Proof.
We have seen that (1) implies (2) and (3) in Lemma 76.19.5. Assume (3). The proof that f is formally smooth is entirely similar to the proof of (1) \Rightarrow (2) of Lemma 76.19.6.
Consider a solid commutative diagram
\xymatrix{ X \ar[d]_ f & T \ar[d]^ i \ar[l]^ a \\ Y & T' \ar[l] \ar@{-->}[lu] }
as in Definition 76.19.1. We will show the dotted arrow exists thereby proving that f is formally smooth. Let \mathcal{F} be the sheaf of sets on (T')_{spaces, {\acute{e}tale}} of Lemma 76.17.4 as in the special case discussed in Remark 76.17.6. Let
\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ T}(a^*\Omega _{X/Y}, \mathcal{C}_{T/T'})
be the sheaf of \mathcal{O}_ T-modules on T_{spaces, {\acute{e}tale}} with action \mathcal{H} \times \mathcal{F} \to \mathcal{F} as in Lemma 76.17.5. The action \mathcal{H} \times \mathcal{F} \to \mathcal{F} turns \mathcal{F} into a pseudo \mathcal{H}-torsor, see Cohomology on Sites, Definition 21.4.1. Our goal is to show that \mathcal{F} is a trivial \mathcal{H}-torsor. There are two steps: (I) To show that \mathcal{F} is a torsor we have to show that \mathcal{F} has étale locally a section. (II) To show that \mathcal{F} is the trivial torsor it suffices to show that H^1(T_{\acute{e}tale}, \mathcal{H}) = 0, see Cohomology on Sites, Lemma 21.4.3.
First we prove (I). To see this consider a diagram (which exists because we are assuming (3))
\xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^ f & Y }
where U and V are schemes, the vertical arrows are étale and surjective, and \psi is formally smooth. By Lemma 76.13.5 the morphism V \to Y is formally étale. Thus by Lemma 76.13.3 the composition U \to Y is formally smooth. Then (I) follows from Lemma 76.13.6 part (4).
Finally we prove (II). By Lemma 76.19.8 we see that \Omega _{U/V} locally projective. Hence \Omega _{X/Y} is locally projective, see Descent on Spaces, Lemma 74.6.5. Hence a^*\Omega _{X/Y} is locally projective, see Properties of Spaces, Lemma 66.31.3. Hence
H^1(T_{\acute{e}tale}, \mathcal{H}) = H^1(T_{\acute{e}tale}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ T}(a^*\Omega _{X/Y}, \mathcal{C}_{T/T'}) = 0
by Lemma 76.19.9 as desired.
\square
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