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Let $K$ be a local field, $G$ a (connected) reductive $K$-group, and $P \le G$ a parabolic subgroup. Is the map $G(K) \rightarrow (G/P)(K)$ necessarily surjective, and, if so, then why?

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    $\begingroup$ @abx: Nope, not so: consider $\mathrm{SL}_n(K) \rightarrow \mathrm{PGL}_n(K)$ (for $K$ of characteristic $0$) for one of many counterexamples. $\endgroup$ – Question Mark Oct 18 '14 at 6:12
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    $\begingroup$ @abx: For a more geometric example consider a smooth projective geometrically connected curve $X_0$ over the finite residue field $k$ of $K$ such that $X_0(k)$ is empty, and let $X$ be a proper flat lift of $X_0$ over $O_K$. The generic fiber $X_K$ is smooth, proper, and geometrically connected curve over $K$ with no $K$-points. $\endgroup$ – user27920 Oct 18 '14 at 12:34
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    $\begingroup$ Even more explicit, how about the map $C\to \text{Spec}(\mathbb{R})$ where $C$ is the smooth projective conic defined by $x^2+y^2+z^2=0$? $\endgroup$ – Daniel Litt Oct 18 '14 at 21:09
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    $\begingroup$ @DanielLitt, there is probably some sort of personality test inherent in the question of whether your example or QuestionMark's is the first one to come to mind. :-) $\endgroup$ – LSpice Jan 13 '15 at 20:41
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    $\begingroup$ @LSpice: That's the case for maps which are quotients by free group actions (as in QuestionMark's example) but the situation is rather more complicated in general. The sense in which my example is "cohomological" is that the curve in question exhibits a Severi-Brauer curve whose associated Brauer class (in $H^2(K, \mu_2)$) is non-trivial. $\endgroup$ – Daniel Litt Jan 15 '15 at 7:32
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The map G(K) to (G/P)(K) is surjective over any field K. Here is a link to an explanation by Brian Conrad. http://math.stanford.edu/~conrad/249CS13Page/handouts/parsurj.pdf

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