5
$\begingroup$

I am reading M. Artin's treatment of the proper base change theorem for étale cohomology in his "Théorèms de représentabilité pur les espaces algébriques", and I have trouble understanding the following remark on page 222:

If $f:X\rightarrow S$ and $g:S'\rightarrow S$ are morphisms of algebraic spaces (or schemes, if you prefer), and if $f':X'\rightarrow S'$, $g':X'\rightarrow X$ denote the base changes of $f$ and $g$, then one can construct for any abelian sheaf $F$ on the big étale site of $X$ the base change morphism $g^*R^qf_*F\rightarrow R^q f'_*(g'^*F)$ (the higher direct images also computed on the big sites). If I understand correctly, Artin claims that if $F$, $R^q f_*F$ and $R^qf'_*(g'^*F)$ are representable on the big étale site of $X$, resp. $S$, resp. $S'$ (i.e. locally constructable), then the base change morphism is an isomorphism.

Why is that? Is that an easy fact?

$\endgroup$

1 Answer 1

3
$\begingroup$

With help from Milne's book on étale cohomology, I figured out how answer the question, although I am not sure that this argument is what Artin had in mind, and I still think that there's an easier argument.

There are morphisms of topoi $\pi_X: X_{ET}\rightarrow X_{et}$ from the topos associated to the big étale site of $X$ to the topos of the small étale site. Similarly for $S$, and $f:X\rightarrow S$ induces $f^s:X_{et}\rightarrow S_{et}$ and $f^b:X_{ET}\rightarrow S_{ET}$, and the obvious diagram commutes, i.e. $f^s\pi_X=\pi_Sf^b$. Given a sheaf $F$ in $S_{ET}$ we get a base change morphism $$ \pi_S^*R^qf^s_*F\rightarrow R^qf_*^b\pi_X^*F$$ Milne calls this "universal base change morphism", for good reasons: Given any morphism $g:S'\rightarrow S$, you also get morphisms of topoi $g^b:S'_{ET}\rightarrow S_{ET}$ and ${g'}^b:X'_{ET}\rightarrow X_{ET}$. Using this to restrict the universal base change morphism to $S'_{ET}$, we get the usual base change morphism for $g$ and $F$. (For this one has to check the commutativity of a few diagrams. All the ingredients can be found, e.g., in great detail in the Stacks Project)

Now, if $F$ is locally constructible, i.e. if the adjunction map $F\rightarrow \pi_X^*\pi_{X,*} F$ is an isomorphism, then it is not hard to check that the universal base change morphism is an isomorphism, and thus every base change morphism is an isorphism.

$\endgroup$

Your Answer

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

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