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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