An etale version of the van Kampen theorem

Question

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).

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