Let $f:X\rightarrow Y$ a morphism between schemes and $Y'\rightarrow Y$ a fpqc morphism
such that the base change $f'$ of $f$ to $Y'$ is formally smooth, does it imply that $f$ is formally smooth?

In the infinitesimal lifting property, can we reduce to easier rings, such local henselian or local for example?
– prochetJun 17 '13 at 9:42

1

Section 1.7 of arxiv.org/abs/math/9812034 contains the claim that formal smoothness is a local property in the fpqc topology (presumably meaning local on the target), and says that Gabber can explain why.
– S. Carnahan♦Jul 29 '13 at 6:22