TITLE: Grothendieck spectral sequence and Mayer-Vietoris sequence
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: 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.