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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