Are formal power series rings over Dedekind domains formally smooth? Let $A$ be a Dedekind domain.
Consider the ring of formal power series $A[[t]]$ over $A$. Now let $B$ be any $A$-algebra, and let $N\subset B$ be a nilpotent ideal. Then, can any homomorphism
$$A[[t]]\rightarrow B/I$$
(of $A$-algebras) be lifted to a homomorphism $A[[t]]\rightarrow B$ (making the natural diagram commute)?
(I feel like the answer is no, though perhaps upon making suitable "continuity" assumptions the answer is yes?)
 A: You are right. $A[[t]]$ is not in general formally smooth over $A$ for the discrete topology, but it is for the $(t)$-adic topology. You need your map takes some power of $t$ to zero.
It is easy to check that if is formally smooth for the (t)-adic topology (e.g., EGA 0$_{IV}$19.3.6). It is not formally smooth for the discrete topology since the differential module is not, in general, free over $A[[t]]$. "It is known" but I do not know why. For instance, if $A=\mathbb Z$, $\Omega_{\mathbb Z[[t]]|\mathbb Z}\otimes_{\mathbb Z}\mathbb Q$ = $\Omega_{\mathbb Q[[t]]|\mathbb Q}$ which is not $\mathbb Q[[t]]$-free, and so $\Omega_{\mathbb Z[[t]]|\mathbb Z}$ is not $\mathbb Z[[t]]$-free. I'm sorry I'm not able to give you a proof nor reference. It must be not so trivial (since for instance if $A$ is a field of positive characteristic it is easy to show that it is finitely generated and flat and therefore free).
I can show you that it is not finitely generated in general, which is only a step. For instance, if $A=\mathbb Z$, $\Omega_{\mathbb Z[[t]]|\mathbb Z}\otimes_{\mathbb Z[[t]]}\mathbb Q((t))$ = $\Omega_{\mathbb Q((t))|\mathbb Q}$ which is not finitely generated since the transcendence degree of $\mathbb Q((t))|\mathbb Q$ is not finite. Therefore $\Omega_{\mathbb Z[[t]]|\mathbb Z}$ is not finitely generated over $\mathbb Z[[t]]$.
