Lemma 27.15.3. In Situation 27.15.1. Suppose U \subset U' \subset U'' \subset S are affine opens. Let A = \mathcal{A}(U), A' = \mathcal{A}(U') and A'' = \mathcal{A}(U''). The composition of the morphisms r : \text{Proj}(A) \to \text{Proj}(A'), and r' : \text{Proj}(A') \to \text{Proj}(A'') of Lemma 27.15.2 gives the morphism r'' : \text{Proj}(A) \to \text{Proj}(A'') of Lemma 27.15.2. A similar statement holds for the isomorphisms \theta .
Proof. This follows from Lemma 27.11.2 since the map A'' \to A is the composition of A'' \to A' and A' \to A. \square
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