## 35.32 Fully faithfulness of the pullback functors

It turns out that the pullback functor between descent data for fpqc-coverings is fully faithful. In other words, morphisms of schemes satisfy fpqc descent. The goal of this section is to prove this. The reader is encouraged instead to prove this him/herself. The key is to use Lemma 35.10.7.

Lemma 35.32.1. A surjective and flat morphism is an epimorphism in the category of schemes.

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

$a^\sharp _ x, b^\sharp _ x : \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$

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. $\square$

Lemma 35.32.2. Let $h : S' \to S$ be a surjective, flat morphism of schemes. The base change functor

$\mathit{Sch}/S \longrightarrow \mathit{Sch}/S', \quad X \longmapsto S' \times _ S X$

is faithful.

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

$\xymatrix{ X_2 \ar[d]^\alpha & S' \times _ S X_2 \ar[l] \ar[d] \ar[r] & X_2 \ar[d]^\beta \\ X_1 & S' \times _ S X_1 \ar[l] \ar[r] & X_1 }$

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 35.32.1. $\square$

Lemma 35.32.3. In the situation of Lemma 35.31.6 assume that $f : X' \to X$ is surjective and flat. Then the pullback functor is faithful.

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 35.32.2. $\square$

Here is the key lemma of this section.

Lemma 35.32.4. In the situation of Lemma 35.31.6 assume

1. $\{ f : X' \to X\}$ is an fpqc covering (for example if $f$ is surjective, flat, and quasi-compact), and

2. $S = S'$.

Then the pullback functor is fully faithful.

Proof. Assumption (1) implies that $f$ is surjective and flat. Hence the pullback functor is faithful by Lemma 35.32.3. 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 35.31.2). By assumption the diagram

$\xymatrix{ V' \times _ S X' \ar[r]_{\varphi '} \ar[d]_{\alpha ' \times \text{id}} & X' \times _ S V' \ar[d]^{\text{id} \times \alpha '} \\ W' \times _ S X' \ar[r]^{\psi '} & X' \times _ S W' }$

commutes. We claim the two compositions

$\xymatrix{ V' \times _ V V' \ar[r]^-{\text{pr}_ i} & V' \ar[r]^{\alpha '} & W' \ar[r] & W } , \quad i = 0, 1$

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 35.31.6) shows that the commutativity of the diagram displayed above says that

$((x_ i, x_ j), \psi (u_ i, x)) = ((x_ i, x_ j), (x, u'_ j))$

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 35.10.7. Namely, $\{ V' \to V\}$ is a fpqc covering as the base change of the morphism $f : X' \to X$. Hence, by Lemma 35.10.7 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

$\xymatrix{ V \times _ S X \ar[r]_{\varphi } \ar[d]_{\alpha \times \text{id}} & X \times _ S V \ar[d]^{\text{id} \times \alpha } \\ W \times _ S X \ar[r]^{\psi } & X \times _ S W }$

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 35.32.2). Hence $\alpha$ is a morphism of descent data $(V, \varphi ) \to (W, \psi )$ as desired. $\square$

The following two lemmas have been obsoleted by the improved exposition of the previous material. But they are still true!

Lemma 35.32.5. Let $X \to S$ be a morphism of schemes. Let $f : X \to X$ be a selfmap of $X$ over $S$. In this case pullback by $f$ is isomorphic to the identity functor on the category of descent data relative to $X \to S$.

Proof. This is clear from Lemma 35.31.6 since it tells us that $f^* \cong \text{id}^*$. $\square$

Lemma 35.32.6. Let $f : X' \to X$ be a morphism of schemes over a base scheme $S$. Assume there exists a morphism $g : X \to X'$ over $S$, for example if $f$ has a section. Then the pullback functor of Lemma 35.31.6 defines an equivalence of categories between the category of descent data relative to $X/S$ and $X'/S$.

Proof. Let $g : X \to X'$ be a morphism over $S$. Lemma 35.32.5 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. $\square$

Lemma 35.32.7. Let $f : X \to X'$ be a morphism of schemes over a base scheme $S$. Assume $X \to S$ is surjective and flat. Then the pullback functor of Lemma 35.31.6 is a faithful functor from the category of descent data relative to $X'/S$ to the category of descent data relative to $X/S$.

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 35.32.6. The second morphism is surjective and flat, hence induces a faithful functor by Lemma 35.32.3. $\square$

Lemma 35.32.8. Let $f : X \to X'$ be a morphism of schemes over a base scheme $S$. Assume $\{ X \to S\}$ is an fpqc covering (for example if $f$ is surjective, flat and quasi-compact). Then the pullback functor of Lemma 35.31.6 is a fully faithful functor from the category of descent data relative to $X'/S$ to the category of descent data relative to $X/S$.

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 35.32.6. The second morphism is an fpqc covering hence induces a fully faithful functor by Lemma 35.32.4. $\square$

Lemma 35.32.9. Let $S$ be a scheme. Let $\mathcal{U} = \{ U_ i \to S\} _{i \in I}$, and $\mathcal{V} = \{ V_ j \to S\} _{j \in J}$, be families of morphisms with target $S$. Let $\alpha : I \to J$, $\text{id} : S \to S$ and $g_ i : U_ i \to V_{\alpha (i)}$ be a morphism of families of maps with fixed target, see Sites, Definition 7.8.1. Assume that for each $j \in J$ the family $\{ g_ i : U_ i \to V_ j\} _{\alpha (i) = j}$ is an fpqc covering of $V_ j$. Then the pullback functor

$\text{descent data relative to } \mathcal{V} \longrightarrow \text{descent data relative to } \mathcal{U}$

of Lemma 35.31.8 is fully faithful.

Proof. Consider the morphism of schemes

$g : X = \coprod \nolimits _{i \in I} U_ i \longrightarrow Y = \coprod \nolimits _{j \in J} V_ j$

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

$\xymatrix{ DD(Y/S) \ar[r] & DD(X/S) \\ DD(\mathcal{V}) \ar[u]^{\text{Lemma }023X} \ar[r] & DD(\mathcal{U}) \ar[u]_{\text{Lemma }023X} }$

This diagram is commutative, see the proof of Lemma 35.31.8. The vertical arrows are equivalences. Hence the lemma follows from Lemma 35.32.4 which shows the top horizontal arrow of the diagram is fully faithful. $\square$

The next lemma shows that, in order to check effectiveness, we may always Zariski refine the given family of morphisms with target $S$.

Lemma 35.32.10. Let $S$ be a scheme. Let $\mathcal{U} = \{ U_ i \to S\} _{i \in I}$, and $\mathcal{V} = \{ V_ j \to S\} _{j \in J}$, be families of morphisms with target $S$. Let $\alpha : I \to J$, $\text{id} : S \to S$ and $g_ i : U_ i \to V_{\alpha (i)}$ be a morphism of families of maps with fixed target, see Sites, Definition 7.8.1. Assume that for each $j \in J$ the family $\{ g_ i : U_ i \to V_ j\} _{\alpha (i) = j}$ is a Zariski covering (see Topologies, Definition 34.3.1) of $V_ j$. Then the pullback functor

$\text{descent data relative to } \mathcal{V} \longrightarrow \text{descent data relative to } \mathcal{U}$

of Lemma 35.31.8 is an equivalence of categories. In particular, the category of schemes over $S$ is equivalent to the category of descent data relative to any Zariski covering of $S$.

Proof. The functor is faithful and fully faithful by Lemma 35.32.9. 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

$c_{ii'} : U_ i \times _{g_ i, V_ j, g_{i'}} U_{i'} \to U_ i \times _ S U_{i'}.$

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 $\mathcal{V}$. $\square$

Lemma 35.32.11. Let $S$ be a scheme. Let $\mathcal{U} = \{ U_ i \to S\} _{i \in I}$, and $\mathcal{V} = \{ V_ j \to S\} _{j \in J}$, be fpqc-coverings of $S$. If $\mathcal{U}$ is a refinement of $\mathcal{V}$, then the pullback functor

$\text{descent data relative to } \mathcal{V} \longrightarrow \text{descent data relative to } \mathcal{U}$

is fully faithful. In particular, the category of schemes over $S$ is identified with a full subcategory of the category of descent data relative to any fpqc-covering of $S$.

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

$\xymatrix{ DD(\mathcal{V}) \ar[rd] \ar[rr] & & DD(\mathcal{U}) \ar[ld] \\ & DD(\mathcal{W}) & }$

Notation as in the proof of Lemma 35.32.9 and commutativity by Lemma 35.31.8 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 35.32.9 as desired. $\square$

Remark 35.32.12. Lemma 35.32.11 says that morphisms of schemes satisfy fpqc descent. In other words, given a scheme $S$ and schemes $X$, $Y$ over $S$ the functor

$(\mathit{Sch}/S)^{opp} \longrightarrow \textit{Sets}, \quad T \longmapsto \mathop{\mathrm{Mor}}\nolimits _ T(X_ T, Y_ T)$

satisfies the sheaf condition for the fpqc topology. The simplest case of this is the following. Suppose that $T \to S$ is a surjective flat morphism of affines. Let $\psi _0 : X_ T \to Y_ T$ be a morphism of schemes over $T$ which is compatible with the canonical descent data. Then there exists a unique morphism $\psi : X \to Y$ whose base change to $T$ is $\psi _0$. In fact this special case follows in a straightforward manner from Lemma 35.32.4. And, in turn, that lemma is a formal consequence of the following two facts: (a) the base change functor by a faithfully flat morphism is faithful, see Lemma 35.32.2 and (b) a scheme satisfies the sheaf condition for the fpqc topology, see Lemma 35.10.7.

Lemma 35.32.13. Let $X \to S$ be a surjective, quasi-compact, flat morphism of schemes. Let $(V, \varphi )$ be a descent datum relative to $X/S$. Suppose that for all $v \in V$ there exists an open subscheme $v \in W \subset V$ such that $\varphi (W \times _ S X) \subset X \times _ S W$ and such that the descent datum $(W, \varphi |_{W \times _ S X})$ is effective. Then $(V, \varphi )$ is effective.

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 35.10.6 to find an open $V_{ij} \subset V_ j$ such that $\alpha _ i^{-1}(W_ i \cap W_ j) = X \times _ S V_{ij}$. 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 35.32.12). 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 $\alpha : X \times _ S U \to V$. $\square$

Comment #3029 by Anon on

Should the section be titled "Full faithfulness..." instead of "Fully faithfulness..."?

Comment #3143 by on

This is just a terrible title anyway, so I am going to leave as is.

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