**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.
$\square$

## Comments (0)