The Stacks project

Proof. This is formal and depends only on functoriality of the pullback functors. We omit the details. Hint: If $s : Y \to X$ is a section, then a quasi-inverse is the functor sending $(\mathcal{F}, \varphi )$ to $s_{small}^{-1}\mathcal{F}$. $\square$

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

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

and the second arrow comes from the map

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. $\square$

Proof. The exact same proof as given in Lemma 59.104.2 works, except the appeal to Lemma 59.55.4 should be replaced by an appeal to Lemma 59.91.5. $\square$

Proof. Formal consequence of the definitions and Lemmas 59.104.2 and 59.104.3. Details omitted. $\square$

Proof. Exactly as in the proof of Lemma 59.104.2 we claim a quasi-inverse is given by the functor sending $(\mathcal{F}, \varphi )$ to

and in order to prove this it suffices to show that $f^{-1}\mathcal{G} \to \mathcal{F}$ is an isomorphism. This we may check locally, hence we may and do assume $Y$ is affine. Then we can find a finite surjective morphism $Z \to Y$ such that there exists an open covering $Z = \bigcup W_ i$ such that $W_ i \to Y$ factors through $X$. See More on Morphisms, Lemma 37.48.6. Applying Lemma 59.104.4 we see that it suffices to prove the lemma after replacing $Y$ by $Z$ and $Z \times _ Y Z$ and $f$ by its base change. Thus we may assume $f$ has sections Zariski locally. Of course, using that the problem is local on $Y$ we reduce to the case where we have a section which is Lemma 59.104.1. $\square$

Proof. We have Lemma 59.104.5 for the morphism $f : \coprod X_ i \to X$. Then a formal argument shows that descent data for $f$ are the same thing as descent data for the covering, compare with Descent, Lemma 35.34.5. Details omitted. $\square$

Proof. We will work with the $2$-fibre product category as constructed in Categories, Example 4.31.3. The functor sends $\mathcal{F}$ to the triple $(f^{-1}\mathcal{F}, i^{-1}\mathcal{F}, c)$ where $c : j^{-1}f^{-1}\mathcal{F} \to g^{-1}i^{-1}\mathcal{F}$ is the canonical isomorphism. We will construct a quasi-inverse functor. Let $(\mathcal{F}', \mathcal{G}, \alpha )$ be an object of the right hand side of the arrow. We obtain an isomorphism

The first equality is Lemma 59.91.5. Using this we obtain maps $i_*\mathcal{G} \to i_*g_*g^{-1}\mathcal{G}$ and $f'_*\mathcal{F}' \to i_*g_*g^{-1}\mathcal{G}$. We set

and we claim that $\mathcal{F}$ is an object of the left hand side of the arrow whose image in the right hand side is isomorphic to the triple we started out with. Let us compute the stalk of $\mathcal{F}$ at a geometric point $\overline{x}$ of $X$.

If $\overline{x}$ is not in $Z$, then on the one hand $\overline{x}$ comes from a unique geometric point $\overline{x}'$ of $X'$ and $\mathcal{F}'_{\overline{x}'} = (f_*\mathcal{F}')_{\overline{x}}$ and on the other hand we have $(i_*\mathcal{G})_{\overline{x}}$ and $(i_*g_*g^{-1}\mathcal{G})_{\overline{x}}$ are singletons. Hence we see that $\mathcal{F}_{\overline{x}}$ equals $\mathcal{F}'_{\overline{x}'}$.

If $\overline{x}$ is in $Z$, i.e., $\overline{x}$ is the image of a geometric point $\overline{z}$ of $Z$, then we obtain $(i_*\mathcal{G})_{\overline{x}} = \mathcal{G}_{\overline{z}}$ and

(by the proper base change for pushforward used above) and similarly

Since we have the identification $E_{\overline{z}} = X'_{\overline{x}}$ and since $\alpha $ defines an isomorphism between the sheaves $\mathcal{F}'|_{X'_{\overline{x}}}$ and $g^{-1}\mathcal{G}|_{E_{\overline{z}}}$ we conclude that we get

in this case.

To finish the proof, we observe that there are canonical maps $i^{-1}\mathcal{F} \to \mathcal{G}$ and $f^{-1}\mathcal{F} \to \mathcal{F}'$ compatible with $\alpha $ which on stalks produce the isomorphisms we saw above. We omit the careful construction of these maps. $\square$

Proof. To prove the lemma we will check conditions (1), (2), and (3) of More on Flatness, Lemma 38.37.13.

Condition (1) holds because we have glueing for sheaves (and Zariski coverings are étale coverings). See Sites, Lemma 7.26.4.

To see condition (2), suppose that $f : X \to Y$ is a surjective, flat, proper morphism of finite presentation over $S$ with $Y$ affine. Then we have descent for $\{ X \to Y\} $ by either Lemma 59.104.5 or Lemma 59.104.3.

Condition (3) follows immediately from the more general Lemma 59.104.7. $\square$