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.



Grothendieck's  section conjecture and base change: restricting sections
$\pi\_1(X\_K)$ 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!


