The Stacks project

Proof. Suppose we have $h : X' \to X$ surjective and flat and $a, b : X \to Y$ morphisms such that $a \circ h = b \circ h$. As $h$ is surjective we see that $a$ and $b$ agree on underlying topological spaces. Pick $x' \in X'$ and set $x = h(x')$ and $y = a(x) = b(x)$. Consider the local ring maps

These become equal when composed with the flat local homomorphism $h^\sharp _{x'} : \mathcal{O}_{X, x} \to \mathcal{O}_{X', x'}$. Since a flat local homomorphism is faithfully flat (Algebra, Lemma 10.39.17) we conclude that $h^\sharp _{x'}$ is injective. Hence $a^\sharp _ x = b^\sharp _ x$ which implies $a = b$ as desired. $\square$

Proof. Let $X_1$, $X_2$ be schemes over $S$. Let $\alpha , \beta : X_2 \to X_1$ be morphisms over $S$. If $\alpha $, $\beta $ base change to the same morphism then we get a commutative diagram as follows

Hence it suffices to show that $S' \times _ S X_2 \to X_2$ is an epimorphism. As the base change of a surjective and flat morphism it is surjective and flat (see Morphisms, Lemmas 29.9.4 and 29.25.8). Hence the lemma follows from Lemma 35.35.1. $\square$

Proof. Let $(V_ i, \varphi _ i)$, $i = 1, 2$ be descent data for $X \to S$. Let $\alpha , \beta : V_1 \to V_2$ be morphisms of descent data. Suppose that $f^*\alpha = f^*\beta $. Our task is to show that $\alpha = \beta $. Note that $\alpha $, $\beta $ are morphisms of schemes over $X$, and that $f^*\alpha $, $f^*\beta $ are simply the base changes of $\alpha $, $\beta $ to morphisms over $X'$. Hence the lemma follows from Lemma 35.35.2. $\square$

Proof. Assumption (1) implies that $f$ is surjective and flat. Hence the pullback functor is faithful by Lemma 35.35.3. Let $(V, \varphi )$ and $(W, \psi )$ be two descent data relative to $X \to S$. Set $(V', \varphi ') = f^*(V, \varphi )$ and $(W', \psi ') = f^*(W, \psi )$. Let $\alpha ' : V' \to W'$ be a morphism of descent data for $X'$ over $S$. We have to show there exists a morphism $\alpha : V \to W$ of descent data for $X$ over $S$ whose pullback is $\alpha '$.

Recall that $V'$ is the base change of $V$ by $f$ and that $\varphi '$ is the base change of $\varphi $ by $f \times f$ (see Remark 35.34.2). By assumption the diagram

commutes. We claim the two compositions

are the same. The reader is advised to prove this themselves rather than read the rest of this paragraph. (Please email if you find a nice clean argument.) Let $v_0, v_1$ be points of $V'$ which map to the same point $v \in V$. Let $x_ i \in X'$ be the image of $v_ i$, and let $x$ be the point of $X$ which is the image of $v$ in $X$. In other words, $v_ i = (x_ i, v)$ in $V' = X' \times _ X V$. Write $\varphi (v, x) = (x, v')$ for some point $v'$ of $V$. This is possible because $\varphi $ is a morphism over $X \times _ S X$. Denote $v_ i' = (x_ i, v')$ which is a point of $V'$. Then a calculation (using the definition of $\varphi '$) shows that $\varphi '(v_ i, x_ j) = (x_ i, v'_ j)$. Denote $w_ i = \alpha '(v_ i)$ and $w'_ i = \alpha '(v_ i')$. Now we may write $w_ i = (x_ i, u_ i)$ for some point $u_ i$ of $W$, and $w_ i' = (x_ i, u'_ i)$ for some point $u_ i'$ of $W$. The claim is equivalent to the assertion: $u_0 = u_1$. A formal calculation using the definition of $\psi '$ (see Lemma 35.34.6) shows that the commutativity of the diagram displayed above says that

as points of $(X' \times _ S X') \times _{X \times _ S X} (X \times _ S W)$ for all $i, j \in \{ 0, 1\} $. This shows that $\psi (u_0, x) = \psi (u_1, x)$ and hence $u_0 = u_1$ by taking $\psi ^{-1}$. This proves the claim because the argument above was formal and we can take scheme points (in other words, we may take $(v_0, v_1) = \text{id}_{V' \times _ V V'}$).

At this point we can use Lemma 35.13.7. Namely, $\{ V' \to V\} $ is a fpqc covering as the base change of the morphism $f : X' \to X$. Hence, by Lemma 35.13.7 the morphism $\alpha ' : V' \to W' \to W$ factors through a unique morphism $\alpha : V \to W$ whose base change is necessarily $\alpha '$. Finally, we see the diagram

commutes because its base change to $X' \times _ S X'$ commutes and the morphism $X' \times _ S X' \to X \times _ S X$ is surjective and flat (use Lemma 35.35.2). Hence $\alpha $ is a morphism of descent data $(V, \varphi ) \to (W, \psi )$ as desired. $\square$

Proof. This is clear from Lemma 35.34.6 since it tells us that $f^* \cong \text{id}^*$. $\square$

Proof. Let $g : X \to X'$ be a morphism over $S$. Lemma 35.35.5 above shows that the functors $f^* \circ g^* = (g \circ f)^*$ and $g^* \circ f^* = (f \circ g)^*$ are isomorphic to the respective identity functors as desired. $\square$

Proof. We may factor $X \to X'$ as $X \to X \times _ S X' \to X'$. The first morphism has a section, hence induces an equivalence of categories of descent data by Lemma 35.35.6. The second morphism is surjective and flat, hence induces a faithful functor by Lemma 35.35.3. $\square$

Proof. We may factor $X \to X'$ as $X \to X \times _ S X' \to X'$. The first morphism has a section, hence induces an equivalence of categories of descent data by Lemma 35.35.6. The second morphism is an fpqc covering hence induces a fully faithful functor by Lemma 35.35.4. $\square$

Proof. Consider the morphism of schemes

over $S$ which on the $i$th component maps into the $\alpha (i)$th component via the morphism $g_{\alpha (i)}$. We claim that $\{ g : X \to Y\} $ is an fpqc covering of schemes. Namely, by Topologies, Lemma 34.9.3 for each $j$ the morphism $\{ \coprod _{\alpha (i) = j} U_ i \to V_ j\} $ is an fpqc covering. Thus for every affine open $V \subset V_ j$ (which we may think of as an affine open of $Y$) we can find finitely many affine opens $W_1, \ldots , W_ n \subset \coprod _{\alpha (i) = j} U_ i$ (which we may think of as affine opens of $X$) such that $V = \bigcup _{i = 1, \ldots , n} g(W_ i)$. This provides enough affine opens of $Y$ which can be covered by finitely many affine opens of $X$ so that Topologies, Lemma 34.9.2 part (3) applies, and the claim follows. Let us write $DD(X/S)$, resp. $DD(\mathcal{U})$ for the category of descent data with respect to $X/S$, resp. $\mathcal{U}$, and similarly for $Y/S$ and $\mathcal{V}$. Consider the diagram

This diagram is commutative, see the proof of Lemma 35.34.8. The vertical arrows are equivalences. Hence the lemma follows from Lemma 35.35.4 which shows the top horizontal arrow of the diagram is fully faithful. $\square$

Proof. The functor is faithful and fully faithful by Lemma 35.35.9. Let us indicate how to prove that it is essentially surjective. Let $(X_ i, \varphi _{ii'})$ be a descent datum relative to $\mathcal{U}$. Fix $j \in J$ and set $I_ j = \{ i \in I \mid \alpha (i) = j\} $. For $i, i' \in I_ j$ note that there is a canonical morphism

Hence we can pullback $\varphi _{ii'}$ by this morphism and set $\psi _{ii'} = c_{ii'}^*\varphi _{ii'}$ for $i, i' \in I_ j$. In this way we obtain a descent datum $(X_ i, \psi _{ii'})$ relative to the Zariski covering $\{ g_ i : U_ i \to V_ j\} _{i \in I_ j}$. Note that $\psi _{ii'}$ is an isomorphism from the open $X_{i, U_ i \times _{V_ j} U_{i'}}$ of $X_ i$ to the corresponding open of $X_{i'}$. It follows from Schemes, Section 26.14 that we may glue $(X_ i, \psi _{ii'})$ into a scheme $Y_ j$ over $V_ j$. Moreover, the morphisms $\varphi _{ii'}$ for $i \in I_ j$ and $i' \in I_{j'}$ glue to a morphism $\varphi _{jj'} : Y_ j \times _ S V_{j'} \to V_ j \times _ S Y_{j'}$ satisfying the cocycle condition (details omitted). Hence we obtain the desired descent datum $(Y_ j, \varphi _{jj'})$ relative to $\mathcal{V}$. $\square$

Proof. Consider the fpqc-covering $\mathcal{W} = \{ U_ i \times _ S V_ j \to S\} _{(i, j) \in I \times J}$ of $S$. It is a refinement of both $\mathcal{U}$ and $\mathcal{V}$. Hence we have a $2$-commutative diagram of functors and categories

Notation as in the proof of Lemma 35.35.9 and commutativity by Lemma 35.34.8 part (3). Hence clearly it suffices to prove the functors $DD(\mathcal{V}) \to DD(\mathcal{W})$ and $DD(\mathcal{U}) \to DD(\mathcal{W})$ are fully faithful. This follows from Lemma 35.35.9 as desired. $\square$

Proof. Let $V = \bigcup W_ i$ be an open covering with $\varphi (W_ i \times _ S X) \subset X \times _ S W_ i$ and such that the descent datum $(W_ i, \varphi |_{W_ i \times _ S X})$ is effective. Let $U_ i \to S$ be a scheme and let $\alpha _ i : (X \times _ S U_ i, can) \to (W_ i, \varphi |_{W_ i \times _ S X})$ be an isomorphism of descent data. For each pair of indices $(i, j)$ consider the open $\alpha _ i^{-1}(W_ i \cap W_ j) \subset X \times _ S U_ i$. Because everything is compatible with descent data and since $\{ X \to S\} $ is an fpqc covering, we may apply Lemma 35.13.6 to find an open $U_{ij} \subset U_ j$ such that $\alpha _ i^{-1}(W_ i \cap W_ j) = X \times _ S U_{ij}$. Now the identity morphism on $W_ i \cap W_ j$ is compatible with descent data, hence comes from a unique morphism $\varphi _{ij} : U_{ij} \to U_{ji}$ over $S$ (see Remark 35.35.12). Then $(U_ i, U_{ij}, \varphi _{ij})$ is a glueing data as in Schemes, Section 26.14 (proof omitted). Thus we may assume there is a scheme $U$ over $S$ such that $U_ i \subset U$ is open, $U_{ij} = U_ i \cap U_ j$ and $\varphi _{ij} = \text{id}_{U_ i \cap U_ j}$, see Schemes, Lemma 26.14.1. Pulling back to $X$ we can use the $\alpha _ i$ to get the desired isomorphism $\alpha : X \times _ S U \to V$. $\square$