Lemma 27.16.1. In Situation 27.15.1. Let $d \geq 1$. Let $F_ d$ 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 $F_ d'$ be the functor associated to $(S', \mathcal{A}')$ above. Then there is a canonical isomorphism
of functors.
Proof. A quadruple $(d, f' : T \to S', \mathcal{L}', \psi ' : (f')^*(\mathcal{A}')^{(d)} \to \bigoplus _{n \geq 0} (\mathcal{L}')^{\otimes n})$ is the same as a quadruple $(d, f, \mathcal{L}, \psi : f^*\mathcal{A}^{(d)} \to \bigoplus _{n \geq 0} \mathcal{L}^{\otimes n})$ together with a factorization of $f$ as $f = g \circ f'$. Namely, the correspondence is $f = g \circ f'$, $\mathcal{L} = \mathcal{L}'$ and $\psi = \psi '$ via the identifications $(f')^*(\mathcal{A}')^{(d)} = (f')^*g^*(\mathcal{A}^{(d)}) = f^*\mathcal{A}^{(d)}$. Hence the lemma. $\square$
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