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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