I'm having trouble understanding the construction of the Leray spectral sequence for continuous maps (not necessarily fibrations). More precisely, given a continuous map $f : X \to Y$ between two topological spaces (say as nice as we want) spaces, I would like to understand how to construct the $E_2$ term $$E_2^{pq} = H^*(Y,\mathcal{H}^*),$$ where $\mathcal{H}^*$ is the pre-sheaf $U \mapsto H^*(f^{-1}(U))$. I know a bit about sheaf cohomology, but honestly I don't really understand how this pre-sheaf appears, or why/if we need to look at its sheafification.

I have several general questions regarding all this, some being even more general:

  1. Why are we always talking about the $E_2$-term and not the $E_1$-term of a spectral sequence?
  2. From where to I need to start in order to construct this $E_2$-term ?
  3. Suppose that there exists an open cover $\{ U_i \}$ of $Y$ such that $\mathcal{H}^*(U_i) = \mathcal{G}^*(U_i)$, where $\mathcal{G}^*$ is another pre-sheaf (for instance the constant one). Under what condition could we have: $$H^*(Y,\mathcal{H}^*) = H^*(Y, \mathcal{G}^*) ?$$

Does anyone have a nice reference to these concepts (important to note that I'm no specialist in algebraic geometry :))

Thanks a lot for your help !

  • $\begingroup$ The Leray spectral sequence arises from a Grothendieck spectral sequence. You have to sheafify your presheaf in order to get the higher direct images. $\endgroup$
    – user19475
    Jun 17, 2018 at 14:57
  • $\begingroup$ Thanks for your comment @TKe. Given that the data of pre-sheave on an open cover does not uniquely determine its sheafification, is their a chance that the 3rd point above might be true, at least in some special cases ? $\endgroup$
    – BrianT
    Jun 17, 2018 at 15:49
  • $\begingroup$ @BrianT I think the 3rd point is really too much to ask for (I think counterexamples can be constructed even for very simple topological spaces $Y$, though I do not yet have a proof). Probably ideas from Cech cohomology will help you to construct a counterexamples (though you need $Y$ to be paracompact for Cech cohomology to coincide with sheaf cohomology) $\endgroup$
    – user74900
    Jun 17, 2018 at 15:59
  • $\begingroup$ Thanks @Aknazar. Actually I think (I’m really not sure) that it might be possible to use Leray’s theorem (saying that with a « nice » cover Czech cohomology is the same as Sheaf cohomology). If my open cover satisfies the conditions of this theorem, do you think it could help ? I’m not sure since I don’t really know how to relate the Leray spectral sequence to Czech cohomology. Thanks again for your help. $\endgroup$
    – BrianT
    Jun 17, 2018 at 17:57

1 Answer 1


Edited Here's a quick slightly expanded answer to the first couple of questions. First, pick a coefficient sheaf $F$, e.g. a constant sheaf for what you seem to want. To compute sheaf cohomology $H^*(X,F)$ choose a $\Gamma$-acyclic (e.g. injective, flasque, or, in good cases, fine) resolution $I^\bullet$ of $F$, then $$H^i(X,F) = H^i(\Gamma(X, I^\bullet))$$ Alternatively, resolve $f_*I^\bullet$ on $Y$ by acyclics to get a double complex $J^{\bullet\bullet}$. (You can either use a Cartan-Eilenberg resolution, or Godement's canonical flasque resolution; see Weibel's Homological Algebra for the first, or Godement's Theorie de Faisceaux for the second.) Then apply $\Gamma(Y,-)$. One of the two spectral sequences for this double complex is $$E_1= H^{p+q}(\Gamma(J^{p\bullet}))\Rightarrow H^{p+q}(Tot(\Gamma(J)))= H^{p+q}(X, F)$$ This can be shown to give the the Leray spectral sequence from the $E_2$ page. So there is an $E_1$ page but as you can see it is complicated and not so useful.

Additional Notes:

  1. I'm really just repeating what TKe said in somewhat more explicit fashion.
  2. Under appropriate conditions (e.g. if X & Y are paracompact), you can do this with Cech cohomology, but it's going to be messy since you need to choose a cover on $Y$ and then refine the preimage.
  3. Your question 3 has a negative answer: let $Y$ be a wedge of two circles, $f$ be the identity, $\{U_i\}= \{Y\}$, $F=\mathbb{Z}$, $G= F\oplus L$, where $L$ is a locally constant sheaf with nontrivial irreducible monodromy. Then $F$ and $G$ have the same sections on the cover but the $H^1$'s differ.
  • $\begingroup$ Thanks @Donu for your answer. I am not very familiar with resolutions of sheaves... How do we obtain a double complex ? And could you be more explicit about the fact that the Leray spectral sequence does not have an $E_2$ page ? Maybe you could give me a reference where this construction is made ? Also, do you know how I could relate what you said to Czech cohomology (it seems the answer to my 3. point might be in Leray’s theorem. Thanks a lot again. $\endgroup$
    – BrianT
    Jun 17, 2018 at 17:52
  • $\begingroup$ OK I expand my answer. $\endgroup$ Jun 17, 2018 at 19:16
  • $\begingroup$ Ok. Thanks a lot for your answer. In your counter example, the open cover does not satisfy Leray’s theorem condition. Do you think it would work if it was the case ? $\endgroup$
    – BrianT
    Jun 17, 2018 at 19:34

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