Let $V$ be a smooth connected algebraic variety over an algebraically closed field $k$. Let $W_1, W_2$ be closed subvarieties of $V$ of positive codimension whose intersection $W_1 \cap W_2$ has codimension at least 2, and let $p$ be a point in $V \backslash (W_1 \cup W_2)$. Then we can form the four etale fundamental groups

$$ \pi_1( V, p ), \pi_1(V \backslash W_1, p), \pi_1(V \backslash W_2, p ), \pi_1(V \backslash (W_1 \cup W_2), p ).$$

There are canonical surjective homomorphisms from $\pi_1(V \backslash W_1, p)$ and $\pi_1(V \backslash W_2, p)$ to $\pi_1(V, p)$, and from $\pi_1(V \backslash (W_1 \cup W_2),p)$ to $\pi_1(V \backslash W_1, p)$ and $\pi_1(V \backslash W_2, p)$, forming a commuting square. (The surjectivity comes from the fact that a connected finite etale cover of a smooth variety remains connected even if one removes a positive codimension piece from the base.) So there is a canonical homomorphism from $\pi_1(V \backslash (W_1 \cup W_2),p)$ to the fibre product $\pi_1(V \backslash W_1, p) \times_{\pi_1(V,p)} \pi_1(V \backslash W_2,p)$.

My question is: is this latter homomorphism necessarily surjective also?

In the case when $k$ has characteristic zero, I believe I can deduce this from the topological van Kampen theorem, after first using the Riemann existence theorem to describe the etale fundamental group as the profinite completion of the topological fundamental group (after passing to a complex model). In the positive characteristic case, it seems to boil down (if I understand the etale fundamental group construction correctly) to verifying the following fact: if one has two finite etale covers of $V \backslash W_1$ and $V \backslash W_2$ respectively that become isomorphic on restriction to $V \backslash (W_1 \cup W_2)$, then they can be "glued" together to create a finite etale cover of $V$ (or of $V \backslash (W_1 \cap W_2)$). By reasoning in analogy with the topological case, this seems very reasonable to me, but I had trouble verifying it rigorously (I could glue together the covers as a prevariety, but then I couldn't establish separability to make the cover a variety again).

  • 1
    $\begingroup$ I think the general answer is in SGA1, and someone could probably find the positive answer there and give it as an official answer. Coincidentally, I was wondering about this a bit last week, and I found papers of J. Stix in my googling. For example, see Corollary 5.3 in: mathi.uni-heidelberg.de/~stix/preprints/STIXSvK.Juli2005.pdf $\endgroup$
    – Marty
    Oct 24, 2012 at 6:25
  • 7
    $\begingroup$ Any etale cover of an open subset of a normal variety $V$ extends canonically to a finite, not necessarily etale, cover of $V$ by considering the normalisation of $V$ in the function field of the cover. Applying this to the open set $V\backslash W_1 \cup W_2$, you get a finite cover of $V$; this is unramified over all of $V$ because it is so over $V \backslash W_1$ and also $V \backslash W_2$. $\endgroup$
    – naf
    Oct 24, 2012 at 6:30
  • 2
    $\begingroup$ By the Purity Theorem, Théorème X.3.4 of SGA 2, the étale fundamental group of $V\setminus (W_1\cup W_2)$ equals the étale fundamental group of $V$. $\endgroup$ Oct 24, 2012 at 10:53
  • 2
    $\begingroup$ @Jason Starr. I think you mean that the fundamental group of $V$ coincides with the fundamental group of $V \setminus (W_1 \cap W_2)$, which does not really answer the question. $\endgroup$ Oct 24, 2012 at 12:27
  • 4
    $\begingroup$ V. Zoonekynd in his "Theoreme de Van Kampen pour les champs algebriques" arxiv.org/abs/math/0111073 refers to SGA1 IV 5. It seems hidden there. Zoonekynd's translation in terms of groupoids might also be helpful. $\endgroup$
    – Niels
    Oct 25, 2012 at 7:47

1 Answer 1


To simplify notation, let me write $U_i$ for $V\smallsetminus W_i$, and $U_{12}$ for $U_1\cap U_2=V\smallsetminus(W_1\cup W_2)$.

Fact: The obvious functor $$(\mathrm{Sch}/V)\longrightarrow (\mathrm{Sch}/U_1) \times_{(\mathrm{Sch}/U_{12})} (\mathrm{Sch}/U_2)$$ is an equivalence. In other words, a $V$-scheme $X$ is the same thing as a $U_1$-scheme $X_1$, a $U_2$-scheme $X_2$, and a $U_{12}$-isomorphism of their restrictions to $U_{12}$.
This is probably somewhere in EGA1. [EDIT: all I could find was section 2.4 of EGA1, relying on (4.1.7) of Chapter 0 (glueing of ringed spaces).]
However, we are dealing here with finite étale schemes, which happen to be affine over the base, so this boils down to the analogous statement for categories of quasicoherent sheaves, which is essentially trivial (plus the fact that ``finite étale'' is a local condition).

If we describe the categories of finite étale covers in terms of $\pi_1$-sets, the above equivalence says that the diagram of groups $$(*)\qquad\begin{array}{rcl} \pi_1(U_{12},p)=:G_{12}& \longrightarrow &G_1:=\pi_1(U_{1},p)\cr \downarrow && \downarrow\cr \pi_1(U_{2},p)=:G_2& \longrightarrow &G:=\pi_1(V,p) \end{array}$$ is cocartesian. In other words, we get the usual van Kampen statement: the natural map $$\pi_1(U_{1},p)\ast_{\pi_1(U_{12},p)}\pi_1(U_{2},p)\longrightarrow \pi_1(V,p)$$ is an isomorphism. [EDIT: the coproduct is in the profinite category, which perhaps makes it hard to describe in general. See Will Savin's comment.]

What we want to prove is that the map "on the other side" $$G_{12}\longrightarrow G_1 \times_G G_2$$ is surjective, given that all the maps in diagram ($\ast$) are surjective.

Identifying $G_i$ ($i=1,2$) with $G_{12}/N_i$, we see from the universal property of the coproduct that $G=G_{12}/N_{1}N_{2}$. [EDIT: clearly this works also in the profinite category: since $N_1$, $N_2$ are both compact normal subgroups, so is $N_1 N_2$, hence $G_{12}/N_{1}N_{2}$ is profinite].

Take any $(x_1,x_2)\in G_1 \times_G G_2$: thus we have $x_i=g_i N_i$ for some $g_i\in G_{12}$, and the fiber product condition says that $g_1=g_2 n_2 n_1$ for some $n_i\in N_i$ (recall that $N_1 N_2=N_2 N_1$). So, $(x_1,x_2)$ is the image of $g_1 n_{1}^{-1}=g_2 n_2\in G_{12}$. QED

  • $\begingroup$ Thanks for this. I think the main thing I was missing was that finiteness was a local condition (and so I started working on separability instead). Time to hunt through EGA1... $\endgroup$
    – Terry Tao
    Oct 25, 2012 at 16:30
  • $\begingroup$ Oh, and thanks also for clarifying that the coproduct is in the profinite category. I was hesitant to pose the "full" version of the etale van Kampen theorem because the amalgamated free product of profinite groups didn't seem to be profinite in general. (It makes me wonder though what the profinite coproduct is, though. Is it just the profinite completion of the amalgamated free product?) $\endgroup$
    – Terry Tao
    Oct 25, 2012 at 16:35
  • 5
    $\begingroup$ Yes. The profinite coproduct is the profinite completion of the topological group coproduct, by the adjunction between the categories of topological groups and profinite groups. Since every pair of maps $A \to C$, $B \to C$ factors through the subgroup of $C$ generated by the image of $A$ and the image of $B$ with the subspace topology, the topological group coproduct is just the free product with the finest group topology such that the inclusions $A \to A * B$ and $B \to A *B$ are continuous. $\endgroup$
    – Will Sawin
    Oct 25, 2012 at 22:07
  • $\begingroup$ @Will: thanks for clarifying this. I was confused because I realized that the profinite completion of the algebraic coproduct wouldn't work. Ultimately, the "van Kampen-like" result is an elementary instance of SGA1, IX, Th. 5.1. The topological coproduct is described there as a "group of Galois type" with generators and relations, just before the statement of the theorem. $\endgroup$ Oct 26, 2012 at 8:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .