The Stacks project

Proof. Choose a diagram as in Lemma 18.7.2. Recall that being a flat module is intrinsic (see Section 18.18 and Definition 18.28.1). Hence it suffices to prove the lemma for the morphism $(h, h^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}_{\mathcal{C}'}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}'), \mathcal{O}_{\mathcal{D}'})$. In other words, we may assume that our sites $\mathcal{C}$ and $\mathcal{D}$ have all finite limits and that $f$ is a morphism of sites induced by a continuous functor $u : \mathcal{D} \to \mathcal{C}$ which commutes with finite limits.

Recall that $f^*\mathcal{F} = \mathcal{O}_\mathcal {C} \otimes _{f^{-1}\mathcal{O}_\mathcal {D}} f^{-1}\mathcal{F}$ (Definition 18.13.1). By Lemma 18.28.13 it suffices to prove that $f^{-1}\mathcal{F}$ is a flat $f^{-1}\mathcal{O}_\mathcal {D}$-module. Combined with the previous paragraph this reduces us to the situation of the next paragraph.

Assume $\mathcal{C}$ and $\mathcal{D}$ are sites which have all finite limits and that $u : \mathcal{D} \to \mathcal{C}$ is a continuous functor which commutes with finite limits. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{D}$ and let $\mathcal{F}$ be a flat $\mathcal{O}$-module. Then $u$ defines a morphism of sites $f : \mathcal{C} \to \mathcal{D}$ (Sites, Proposition 7.14.7). To show: $f^{-1}\mathcal{F}$ is a flat $f^{-1}\mathcal{O}$-module. Let $U$ be an object of $\mathcal{C}$ and let

be a complex of $f^{-1}\mathcal{O}|_ U$-modules. Our goal is to construct a factorization of $(s_1, \ldots , s_ n)$ on the members of a covering of $U$ as in Lemma 18.28.14 part (2). Consider the elements $s_ a \in f^{-1}\mathcal{F}(U)$ and $f_ a \in f^{-1}\mathcal{O}(U)$. Since $f^{-1}\mathcal{F}$, resp. $f^{-1}\mathcal{O}$ is the sheafification of $u_ p\mathcal{F}$ we may, after replacing $U$ by the members of a covering, assume that $s_ a$ is the image of an element $s'_ a \in u_ p\mathcal{F}(U)$ and $f_ a$ is the image of an element $f'_ a \in u_ p\mathcal{O}(U)$. Then after another replacement of $U$ by the members of a covering we may assume that $\sum f'_ as'_ a$ is zero in $u_ p\mathcal{F}(U)$. Recall that the category $(\mathcal{I}_ U^ u)^{opp}$ is directed (Sites, Lemma 7.5.2) and that $u_ p\mathcal{F}(U) = \mathop{\mathrm{colim}}\nolimits _{(\mathcal{I}_ U^ u)^{opp}} \mathcal{F}(V)$ and $u_ p\mathcal{O}(U) = \mathop{\mathrm{colim}}\nolimits _{(\mathcal{I}_ U^ u)^{opp}} \mathcal{O}(V)$. Hence we may assume there is a pair $(V, \phi ) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}_ U^ u)$ where $V$ is an object of $\mathcal{D}$ and $\phi $ is a morphism $\phi : U \to u(V)$ of $\mathcal{D}$ and elements $s''_ a \in \mathcal{F}(V)$ and $f''_ a \in \mathcal{O}(V)$ whose images in $u_ p\mathcal{F}(U)$ and $u_ p\mathcal{O}(U)$ are equal to $s'_ a$ and $f'_ a$ and such that $\sum f''_ a s''_ a = 0$ in $\mathcal{F}(V)$. Then we obtain a complex

and we can apply the other direction of Lemma 18.28.14 to see there exists a covering $\{ V_ i \to V\} $ of $\mathcal{D}$ and for each $i$ a factorization

of $(s''_1, \ldots , s''_ n)|_{V_ i}$ such that $B_ i \circ (f''_1, \ldots , f''_ n)|_{V_ i} = 0$. Set $U_ i = U \times _{\phi , u(V)} u(V_ i)$, denote $B_ i \in \text{Mat}(l_ i \times n, f^{-1}\mathcal{O}(U_ i))$ the image of $B''_ i$, and denote $t_{ij} \in f^{-1}\mathcal{F}(U_ i)$ the image of $t''_{ij}$. Then we get a factorization

of $(s_1, \ldots , s_ n)|_{U_ i}$ such that $B_ i \circ (f_1, \ldots , f_ n)|_{U_ i} = 0$. This finishes the proof. $\square$