Lemma 27.3.2. In Situation 27.3.1. Suppose $U \subset U' \subset S$ are affine opens. Let $A = \mathcal{A}(U)$ and $A' = \mathcal{A}(U')$. The map of rings $A' \to A$ induces a morphism $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A')$, and the diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(A) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(A') \ar[d] \\ U \ar[r] & U' } \]

is cartesian.

**Proof.**
Let $R = \mathcal{O}_ S(U)$ and $R' = \mathcal{O}_ S(U')$. Note that the map $R \otimes _{R'} A' \to A$ is an isomorphism as $\mathcal{A}$ is quasi-coherent (see Schemes, Lemma 26.7.3 for example). The result follows from the description of the fibre product of affine schemes in Schemes, Lemma 26.6.7.
$\square$

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