Lemma 29.11.7. The composition of affine morphisms is affine.

**Proof.**
Let $f : X \to Y$ and $g : Y \to Z$ be affine morphisms. Let $U \subset Z$ be affine open. Then $g^{-1}(U)$ is affine by assumption on $g$. Whereupon $f^{-1}(g^{-1}(U))$ is affine by assumption on $f$. Hence $(g \circ f)^{-1}(U)$ is affine.
$\square$

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