2010-08-29 15:49:41Z UTC

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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!

answered Aug 29 2010 at 15:49

Keerthi Madapusi Pera
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In fact, Keerthi Madapusi's idea works for the topological fundamental group. The point is that the statement holds for analytic spaces; by considering the universal cover, which is again an analytic space, it reduces to the statement that removing an analytic subspace of codimension 2 from a normal analytic space does not disconnect it, which follows easily from the Serre's characterization of normal local rings, by using local cohomology. – Angelo Aug 29 2010 at 17:47

Given Keerti's answer + comments by Angelo & BCnrd, I think I'm off the hook :) – Donu Arapura Aug 29 2010 at 18:02

Dear Angelo: can we really bypass invoking the Riemann Extension Theorem for bounded analytic functions on normal analytic spaces? It's only the bounded analytic functions that extend, not all of them, so how can local cohomology or other purely algebraic methods detect this? I am very interested to hear the idea behind how such an argument might go. (Note: I've never really looked into gap sheaves, if that is somehow relevant.) If there is an "algebraic" proof of the Extension Theorem, I'd be happy to see that too. – BCnrd Aug 29 2010 at 21:15

Dmitri · Answer 2 · 2010-09-01 20:51:58Z UTC

I would like to risk an answer that does not use the language of algebraic geometry. For a pair (complex analytic variety $X$; closed anlalitic 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 sinuglarities 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 homothoped into $X\setminus Y$. Since $Y$ it is contained in the simplicial subcomplex of codimesnion $2$ it is enougth to show that any loop in $X$ can be homothoped so it does not touch any simplex of codim $2$, but this is true for every $PL$ space that is a manifold in codim $2$.

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