Let $G$ be à profinite group and $\widehat{G}$ his profinite completion. I want to show (I hope it's true) that $\widehat{G}$ is strongly complete that is all normal subgroups of finite index are open sets.
For that I take $\widehat{N}$ a finite index normal subgroup of $\widehat{G}$ and try to find $N$ normal subgroup of finite index of $G$ such that $\widehat{N}=\ker\left(\varphi_N:\widehat{G}\to G/N\right)$. I tried $N=\theta^{-1}\left(\widehat{N}\right)$ where $\theta:G\to\widehat{G}$ but could not conclude.
Not rely with this question I think: I don't want a different definition of strongly complete but I want to check that somewath ($\widehat{G}$ here) is strongly complete.
