# Base change in crystalline cohomology?

Does one have a base change theorem in crystalline cohomology like in étale cohomology?

Suppose one has the following cartesian diagram


Where $U,V, X, Y$ are all smooth proper scheme over $\mathbb{Z}_p$ and $h$ is embedding. Assume $g$ is a good enough map (smooth, proper, etc..). Consier the map $$h^* R^1(g_*)_{cry}(\mathcal{O}_{V})\to R^1 (f_{*})_{cry}(\mathcal{O}_U)$$ Is the above map an isomorphism (but not in the sense of derived categories)?

You finish your question by insisting that the isomorphism be such but "not in the sense of derived categories", which I do not understand completely, since both objects you have at hands naturally live in a derived category. If you are happy with that, Corollary 7.12 of Berthelot&Ogus' "Notes on Crystalline Cohomology" says that if you assume that

• $g$ is quasi-separated and smooth;
• $Y$ is quasi-compact; and, most important
• $U=X\times_Y V$,

then indeed $$\mathbb{L}h_\mathrm{cris}^*\mathbb{R}g_{\mathrm{cris}*}\mathcal{O}_V\to \mathbb{R}f_{\mathrm{cris}*}\mathcal{O}_U$$ is an isomorphism in the derived category of crystals on $X$ (a priori both terms live in the derived category of $\mathcal{O}_{X/\mathbb{Z}_p}$-modules, but Corollary 7.11 tells you that they are actually in the derived category of crystals under the above assumptions).

• @ Filippo, I mean $\mathbb{R}g_{cris*}\mathcal{O}_V$ is a complex of $\mathcal{O}_{Y/\mathbb{Z}_p}$ modules (up to quasi-isomorphic), one takes the first cohomology sheaf and regard it as a $\mathcal{O}_{Y/\mathbb{Z}_p}$ module (crystal?), and pulled back by the map $h^*_{cris}$. Then as a crystalline sheaf (crystal) is it isomorphic to the right hand side of the above map which is also consider as the first cohomology sheaf. Does it make sense?
– Lan
Jan 31, 2014 at 0:52
• Not really, in general. The problem is that objects in the derived category are not actual complexes, they can be represented by complexes but not uniquely (to form the derived category you need to localise). The result is that you lose the right of performing honest operations like "pulling back" and are left with more involved functors, like $\mathbb{L}h^*$, whose image is not a sheaf in general but can be represented by a complex of sheaves, again. Somehow, honest sheaves as complexes concentrated in one degree get lost among all possible beasts... Jan 31, 2014 at 1:31
• I am sorry to response so late, I was on travel.
– Lan
Feb 2, 2014 at 15:23