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

non-emptyclosed set meets $\pi=0$, since in an adic ring the topological nilpotents lie in the Jacobson radical). – user36938 Jul 24 '13 at 11:26