Lemma 37.61.8. Let
\xymatrix{ X \ar[rr]_ f \ar[rd] & & Y \ar[ld] \\ & S }
be a commutative diagram of morphisms of schemes. Assume Y \to S smooth and X \to S perfect. Then f : X \to Y is perfect.
Proof. We can factor f as the composition
X \longrightarrow X \times _ S Y \longrightarrow Y
where the first morphism is the map i = (1, f) and the second morphism is the projection. Since Y \to S is flat, see Morphisms, Lemma 29.34.9, we see that X \times _ S Y \to Y is perfect by Lemma 37.61.3. As Y \to S is smooth, also X \times _ S Y \to X is smooth, see Morphisms, Lemma 29.34.5. Hence i is a section of a smooth morphism, therefore i is a regular immersion, see Divisors, Lemma 31.22.8. This implies that i is perfect, see Lemma 37.61.7. We conclude that f is perfect because the composition of perfect morphisms is perfect, see Lemma 37.61.4. \square
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