Lemma 10.138.5. Let $R \to S$ be a ring map. Let $P \to S$ be a surjective $R$-algebra map from a polynomial ring $P$ onto $S$. Denote $J \subset P$ the kernel. Then $R \to S$ is formally smooth if and only if there exists an $R$-algebra map $\sigma : S \to P/J^2$ which is a right inverse to the surjection $P/J^2 \to S$.
Proof. Assume $R \to S$ is formally smooth. Consider the commutative diagram
\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & P/J \\ R \ar[r] \ar[u] & P/J^2\ar[u] } \]
By assumption the dotted arrow exists. This proves that $\sigma $ exists.
Conversely, suppose we have a $\sigma $ as in the lemma. Let a solid diagram
\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] } \]
as in Definition 10.138.1 be given. Because $P$ is formally smooth by Lemma 10.138.4, there exists an $R$-algebra homomorphism $\psi : P \to A$ which lifts the map $P \to S \to A/I$. Clearly $\psi (J) \subset I$ and since $I^2 = 0$ we conclude that $\psi (J^2) = 0$. Hence $\psi $ factors as $\overline{\psi } : P/J^2 \to A$. The desired dotted arrow is the composition $\overline{\psi } \circ \sigma : S \to A$. $\square$
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