Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 8 down vote accepted

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 $E_1^{0,j}=H^j(U',F)\oplus H^j(U'',F)$.

share|improve this answer
    
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? –  user7316 Jul 5 '10 at 11:56
    
fs1504 -- I've added some details. –  algori Jul 5 '10 at 22:29
    
Here you can copy the tilde: Théorie –  Peter Jan 17 at 15:07

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.