Lemma 26.4.1. In Situation 26.3.1. Let $F$ be the functor associated to $(S, \mathcal{A})$ above. Let $g : S' \to S$ be a morphism of schemes. Set $\mathcal{A}' = g^*\mathcal{A}$. Let $F'$ be the functor associated to $(S', \mathcal{A}')$ above. Then there is a canonical isomorphism

\[ F' \cong h_{S'} \times _{h_ S} F \]

of functors.

**Proof.**
A pair $(f' : T \to S', \varphi ' : (f')^*\mathcal{A}' \to \mathcal{O}_ T)$ is the same as a pair $(f, \varphi : f^*\mathcal{A} \to \mathcal{O}_ T)$ together with a factorization of $f$ as $f = g \circ f'$. Namely with this notation we have $(f')^* \mathcal{A}' = (f')^*g^*\mathcal{A} = f^*\mathcal{A}$. Hence the lemma.
$\square$

## Comments (1)

Comment #3837 by anonymous on

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