Let $A \subset X$ and $B \subset Y$ be topological spaces, with $A$ and $B$ closed, and $f: X \to Y$ be a continuous functions such that $f(A) \subset B$.

The Leray spectral sequence (with complex coefficients) associated with $f$ has $E_2$-term $$E_2^{pq} = H^p(Y,\mathcal{J}^q),$$ where $\mathcal{J}^q$ is the sheafification of the presheaf $$U \to H^q(f^{-1}(U), \mathbb{C}), \quad \forall U \subset Y \text{ open},$$ and converges to $H^*(X,\mathbb{C})$.

In a similar manner, one can construct the Leray spectral sequence associated with $f: (X,A) \to (Y,B)$, seen as a continuous function between pairs. My guess is that it should have $E_2$-term $$E_2^{pq} = H^p(Y,B ; \mathcal{J}^q),$$ where we are now considering local cohomology.

My problem is that I'm not sure to see what the sheaf $\mathcal{J}^q$ looks like in this case. A natural guess would be that it associates to any open set $U \subset Y$ the relative cohomology group $H^q(f^{-1}(U), f^{-1}(U \cap B) ; \mathbb{C})$. In such case, we would have $$\mathcal{J}^q(U) = 0$$ for any open set $U \subset X$ such that $U \subset B$. In particular, the stalk above any point of $B$ would be $0$, which seems very strange.

Could someone help ?

Thanks a lot !


1 Answer 1


Let me first consider the easier case of the Serre spectral sequence. So we assume $f:X \to Y$ is a fibration with fiber $F$, and that $f^{-1}(B)=A$. Then there is a "relative" Serre spectral sequence $$ E_2^{pq} = H^p(Y,B;H^q(F,\mathbf Z)) \implies H^{p+q}(X,A;\mathbf Z).$$ The proof of this can be found in many places (I'm quite sure it's in Spanier for example). It can be proven by a modification of the standard cellular proof of the Serre spectral sequence.

Now suppose you want to work in a sheaf-theoretic world instead, to get the genuine Leray spectral sequence. If you've been fully "six-functor"-ified then relative cohomology is somehow not so natural: rather than considering the relative cohomology groups $H^\ast(X,A;F)$ one can instead consider $H^\ast(X,j_!j^\ast F)$ where $j \colon (X\setminus A) \to X$ is the open inclusion. So to get Leray spectral sequence converging to the relative groups $H(X,A;F)$ just apply the usual Leray spectral sequence to the sheaf $j_!j^\ast F$ instead.

  • $\begingroup$ Thanks a lot @Dan. If I understand you well, the $E_2$-page of the Leray spectral sequence would, in the case, be: $$E_2^{pq} = H^p(Y, R^q f_* (j_{!} j^* F)),$$ where $f : X \to Y$ is our continuous map sending $A$ to $B$. How is it that $B$ doesn't appear anywhere here ? Thanks a lot. $\endgroup$
    – BrianT
    Jun 25, 2018 at 9:07
  • $\begingroup$ Well, the spectral sequence you get doesn't depend on B. It depends on X, Y and A though. Let me remark that there can't be a sequence exactly like the one you wrote, where you can take for B any closed subset containing the image of A. For then you could take B=Y and you'd get the E2-page identically zero, but converging to something nontrivial... $\endgroup$ Jun 25, 2018 at 17:11
  • $\begingroup$ Thank you very much for your help @Dan, things are much clearer now. $\endgroup$
    – BrianT
    Jun 25, 2018 at 17:22

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