Lemma 58.94.4. In Lemma 58.94.2 assume $f$ is flat, locally of finite presentation, and surjective. Then the functor

$\mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale}) \longrightarrow \left\{ (\mathcal{G}, \mathcal{H}, \alpha ) \middle | \begin{matrix} \mathcal{G} \in \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale}),\ \mathcal{H} \in \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/Y)_{fppf}), \\ \alpha : a_ X^{-1}\mathcal{G} \to f_{big, fppf}^{-1}\mathcal{H} \text{ an isomorphism} \end{matrix} \right\}$

sending $\mathcal{F}$ to $(f_{small}^{-1}\mathcal{F}, a_ Y^{-1}\mathcal{F}, can)$ is an equivalence.

Proof. The functor $a_ X^{-1}$ is fully faithful (as $a_{X, *}a_ X^{-1} = \text{id}$ by Lemma 58.94.1). Hence the forgetful functor $(\mathcal{G}, \mathcal{H}, \alpha ) \mapsto \mathcal{H}$ identifies the category of triples with a full subcategory of $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/Y)_{fppf})$. Moreover, the functor $a_ Y^{-1}$ is fully faithful, hence the functor in the lemma is fully faithful as well.

Suppose that we have an étale covering $\{ Y_ i \to Y\}$. Let $f_ i : X_ i \to Y_ i$ be the base change of $f$. Denote $f_{ij} = f_ i \times f_ j : X_ i \times _ X X_ j \to Y_ i \times _ Y Y_ j$. Claim: if the lemma is true for $f_ i$ and $f_{ij}$ for all $i, j$, then the lemma is true for $f$. To see this, note that the given étale covering determines an étale covering of the final object in each of the four sites $Y_{\acute{e}tale}, X_{\acute{e}tale}, (\mathit{Sch}/Y)_{fppf}, (\mathit{Sch}/X)_{fppf}$. Thus the category of sheaves is equivalent to the category of glueing data for this covering (Sites, Lemma 7.26.5) in each of the four cases. A huge commutative diagram of categories then finishes the proof of the claim. We omit the details. The claim shows that we may work étale locally on $Y$.

Note that $\{ X \to Y\}$ is an fppf covering. Working étale locally on $Y$, we may assume there exists a morphism $s : X' \to X$ such that the composition $f' = f \circ s : X' \to Y$ is surjective finite locally free, see More on Morphisms, Lemma 37.43.1. Claim: if the lemma is true for $f'$, then it is true for $f$. Namely, given a triple $(\mathcal{G}, \mathcal{H}, \alpha )$ for $f$, we can pullback by $s$ to get a triple $(s_{small}^{-1}\mathcal{G}, \mathcal{H}, s_{big, fppf}^{-1}\alpha )$ for $f'$. A solution for this triple gives a sheaf $\mathcal{F}$ on $Y_{\acute{e}tale}$ with $a_ Y^{-1}\mathcal{F} = \mathcal{H}$. By the first paragraph of the proof this means the triple is in the essential image. This reduces us to the case described in the next paragraph.

Assume $f$ is surjective finite locally free. Let $(\mathcal{G}, \mathcal{H}, \alpha )$ be a triple. In this case consider the triple

$(\mathcal{G}_1, \mathcal{H}_1, \alpha _1) = (f_{small}^{-1}f_{small, *}\mathcal{G}, f_{big, fppf, *}f_{big, fppf}^{-1}\mathcal{H}, \alpha _1)$

where $\alpha _1$ comes from the identifications

\begin{align*} a_ X^{-1}f_{small}^{-1}f_{small, *}\mathcal{G} & = f_{big, fppf}^{-1}a_ Y^{-1}f_{small, *}\mathcal{G} \\ & = f_{big, fppf}^{-1}f_{big, fppf, *}a_ X^{-1}\mathcal{G} \\ & \to f_{big, fppf}^{-1}f_{big, fppf, *}f_{big, fppf}^{-1}\mathcal{H} \end{align*}

where the third equality is Lemma 58.94.3 and the arrow is given by $\alpha$. This triple is in the image of our functor because $\mathcal{F}_1 = f_{small, *}\mathcal{F}$ is a solution (to see this use Lemma 58.94.3 again; details omitted). There is a canonical map of triples

$(\mathcal{G}, \mathcal{H}, \alpha ) \to (\mathcal{G}_1, \mathcal{H}_1, \alpha _1)$

which uses the unit $\text{id} \to f_{big, fppf, *}f_{big, fppf}^{-1}$ on the second entry (it is enough to prescribe morphisms on the second entry by the first paragraph of the proof). Since $\{ f : X \to Y\}$ is an fppf covering the map $\mathcal{H} \to \mathcal{H}_1$ is injective (details omitted). Set

$\mathcal{G}_2 = \mathcal{G}_1 \amalg _\mathcal {G} \mathcal{G}_1\quad \mathcal{H}_2 = \mathcal{H}_1 \amalg _\mathcal {H} \mathcal{H}_1$

and let $\alpha _2$ be the induced isomorphism (pullback functors are exact, so this makes sense). Then $\mathcal{H}$ is the equalizer of the two maps $\mathcal{H}_1 \to \mathcal{H}_2$. Repeating the arguments above for the triple $(\mathcal{G}_2, \mathcal{H}_2, \alpha _2)$ we find an injective morphism of triples

$(\mathcal{G}_2, \mathcal{H}_2, \alpha _2) \to (\mathcal{G}_3, \mathcal{H}_3, \alpha _3)$

such that this last triple is in the image of our functor. Say it corresponds to $\mathcal{F}_3$ in $\mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale})$. By fully faithfulness we obtain two maps $\mathcal{F}_1 \to \mathcal{F}_3$ and we can let $\mathcal{F}$ be the equalizer of these two maps. By exactness of the pullback functors involved we find that $a_ Y^{-1}\mathcal{F} = \mathcal{H}$ as desired. $\square$

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