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Fix error in proof lemma

Thanks to 李一笑 who wrote

"In the proof of lemma 0241, one applies $\phi'$ to $(v_i,x_j)$, which
may not be a well defined $V' \times_{S'} X'$-point."

Unfortunately, the lemma wasn't fixable and we needed to change the
statement of the lemmma as well.

A counter example to the original statement of the lemma is for example
given by taking a nonempty scheme S and taking both X and S' to be two
disjoint copies of S mapping to S and finally taking both V and W to be
two copies of X mapping to X with the descent datum that produces
again two copies of S mapping to S when you descend V and W along X
mapping to S. Then you see that a morphism of descent data from V' to W'
just comes down to a morphism of two copies of S' over S' to two copies
of S' over S' which doesn't always descend to a corresponding morphism
over S... (This is just to remind me what went wrong; when I wrote this
I must have thought that the condition that X' x_{S'} X' ---> X x_S X
would prevent this kind of thing, but it was obviously garbage.)
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aisejohan committed Mar 19, 2020
1 parent 568a88e commit 3ab41432f8fbfb4286bc196ced2af82c66c45f40
Showing with 9 additions and 10 deletions.
  1. +9 −10 descent.tex
@@ -7796,8 +7796,7 @@ \section{Fully faithfulness of the pullback functors}
\begin{enumerate}
\item $\{f : X' \to X\}$ is an fpqc covering (for example if $f$ is
surjective, flat, and quasi-compact), and
\item $f \times f : X' \times_{S'} X' \to X \times_S X$ is
surjective and flat\footnote{This follows from (1) if $S = S'$.}.
\item $S = S'$.
\end{enumerate}
Then the pullback functor is fully faithful.
\end{lemma}
@@ -7809,7 +7808,7 @@ \section{Fully faithfulness of the pullback functors}
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 $S'$.
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 $\alpha'$.

@@ -7820,10 +7819,10 @@ \section{Fully faithfulness of the pullback functors}
By assumption the diagram
$$
\xymatrix{
V' \times_{S'} X' \ar[r]_{\varphi'} \ar[d]_{\alpha' \times \text{id}} &
X' \times_{S'} V' \ar[d]^{\text{id} \times \alpha'} \\
W' \times_{S'} X' \ar[r]^{\psi'} &
X' \times_{S'} W'
V' \times_S X' \ar[r]_{\varphi'} \ar[d]_{\alpha' \times \text{id}} &
X' \times_S V' \ar[d]^{\text{id} \times \alpha'} \\
W' \times_S X' \ar[r]^{\psi'} &
X' \times_S W'
}
$$
commutes. We claim the two compositions
@@ -7860,7 +7859,7 @@ \section{Fully faithfulness of the pullback functors}
((x_i, x_j), \psi(u_i, x)) = ((x_i, x_j), (x, u'_j))
$$
as points of
$(X' \times_{S'} X') \times_{X \times_S X} (X \times_S W)$
$(X' \times_S X') \times_{X \times_S X} (X \times_S W)$
for all $i, j \in \{0, 1\}$. This shows that $\psi(u_0, x) = \psi(u_1, x)$
and hence $u_0 = u_1$ by taking $\psi^{-1}$.
This proves the claim because the argument above was formal
@@ -7884,8 +7883,8 @@ \section{Fully faithfulness of the pullback functors}
W \times_S X \ar[r]^{\psi} & X \times_S W
}
$$
commutes because its base change to $X' \times_{S'} X'$
commutes and the morphism $X' \times_{S'} X' \to X \times_S X$
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 \ref{lemma-ff-base-change-faithful}).
Hence $\alpha$ is a morphism of descent data
$(V, \varphi) \to (W, \psi)$ as desired.

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