MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $p:P\to S$ (and $p':P\to S$) be proper smooth morphisms of 'nice' schemes (one may assume that $S$ is a complex variety). It is well-known that the fibres of $p$ (and $p'$) over all geometric points of $S$ have isomorphic etale cohomology. Now, let $f:P\to P'$ be an $S$-morphism (that is certainly proper, but not necessarily smooth). My question is: is $f$ necessarily compatible with the isomorphisms mentioned?

It suffices to consides two points $s_{0,1}$ of $S$, $s_1$ is a generic point. As far as I understand the proof, the isomorphisms mentioned are obtained by applying cohomology to the base change via $p$ and $p'$ of the natural morphisms $s_i\to Spec O^h_{s_0}$ ($i=0,1$, $s_i$ are the corresponding spectra of fields, $O^h_{s_0}$ is the henselization of $S$ at $s_0$). If this is correct, then the answer to my question should be positive for obvious reasons. I have serious doubts here since this seems to contradict some other well known facts.

Any hints or references would be very welcome!

share|cite|improve this question
It looks correct; what would it contradict? Let $U$ be a contractible subset of $S$ By Ehresmann fibration theorem, $P_U$ has the homotopy type of the fiber $P_s=p^{-1}(s)$ for every $s\in S$ and the injection $P_s\to P_U$ induces an isomorphism on cohomology. Similarly $P'_s\to P'_U$ induces an isomorphism on cohomology. These isomorphisms are those used to compare the various cohomologies of the fibers (for $s\in U$). Then the morphisms $P_s\to P'_s$ and $P_U\to P'_U$ induce the same morphism on cohomology. – ACL Jan 3 '13 at 11:26

The affirmative answer involves just going back to how the comparison isomorphism between cohomologies of different geometric fibers are defined by means of specialization maps for higher direct image sheaves on the base space, making sure we are using the "same" strict henselizations to make the specialization maps on the two sides (for $p$ and $p'$), and noting that such specialization maps for a fixed strict henselization are functorial in the sheaf on the base space.

To be precise, let $p:P \rightarrow S$ and $p':P' \rightarrow S$ be scheme morphisms and $f:P' \rightarrow P$ an $S$-morphism. Choose an etale abelian sheaf $F$ on $P$. For its pullback $F' = f^{\ast}F$ on $P'$ there are naturally induced "$f$-pullback" maps $\theta^i_F: R^i p_{\ast}(F) \rightarrow R^i p'_{\ast}(F')$ (in the style of "base change" morphisms, applied to the typically non-cartesian but commutative square with bottom side the identity map of $S$, top side $f$, and left and right sides $p'$ and $p$ respectively). Having set up this notation, put it to the side for a moment.

For any etale abelian sheaf $G$ on $S$ and points $s, \eta \in S$ with $s$ a specialization of $\eta$, upon choosing a geometric point $\overline{s}$ over $s$ and a geometric point $\overline{\eta}$ over $\eta$ on Spec($O_{S,\overline{s}}^{\rm{sh}}$)` we have a natural map ${\rm{sp}}_{\overline{s},\overline{\eta},G}:G_{\overline{s}} \rightarrow G_{\overline{\eta}}$. In the above setting, if $p$ is a smooth proper surjection and $F$ is lcc with torsion orders invertible on $P$ then $R^i p_{\ast}F$ is an lcc sheaf (smooth and proper base change theorems) and this specialization map for $G = R^i p_{\ast}(F)$ on $S$ is an isomorphism. That isomorphism is by definition the comparison isomorphism between cohomologies of $F$ on the geometric fibers at $\overline{s}$ and $\overline{\eta}$ (isomorphism depending on the "choice" of strict henselization).

So ultimately the affirmative answer to your question (implicitly assuming we are using a single such choice of strict henselization underlying the fibral comparison isomorphisms for both $p$ and $p'$!) amounts to the observations that (i) the formation of the specialization map ${\rm{sp}}_{\overline{s},\overline{\eta},G}$ for general $G$ is functorial in $G$ with respect to any morphism between abelian etale sheaves on $S$ (such as the maps $\theta^i_F$), (ii) in the special case that $S$ is a geometric point, $\theta^i_F$ is the natural pullback map in degree-$i$ cohomologies associated to $F$ and $F'$.

As this argument makes plain, we have to use the same strict henselization on both sides or else it all breaks down. In terms of ACL's comment above in the topological analogue over the complex numbers (using constant coefficients), this corresponds to identifying the cohomologies on fibers over the various points of the contractible $U$ via their pullback identifications with the cohomologies on the total spaces $P_U$ and $P'_U$ over $U$ (i.e., the choice of strict henselization in the algebraic theory plays the role of $U$ in the topological case). That is, if we use a different contractible open $V$ in $S$ containing the same pair of points $s_1, s_2 \in S$ then we generally get a different isomorphism between cohomologies on the $s_i$-fibers (for both $P$ and $P'$), and we certainly wouldn't claim any compatibility for such comparison isomorphisms if we use $U$ for the definition of fibral comparison for $P$ and use $V$ for the definition of fibral comparison for $P'$.

share|cite|improve this answer
Thank you! Could you tell me, where can I read more about specialization maps (and about your observation (ii))? – Mikhail Bondarko Jan 3 '13 at 17:37
The specialization maps are explained very nicely in the book of Freitag and Kiehl (and in many other books on etale cohomology). As for (ii), it is just carefully thinking through how $\theta^i_F$ is defined and how pullback in sheaf cohomology is defined (using $\delta$-functorial considerations, allowing for rather general $F$ and hence degree-shifting, etc.), so I don't have a reference to point too (there many such compatibilities and concrete incarnations that one is confronted with when learning etale cohomology, and I don't know what to say other than that one has to think about it). – user30379 Jan 3 '13 at 19:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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