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 10.127.4 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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