Remark 29.11.4. We can also argue directly that (2) implies (1) in Lemma 29.11.3 above as follows. Assume $S = \bigcup W_ j$ is an affine open covering such that each $f^{-1}(W_ j)$ is affine. First argue that $\mathcal{A} = f_*\mathcal{O}_ X$ is quasi-coherent as in the proof above. Let $\mathop{\mathrm{Spec}}(R) = V \subset S$ be affine open. We have to show that $f^{-1}(V)$ is affine. Set $A = \mathcal{A}(V) = f_*\mathcal{O}_ X(V) = \mathcal{O}_ X(f^{-1}(V))$. By Schemes, Lemma 26.6.4 there is a canonical morphism $\psi : f^{-1}(V) \to \mathop{\mathrm{Spec}}(A)$ over $\mathop{\mathrm{Spec}}(R) = V$. By Schemes, Lemma 26.11.6 there exists an integer $n \geq 0$, a standard open covering $V = \bigcup _{i = 1, \ldots , n} D(h_ i)$, $h_ i \in R$, and a map $a : \{ 1, \ldots , n\} \to J$ such that each $D(h_ i)$ is also a standard open of the affine scheme $W_{a(i)}$. The inverse image of a standard open under a morphism of affine schemes is standard open, see Algebra, Lemma 10.17.4. Hence we see that $f^{-1}(D(h_ i))$ is a standard open of $f^{-1}(W_{a(i)})$, in particular that $f^{-1}(D(h_ i))$ is affine. Because $\mathcal{A}$ is quasi-coherent we have $A_{h_ i} = \mathcal{A}(D(h_ i)) = \mathcal{O}_ X(f^{-1}(D(h_ i)))$, so $f^{-1}(D(h_ i))$ is the spectrum of $A_{h_ i}$. It follows that the morphism $\psi $ induces an isomorphism of the open $f^{-1}(D(h_ i))$ with the open $\mathop{\mathrm{Spec}}(A_{h_ i})$ of $\mathop{\mathrm{Spec}}(A)$. Since $f^{-1}(V) = \bigcup f^{-1}(D(h_ i))$ and $\mathop{\mathrm{Spec}}(A) = \bigcup \mathop{\mathrm{Spec}}(A_{h_ i})$ we win.

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