Lemma 37.58.8. Let

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.

Lemma 37.58.8. 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 37.58.3. 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 37.58.7. We conclude that $f$ is perfect because the composition of perfect morphisms is perfect, see Lemma 37.58.4. $\square$

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