Take $A$ to be any ring, and $I \subset A$ an ideal. I'd like to prove that $\hat{A}$ is complete with respect to $I$. I think this is true since it comes with universal property of inverse limit. So I take the two inverse system $(\hat{A},f_{n,m})$ where $f_{n,m}$ are the map of the inverse system of $\hat{A}$ and $(\hat{\hat{A}},\pi_j)$ where the $\pi_j : \hat{\hat{A}} \to \hat{A}/I^n$ are just the projection on one factor.
On the other hand the diagram of inverse limit here (general definition) can be inverted, i.e swapping $pi_j$ and $f_{n,m}$ in order to create an inverse of the map from $ \hat{A} \to \hat{\hat{A}}$. It this true?
What's not clear to me is the following, I made an abuse of notation in writing $\pi_j : \hat{\hat{A}} \to \hat{A}/I^n$ since $I^n$ is an ideal of $A$ not an ideal of $\hat{A}$ hence I'm not allowed to do it, it certanly is if $A$ is noetherian local but this seems to much to me. Is this true understanding $I^n$ as the extension of $I^n$ in $\hat{A}$? i.e taking the completion with respect to $(u(I^n)) \subset \hat{A}$, where $u: A \to \hat{A}?$
Edit: Also (But I think is not needed) in the Noetherian case we have that $(u(I^n)) = I^n \hat{A} = \hat{I^n}$