The Stacks project

Lemma 35.35.4. In the situation of Lemma 35.34.6 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 35.35.3. 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

The maps can be written as , which are clearly equal to .

Comment #4993 by Li Yixiao on

emm I made a trivial mistake...

There are also:

  • 2 comment(s) on Section 35.35: Fully faithfulness of the pullback functors

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0241. Beware of the difference between the letter 'O' and the digit '0'.