Lemma 85.11.1. With notation as above. The morphism a : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D}) is flat if and only if a_ n : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n), \mathcal{O}_ n) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D}) is flat for n \geq 0.
Proof. Since g_ n : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n), \mathcal{O}_ n) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}), \mathcal{O}) is flat, we see that if a is flat, then a_ n = a \circ g_ n is flat as a composition. Conversely, suppose that a_ n is flat for all n. We have to check that \mathcal{O} is flat as a sheaf of a^{-1}\mathcal{O}_\mathcal {D}-modules. Let \mathcal{F} \to \mathcal{G} be an injective map of a^{-1}\mathcal{O}_\mathcal {D}-modules. We have to show that
is injective. We can check this on \mathcal{C}_ n, i.e., after applying g_ n^{-1}. Since g_ n^* = g_ n^{-1} because g_ n^{-1}\mathcal{O} = \mathcal{O}_ n we obtain
which is injective because g_ n^{-1}a^{-1}\mathcal{O}_\mathcal {D} = a_ n^{-1}\mathcal{O}_\mathcal {D} and we assume a_ n was flat. \square
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