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Suppose $U'\cup U''=X$ is an open cover $U$ of a topological space $X$ and $F$ is a sheaf on $X$ with values in abelian groups. There is a special instance of the Grothendieck spectral sequence relating Cech to sheaf cohomology:

$$E_2^{p,q}=\tilde{H}^p(U,H^q(-,F))\Rightarrow H^{p+q}(X,F)$$

I would like to see, how this implies the Mayer-Vietoris sequence for this easy cover $U$. Drawing the $E_2$-page, I get so far that only the first two columns $p=0,1$ are non-zero. Therefore this page equals the $E_\infty$-page.

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2 Answers 2

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Recall that the Cech-to-derived functor spectral sequence is constructed as follows. We start with a sheaf $F$ and an open cover $\mathfrak{U}$. Then we can write the Cech resolution of the sheaf; take an injective (or Godement or...) resolution thereof to get a double complex. Let $C^{\ast,\ast}$ be the resulting complex of global sections and take the filtration $F^i=\bigoplus C^{\geq i,\ast}$. See e.g. Godement, Th\'eorie des faisceaux, 5.2. The rows of the $E_1$ sheet are precisely the Cech cochain complexes constructed from the open cover $\mathfrak{U}$ and the presheaves $U\mapsto H^i(U,F)$ (see Godement, ibid, just before theorem 5.2.4).

If $\mathfrak{U}$ has just two elements, $U$ and $U''$, then the $E_1$ term has two columns, the 0-th and the 1-st ones. Applying e.g. theorem 4.6.1 from Godement, ibid, one gets the long exact sequence

$$\cdots\to E_1^{1,i-1}\to H^i(X,F)\to E_1^{0,i}\to E^{1,i}_1\to\cdots$$

where the last arrow is the $d_1$ differential, $E_1^{1,j}=H^j(U'\cap U'',F)$ and $E_1^{0,j}=H^j(U',F)\oplus H^j(U'',F)$.

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  • $\begingroup$ How does the $E_1$ page of the above spectral sequence look like? Does it exist or does one have to consider the double complex spectral sequences? $\endgroup$
    – user7316
    Jul 5, 2010 at 11:56
  • $\begingroup$ fs1504 -- I've added some details. $\endgroup$
    – algori
    Jul 5, 2010 at 22:29
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Here is a slightly different argument than algori's, not using the construction of the Čech-to-derived functor spectral sequence and only using $E_2$ terms, not $E_1$ terms.

As you say, the spectral sequence is given by $$ E_2^{pq} = \check H^p(\mathcal{U},H^q(-,F)) \Longrightarrow H^{p+q}(X,F). $$ Since the covering $\mathcal{U}$ only consists of two open sets – $A$ and $B$, say – the $E_2$ page looks like this:

  • $E_2^{pq}$ is zero for $p \geq 2$.
  • $E_2^{0q}$ equals the kernel of the map $H^q(A,F) \oplus H^q(B,F) \to H^q(A \cap B,F)$ which sends $(s,t)$ to $t|_{A \cap B} - s|_{A \cap B}$. (Use the alternating Čech complex.)
  • $E_2^{1q}$ is the cokernel of that map.

Therefore the spectral sequence degenerates on the $E_2$ page. Now consider, for any $n \geq 0$, the canonical short exact sequence $$ 0 \longrightarrow F^1 E_\infty^n \longrightarrow E_\infty^n \longrightarrow E_\infty^n/F^1 E_\infty^n \longrightarrow 0. $$ Since $F^2 E_\infty^n = F^3 E_\infty^n = \cdots = 0$ and $F^0 E_\infty^n = F^{-1} E_\infty^n = \cdots = E_\infty^n$ (by the vanishing of most columns, see for instance these notes by Matthew Greenberg), we can express the outer terms of this sequence in $E_2$ terms: $$ 0 \longrightarrow E_2^{1,n-1} \longrightarrow E_\infty^n \longrightarrow E_2^{0,n} \longrightarrow 0, $$ i.e. $$ 0 \longrightarrow \mathrm{cok}(H^{n-1}(A) \oplus H^{n-1}(B) \to H^{n-1}(A \cap B)) \longrightarrow H^n(X) \longrightarrow \mathrm{ker}(H^n(A) \oplus H^n(B) \to H^n(A \cap B)) \longrightarrow 0.$$ We can splice these short exact sequences to obtain the long exact Mayer–Vietoris sequence.

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