Lemma 27.3.3. In Situation 27.3.1. 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 \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A'), and \mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A'') of Lemma 27.3.2 gives the morphism \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A'') of Lemma 27.3.2.
Proof. This follows as the map A'' \to A is the composition of A'' \to A' and A' \to A (because \mathcal{A} is a sheaf). \square
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