Lemma 59.104.2. Let $f : X \to Y$ be a surjective integral morphism of schemes. The functor

is an equivalence of categories.

Lemma 59.104.2. Let $f : X \to Y$ be a surjective integral morphism of schemes. The functor

\[ \mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale}) \longrightarrow \text{descent data for étale sheaves wrt }\{ X \to Y\} \]

is an equivalence of categories.

**Proof.**
In this proof we drop the subscript ${}_{small}$ from our pullback and pushforward functors. Denote $X_1 = X \times _ Y X$ and denote $f_1 : X_1 \to Y$ the morphism $f \circ \text{pr}_0 = f \circ \text{pr}_1$. Let $(\mathcal{F}, \varphi )$ be a descent datum for $\{ X \to Y\} $. Let us set $\mathcal{F}_1 = \text{pr}_0^{-1}\mathcal{F}$. We may think of $\varphi $ as defining an isomorphism $\mathcal{F}_1 \to \text{pr}_1^{-1}\mathcal{F}$. We claim that the rule which sends a descent datum $(\mathcal{F}, \varphi )$ to the sheaf

\[ \mathcal{G} = \text{Equalizer}\left( \xymatrix{ f_*\mathcal{F} \ar@<1ex>[r] \ar@<-1ex>[r] & f_{1, *}\mathcal{F}_1 } \right) \]

is a quasi-inverse to the functor in the statement of the lemma. The first of the two arrows comes from the map

\[ f_*\mathcal{F} \to f_*\text{pr}_{0, *}\text{pr}_0^{-1}\mathcal{F} = f_{1, *}\mathcal{F}_1 \]

and the second arrow comes from the map

\[ f_*\mathcal{F} \to f_* \text{pr}_{1, *}\text{pr}_1^{-1}\mathcal{F} \xleftarrow {\varphi } f_* \text{pr}_{0, *} \text{pr}_0^{-1}\mathcal{F} = f_{1, *}\mathcal{F}_1 \]

where the arrow pointing left is invertible. To prove this works we have to show that the canonical map $f^{-1}\mathcal{G} \to \mathcal{F}$ is an isomorphism; details omitted. In order to prove this it suffices to check after pulling back by any collection of morphisms $\mathop{\mathrm{Spec}}(k) \to Y$ where $k$ is an algebraically closed field. Namely, the corresponding base changes $X_ k \to X$ are jointly surjective and we can check whether a map of sheaves on $X_{\acute{e}tale}$ is an isomorphism by looking at stalks on geometric points, see Theorem 59.29.10. By Lemma 59.55.4 the construction of $\mathcal{G}$ from the descent datum $(\mathcal{F}, \varphi )$ commutes with any base change. Thus we may assume $Y$ is the spectrum of an algebraically closed point (note that base change preserves the properties of the morphism $f$, see Morphisms, Lemma 29.9.4 and 29.44.6). In this case the morphism $X \to Y$ has a section, so we know that the functor is an equivalence by Lemma 59.104.1. However, the reader may show that the functor is an equivalence if and only if the construction above is a quasi-inverse; details omitted. This finishes the proof. $\square$

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