A morphism of set-valued functors $\eta: F \to G$ on $\mathcal{C}$ is called smooth if for all epimorphisms $B \to A$, the natural morphism $F(B) \to F(A) \times\_{G(A)} G(B)$ is surjective.


Obviusly "smooth => formally" smooth for $\mathcal{C} = \mathrm{Sch}$.


Now my question: Does the converse hold?


My thoughts: 
1. Assume the morphism is of finite presentation.


1. Assume it is in the local standard form: an étale morphism followed by an affine projection
2. It is clear that an affine projection is smooth in the above sense, so we have reduced the problem to étale morphisms.



formally smooth => smooth
In the case where $F\to G$ is representable by an epimorphism $C\to D$ of rings, then your smoothness condition implies that $C\to D$ has a section. (Take $A\to B$ to be the given map $C\to D$.) But it is not hard to find a formally smooth epimorphism that does not admit a section. Any localization will do. For instance, you can take $D$ to be the zero ring and $C$ any ring but the zero ring.


If you visualize the geometry here, it's pretty clear that your smoothness condition is much stronger than anything normally called smoothness. 


