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

$\mathcal{F} \otimes _{a^{-1}\mathcal{O}_\mathcal {D}} \mathcal{O} \to \mathcal{G} \otimes _{a^{-1}\mathcal{O}_\mathcal {D}} \mathcal{O}$

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

$g_ n^{-1}\mathcal{F} \otimes _{g_ n^{-1}a^{-1}\mathcal{O}_\mathcal {D}} \mathcal{O}_ n \to g_ n^{-1}\mathcal{G} \otimes _{g_ n^{-1}a^{-1}\mathcal{O}_\mathcal {D}} \mathcal{O}_ n$

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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