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Let $X$ be a proper smooth scheme over a field $k$ of characteristic zero (well you can naturally weaken the assumption to normal integral scheme over some "nice" base like $\mathbb{Z}$, $\mathbb{F}_q$, etc.), and $x_1,\dots,x_n$ finitely many points in $X(k)$. Then one gets exact sequences $$1\rightarrow\pi_1(X\otimes_k\bar{k},\bar{x_i})\rightarrow\pi_1(X,\bar{x_i})\rightarrow Gal(\bar{k}/k) \rightarrow 1$$ where $\bar{x_i}$ is the geometric point $Spec(\bar{k})\rightarrow X$ lying over $x_i$. Note that $x_i\in X(k)$ splits the above sequences by maps $$\rho_i:Gal(\bar{k}/k)\simeq\pi_1(\bar{x_i},x_i)\rightarrow \pi_1(X,\bar{x_i})$$.

It is known that different $\pi_1(X,\bar{x_i})$ are isomorphic to each other, however the images of different $\rho_i$ might not be taken onto one another under these isomorphisms. Even one fixes one $k$-rational point, there might be more that one way to split the exact sequence of fundamental group.

If we assume $\pi_1(X\otimes_k\bar{k},\bar{x_i})$ to be topologically finitely generated (which is the case if the topological fundamental group of $X(\mathbb{C})_{an}$ is finitely generated, like the case of Riemann surfaces, etc.), then can one find finitely many sections $\rho_i$ as above such that when mapping them isomorphically onto a single $\pi_1(X,\bar{x_1})$, their images generate the whole $\pi_1$ topologically? or can one even reduce to the case of one single rational point with different splittings of the exact sequences?

Moreover, given a topologically finitely generated closed subgroup $H$ of $\pi_1(X,\bar{x})$, can one find a closed subscheme $Y\subset X$ containing $\bar{x}$ such that the image of $\pi_1(Y,\bar{x})\rightarrow\pi_1(X,\bar{x})$ is eactly the given group $H$? The example I have in mind comes from compact Shimura varieties and their special subvarieties, where a compact Shimura variety is more or less of the form $X/\Gamma$ where $X$ is an hermitian symmetric domain whose automorphism group is given as the real points of some rationally defined linear algebraic group (of $\mathbb{Q}$-rank zero in this example), so that its topologicaal fundamental group is given by, say, some torsion free arithmetic lattice in the linear group mentioned, and special subvarieties are those subvarieties corresponding to some Shimura subdata (of dimension $>0$ indicated by the following remark of YBL).


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Regarding your last question, isn't $\pi_1(\mathbb{G}_m\otimes \bar{k},1) = \widehat{\mathbb{Z}}$ already a counterexample? – AFK Mar 16 '11 at 3:49
I am not sure I understand the first question. If I do, the answer is clearly no if the field is algebraically closed! – Laurent Moret-Bailly Mar 16 '11 at 9:37

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