Lemma 32.8.2. Notation and assumptions as in Situation 32.8.1. If $f$ is affine, then there exists an index $i \geq 0$ such that $f_ i$ is affine.

Proof. Let $Y_0 = \bigcup _{j = 1, \ldots , m} V_{j, 0}$ be a finite affine open covering. Set $U_{j, 0} = f_0^{-1}(V_{j, 0})$. For $i \geq 0$ we denote $V_{j, i}$ the inverse image of $V_{j, 0}$ in $Y_ i$ and $U_{j, i} = f_ i^{-1}(V_{j, i})$. Similarly we have $U_ j = f^{-1}(V_ j)$. Then $U_ j = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_{j, i}$ (see Lemma 32.2.2). Since $U_ j$ is affine by assumption we see that each $U_{j, i}$ is affine for $i$ large enough, see Lemma 32.4.13. As there are finitely many $j$ we can pick an $i$ which works for all $j$. Thus $f_ i$ is affine for $i$ large enough, see Morphisms, Lemma 29.11.3. $\square$

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