Lemma 16.3.5. Let $R \to \Lambda$ be a ring map. If $\Lambda$ is a filtered colimit of smooth $R$-algebras, then $\Lambda$ is a filtered colimit of standard smooth $R$-algebras.

Proof. Let $A \to \Lambda$ be an $R$-algebra map with $A$ of finite presentation over $R$. According to Algebra, Lemma 10.126.4 we have to factor this map through a standard smooth algebra, and we know we can factor it as $A \to B \to \Lambda$ with $B$ smooth over $R$. Choose an $R$-algebra map $B \to C$ with a retraction $C \to B$ such that $C$ is standard smooth over $R$, see Lemma 16.3.4. Then the desired factorization is $A \to B \to C \to B \to \Lambda$. $\square$

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