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Let $X$ be a smooth projective geometrically connected curve over $\mathbf{Q}$ of genus at least two. Fix an algebraic closure $\overline{\mathbf{Q}}$ of $\mathbf{Q}$ and let $G_{\mathbf{Q}}$ be the absolute Galois group of $\mathbf{Q}$. Moreover, fix a rational base point $x$ in $X(\mathbf{Q})$.

Let $\sigma:G_{\mathbf{Q}}\to \pi_1(X)$ be a section of the exact sequence of groups $$ 1\to \pi_1(X_{\overline{\mathbf{Q}}})\to \pi_1(X) \to G_{\mathbf{Q}}\to 1.$$

Let $K\subset \overline{\mathbf{Q}}$ be a finite field extension of $\mathbf{Q}$. (I don't want to assume $K$ to be Galois, but please do if this helps.)

Let $G_K$ be the absolute Galois group of $K$. Then $G_K$ is an open subgroup of $G_{\mathbf{Q}}$. Note that $\pi_1(X_K)$ (with the same base point $x$) injects into $\pi_1(X)$.

Question. Does there exist a section $\sigma^\prime:G_{\mathbf{Q}}\to \pi_1(X)$ which is a $\pi_1(X_{\overline{\mathbf{Q}}})$-conjugate of $\sigma$ such that the image of $\sigma^\prime|_{G_K}$ lies in $\pi_1(X_K)$?

Motivation. If $a\in X(\mathbf{Q})$, then $a\in X(K)$. Thus, if $\sigma$ is a section associated to $a$, then the answer to the above question is positive. My question is really about sections that a priori do not come from a rational point.

Note. I always use the base point $x$ to define the fundamental group and we can replace $\mathbf{Q}$ by any number field.

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up vote 5 down vote accepted

$\pi_1(X_K)$ is just the subgroup of $\pi_1(X)$ that is the inverse image of $G_K$. So, since the map is a section, its restriction immediately factors through - we do not even have to conjugate!

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Thanks a lot. I was clearly confused. – Jan Hendrik Nov 3 '12 at 19:37

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