18
$\begingroup$

Let $X$ be a topological space and $U$ an open cover of $X$.

In this thread Angelo explained beautifully how presheaf cohomology (Cech cohomology) relates to sheaf cohomology:

The zeroth Cech cohomology functor $\tilde H^0(U,-):Pre(X)\to Ab$ from presheaves on $X$ to abelian groups is left exact and its right derived functors coincide with the cohomology $\tilde H^n(U,-)$ of the chech complex. So one may interpret Cech cohomology as derived presheaf cohomology.

On the other hand the inclusion $i:Shv(X)\to Pre(X)$ of sheaves on $X$ into presheaves is left exact and the diagram [ \begin{array}{rcl} Shv(X)&\xrightarrow{\Gamma_X}&Ab\\ i\searrow&&\nearrow \tilde H^0(U,-)\\ &Pre(X)& \end{array} ] commutes. Let $F\in Shv(X)$ be a sheaf. The derived functors $H^n(X,F)$ of the left exact functor $\Gamma_X$ are called sheaf cohomology. They are in general not equal to the derived functors $\tilde H^n(U,-)$ (Cech cohomology). The relation between the two is the spectral sequence

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

induced by the Grothendieck spectral sequence.

  • What is the general picture behind this?
  • In this thread it is explained why the presheaf $H^q(-,F)$ in the spectral sequence sheafifies to zero for $q\geq 1$. How can one interpret this?
  • For $X$ locally contractible and $F$ the sheaf of localy constant functions, sheaf cohomology equals singular cohomology and for a cover $U$ with two open sets the spectral sequence is just the Mayer Vietoris sequence. How does this fit into the general picture?
$\endgroup$
1
  • 1
    $\begingroup$ I don't really understand the questions (what kind of general picture would you like to have ? categorical one ?) but I would advice you to read [this note][1]. [1]: therisingsea.org/notes/SpectralSequences.pdf $\endgroup$
    – DamienC
    Aug 11, 2010 at 7:51

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.