5
$\begingroup$

I was unable to find a suitable answer for the following question:

Once one learns that singular cohomology is the same as cohomology with coefficients in locally constant sheaf, it is natural to try and understand the meaning of LES for relative cohomology $\rightarrow H^i(X,U) \rightarrow H^i(X) \rightarrow H^i(U) \rightarrow H^{i+1}(X,U)\rightarrow$

I am curious to what extent it can be interpreted in terms of sheaves on X using functors $j^*,j_*,j_!$?

It seems that there might be some subtleties if $X\setminus U$ is something like a closed subset in Zariski topology as opposed to "thick" closed subsets...

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

Suppose $j:U \to X$ is an open inclusion, and $i: Z \to X$ is the inclusion of the closed complement. Then we have a short exact sequence $$0 \to j_! \Bbb Z \to \Bbb Z \to i_* \Bbb Z \to 0.$$ I'm going to apply $Ext^*_X(-,{\cal F})$ to this and get a long exact sequence.

Note that $j_!$ is exact with exact right adjoint $j^*$, so that we have an isomorphism $$ Ext^p_X(j_! \Bbb Z, {\cal F}) \cong Ext^p_U(\Bbb Z, j^* {\cal F}) = H^p(U, j^* {\cal F}). $$ Moreover, $Hom(i_* \Bbb Z, -)$ is the functor of "sections supported on Z" (a composite of $i^!$ with the global section functor), whose derived functors are called cohomology groups with support on Z and denoted $H^p_Z(X, {\cal F})$.

Therefore, the long exact sequence takes the form $$ \cdots \to H^p_Z(X, {\cal F}) \to H^p(X,{\cal F}) \to H^p(U, j^*{\cal F}) \to \cdots $$ The link above will take you to the stacks project where you can find more comprehensive answers to your questions.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Thank you! I believe it might be the best answer to my question. In the resulting LES can one tautologically replace $H^p_Z(X, \cal F)$ by something of sort $H^p(X, i_*Ri^!\cal F)$? I also wonder whether anything similar is true for the RHS term $H^p(U,j^*\cal F)$, again can we see it as a sheaf on X? $\endgroup$
    – user71111
    Commented Apr 29, 2015 at 1:12
  • 2
    $\begingroup$ @user71111 Yes, the cohomology supported on $Z$ is the cohomology group you mention (because $\Bbb Z = i^* \Bbb Z$, and both $i^*$ and $i_*$ are exact). Similarly I think you can calculate $H^* (U,j^*{\cal F})$ by the cohomology of $X$ with coefficients in $(Rj_*) (j^* {\cal F})$ (because $j_*$ and $j^*$ have exact left adjoints $j^*$ and $j_!$ respectively). $\endgroup$
    – Tyler Lawson
    Commented Apr 29, 2015 at 15:43
  • 1
    $\begingroup$ I just wanted to add that there is a nice exposition of @Tyler Lawson 's answer in J. Milne's lectures on étale cohomology: jmilne.org/math/CourseNotes/LEC.pdf in section 9. $\endgroup$
    – user71111
    Commented May 10, 2015 at 18:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .