## Special case of Leray spectral sequence

I am looking for a reference for what is stated in Srinivasan's book "Representations of Finite Chevalley Groups", which is apparently a special case of Leray spectral sequence. I'll quote the interesting part :

Let $\phi$ be a torsion sheaf on $X$, $f : X \to Y$ a morphism of schemes. Then we have a Leray spectral sequence $E_2^{pq} = H^p_c(Y, R^q f_{!} \phi ) \Rightarrow H^{p+q}_c(X, \phi)$. The way we use this is as follows.

Suppose that all the fibres of $f$ are isomorphic to a fixed scheme $Z$ such that $H^q_c(Z, \phi) = 0$ except for $q= q_0$. Then $H^p_c(Y, R^{q_0} f_{!} \phi ) \simeq H^{p+q}_c(X, \phi)$.

The reference I'm looking for is for the second part of what I quoted.

Since I just happened to work with spectral sequences, maybe this statement is obviously equivalent to the definition of convergence of spectral sequences, in which case I'm sorry for asking.

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Here is a small remark: this statement is false without additional assumptions. Take e.g. the map $\mathbb{A}^1\setminus\{0\}\sqcup \{0\}\to \mathbb{A}^1$. All closed fibers are points, but the cohomology of the target is not the same as the cohomology of the source. – algori May 31 2011 at 15:26
.. However, it is true, as Dan says in his answer, that if the sheaf $R^{q_0}f_! \phi$ is constant, then everything works fine. – algori May 31 2011 at 15:29

## 1 Answer

The assumption on the fibers is equivalent to that $R^qf_! \phi$ vanishes unless $q = q_0$. Hence there is only one nonzero row in the $E_2$ page of the spectral sequence, which happens so often that there is a special name for this situation: the spectral sequence collapses at $E_2$. Collapsing is stronger than both degenerating and converging. In particular, if you know that a spectral sequence collapses, then you can read off the actual cohomology groups that the sequence converges to and not just their respective associated gradeds.

All this follows because no maps after the $E_1$ page send the nonzero row to itself, so there can be no nonzero differentials; moreover, the filtration on each $H^i$ of the $H^\ast$ that the sequence converges to has at most one step (corresponding to the nonzero row), so each $H^i$ is simply given by the corresponding entry in the nonzero row. This should be explained in any reference on spectral sequence, e.g. Weibel's "An introduction to homological algebra".

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