Lemma 35.34.4. In the situation of Lemma 35.33.6 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 35.34.3. 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 35.33.2). 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 35.33.6) 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 35.12.7. Namely, $\{ V' \to V\}$ is a fpqc covering as the base change of the morphism $f : X' \to X$. Hence, by Lemma 35.12.7 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 35.34.2). Hence $\alpha$ is a morphism of descent data $(V, \varphi ) \to (W, \psi )$ as desired. $\square$

Comments (2)

Comment #4992 by Li Yixiao on

The maps $V' \times _{V} V' \to V' \to W' \to W$ can be written as $V \times _ X X' \times _ X X' \to V \times _ X X' \to W \times _ X X' \to W$, which are clearly equal to $V \times _ X (X' \times _X X') \to V \to W$.

Comment #4993 by Li Yixiao on

emm I made a trivial mistake...

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• 2 comment(s) on Section 35.34: Fully faithfulness of the pullback functors

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