Lemma 35.35.4. In the situation of Lemma 35.34.6 assume
\{ f : X' \to X\} is an fpqc covering (for example if f is surjective, flat, and quasi-compact), and
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 35.35.3. 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
\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 35.34.6) 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 35.13.7. 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
\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 35.35.2). Hence \alpha is a morphism of descent data (V, \varphi ) \to (W, \psi ) as desired. \square
Comments (2)
Comment #4992 by Li Yixiao on
Comment #4993 by Li Yixiao on
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