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).



An etale version of the van Kampen theorem
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


