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This is a rewrite of a previous question, which was in turn a follow up question to Leray-Hirsch principle for étale cohomology The motivation is to clarify some points in Torsten Ekedahl's answer there. (Some time ago I also posted a brief version as a comment there.)

Let $X\to Y$ be a surjective morphism of connected algebraic varieties over an algebraically closed field $k$ and let $F$ be a constructible sheaf on $X$ whose stalks are of order prime to $char (k)$. For any closed geometric point $y\in Y$ we have the natural base change map $(R^i f_*F)_y\to H^i(X_y,F)$ where $X_y$ is the fiber over $y$.

Can one deduce that all $R^i f_{\ast} F,i>0$ are zero assuming that for each closed $y$ we have $H^i(X_y,F)=0,i>0$? If the answer is yes, I would be interested in knowing whether there is an analog of this for the $l$-adic cohomology with $\mathbf{Q}_l$ coefficients.

[upd: nope, as per Dustin's comment below.]

Note that in the \'etale case Proposition 4.12, Chapter VI in Milne, \'Etale cohomology gives a stronger conclusion ($H^{\ast}(X,F')\to H^{\ast}(Y,f^{-1}F')$ is an iso for any $F'$) under a stronger hypothesis that the $H^{\ast}(X_y,F'')$ vanishes in positive degrees for any geometric point $y$, which may or may not be closed, and for all $F''$.

So here is a side-question: does it suffice to check the vanishing for closed points?

At first I thought the answer would be yes, which is why I accepted Torsten's answer to the above question, but then I realized I don't know how to prove this.

[upd: .. and for a good reason since it's false.]

Remark: if $f$ is proper, the statement is true by the proper base change theorem, but as pointed out by Dustin Clausen in the above thread, assuming $f$ smooth does not help. In fact, if $f:\mathbf{A}^2\setminus 0 \to \mathbf{A}^1$ is the projection to one of the coordinate axes, then (unless I'm mistaken) $R^3 f_*\underline{A}$ is non-zero at the origin (where $A$ is a finite group of order prime to $char(k)$), but the cohomology of the preimage of the origin vanishes in degrees $>1$.

Here is some more motivation and one more question. If $G$ is a Lie group that acts on a sufficiently nice topological space $X$, then the stack cohomology of the quotient is simply the cohomology of the Borel construction $X\times EG/G$ where $EG$ is the universal $G$-bundle and the action is diagonal. The Borel construction is mapped to $X/G$ (assuming the topological quotient is reasonable) and the fiber over $[x]\in X/G,x\in X$ is the classifying space of the stabilizer of $x$. So if the action is with finite stabilizers, $H^{\ast}(X/G,\mathbf{Q})$ is isomorphic to the cohomology of the Borel quotient. One can try to mimick this in the \'etale setting, but there are some technical difficulties: in the topological case there are some tools (maximal compact subgroup, the existence of slices for compact group actions, ...) which don't seem to have easy analogs.

However, I would still guess that if $G$ is an algebraic group that acts with finite stabilizers on an algebraic variety $X$ (I'd be willing to assume it smooth but not affine), the cohomology of the quotient with $\mathbf{Q}_l$ coefficients should be the same as the cohomology of the Borel construction and it seems plausible that someone has looked into this before.

So I would like to ask: is there a reference for that?

(Note that although $EG$ does not exist as an algebraic variety, it can be approximated by algebraic varieties: e.g. if $G=GL_n(k)$, one can take the spaces of $n$-frame bundles in $k^N,N\to\infty$.)

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Dear algori, note that I extended my answer to the question you refer to to give a proof of the comparison with the cohomology of the quotient and the Borel quotient (which is just the topological version of the stack quotient) which proves it directly bypassing your other questions here. I am not sure if you missed that or if there is something you do not like with it (in which case I will try to answer any problems you find with it). Your other questions are of course legitimate in themselves. – Torsten Ekedahl Sep 20 '10 at 20:18
Dear Torsten -- thanks a lot, I must have missed it. Will take a look at the addendum. – algori Sep 20 '10 at 20:24
I think the answer to the first question is no: let X be the blow-up of A^2 at the origin, except remove a point in the exceptional divisor. Then take X --> A^2 and let F be constant. The problem is that both the base and the fibers of this map are contractible, but the total space has higher homology. – Dustin Clausen Sep 20 '10 at 21:34
Dustin -- thanks! I agree with that. – algori Sep 20 '10 at 22:00
algori -- glad you agree; i was kind of unsure because it seems to contradict the Leray-Hirsch statement made in the linked thread. what's your take? – Dustin Clausen Sep 21 '10 at 1:13

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