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Let $f:X\rightarrow S= Spec(k[[\pi]])$ a finite type faithfully flat morphism. Let $U\subset X$ be an open subset such that $U$ is smooth and surjective on $S$.

We consider the $k$-scheme $X(k[[\pi]])$ such that for any $k$-algebra $R$, the $R$-points are given by $X(R[[\pi]])$ and we also consider the $k$-scheme $U(k[[\pi]])$

For all $n$, we have a morphism $X(k[[\pi]])\rightarrow X(k[\pi]/\pi^{n})$.

Do we have that $U(k[[\pi]])=X(k[[\pi]])\times_{X(k[\pi]/\pi^{n})}U(k[\pi]/\pi^{n})$?

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What does "scheme of sections" mean? (As with your earlier question on affine Grassmannians, you seem to denote some scheme by its set of $k$-points, which can be really confusing.) Does your question mean to ask if for any $k$-algebra $R$ a $k[\![\pi]\!]$-morphism $f:{\rm{Spec}}(R[\![\pi]\!])\rightarrow X$ factors through $U$ when it does so mod $\pi^n$? If so, it is true because the only open subset of Spec($R[\![\pi]\!])$ containing $\{\pi=0\}$ is the entire space (as any non-empty closed set meets $\pi=0$, since in an adic ring the topological nilpotents lie in the Jacobson radical). –  user36938 Jul 24 '13 at 11:26
    
I added the definition for $X(k[[\pi]])$. Yes that what I asked, thx. –  prochet Jul 24 '13 at 12:26
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