This is the lemma 4.25 of Vistoli's note

Let $S$ be a scheme, $\mathscr{F} \to \mathscr{S}ch/S$ a fibred category. Then $\mathscr{F}$ is a stack over the fpqc site on $S$ iff
(1) $\mathscr{F}$ is a stack over the Zariski site on $S$, and
(2) For every fpqc morphism $V \to U$ over $S$, with $U,V$ affine, $\mathscr{F}(U) \to \mathscr{F}(V \to U)$ is equivalent.

I'm trouble in the last line at step5.

For the notation, see the pdf.

Let $f : V \to U$ be an fpqc morphism (in the sense of this note. In particular is not quasi-compact.) over $S$, and $(\eta, \phi) \in\mathscr{F}(V \to U)$. Then by Zariski descent and by the step 4, we have $\xi \in \mathscr{F}(U)$ and an isomorphism $\beta : f^* \xi \cong \eta$ in $\mathscr{F}(V)$.
Is this morphism in actually a morphism of descent data?
i.e., does this diagram $\require{AMScd}$ \begin{CD} p_2^* f^* \xi @>{p_2^* \beta}>> p_2^* \eta\\ @V{=}VV @V{\phi}VV\\ p_1^*f^*\xi @>{p_1^* \beta}>> p_1^* \eta \end{CD}

commute? (where $p_i : V \times_U V \to V$ is the $i$-th projection.)

By the construction of $\beta$, this diagram commutes on $V_i \times_{U_i} V_i$.
To show the commutativity on the whole of $V \times_U V$, by Zariski descent, we must show the commutativity on $V_i \times_U V_j$ for distinct $i,j$. But I can't.

As in step 4, if $\mathscr{F}(U_i \cup U_j) \to \mathscr{F}(V_i \cup V_j \to U_i \cup U_j)$ is equivalence, then I think the diagram is commutative. So I reduce this to the case of quasi-compact $U$.

In stack project, the aouthor ommits this important part.
And in Olsson's "Algebraic spaces and stacks" (theorem 4.3.12), the author uses (essentially) same argument, but dosn't mention this commutativity.
And in Lei Fu's "etale cohomology theory" (theorem 1.6.1), the author leaves so many parts (the whole of the lemma 4.25 of Vistoli's note) as exercises...

Any help will be appreciated!


1 Answer 1


I have understood.

First by the construction, the diagram commutes on $V' \times_{U'} V'$ for every affine open $U'$ of $U$ and its inverse image $V'$ in $V$. Now $V \times_U V = \cup V' \times_{U'} V'$, where $U'$ runs through over affine opens of $U$. So by Zariski descent, the diagram commutes on whole of $V \times_U V$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.