Question

Is there a relative version of sheaf cohomology? EDIT: I rather mean the cohomology of pairs.

Answer 1

It turns out that my previous answer dealt with the wrong question. The answer to the new question is also yes: local cohomology $H_Z(X,\mathcal F)$ corresponds to cohomology of the pair $(X,X\setminus Z)$.

Answer 2

To elaborate on Kevin's comment: If $f: X \to S$, and $\mathcal F$ is a sheaf on $X$, then $f_*\mathcal F$ is the sheaf on $S$ defined by $H^0(U,f_*\mathcal F) := H^0(f^{-1}(U),\mathcal F).$

Taking the derived functors of $f_*$ gives functors $R^if_*$, and it turns out (fairly easily) that $R^if_*(\mathcal F)(U)$ is the sheaf associated to the presheaf $U \mapsto H^i(f^{-1}(U),\mathcal F)$. If $i > 0,$ then this presheaf may not be a sheaf (unlike the $i = 0$ case), and this is related to the fact that it can be a little subtle to compute the stalks of $R^if_*\mathcal F$ in general; for example, it need not be the case in general that the stalk $(R^if_*\mathcal F)_s$ is equal to $H^i(f^{-1}(s),\mathcal F)$. (E.g. think about the case when $f$ is the inclusion of a punctured disk into a disk, $\mathcal F$ is the constant sheaf ${\mathbb Z}$, and $s$ is the centre of the disk (so that $f^{-1}(s)$ is empty).)

In other words, $R^if_*\mathcal F$ does not always literally interpolate the cohomology of the fibres.

There is one case where one knows that $R^if_*\mathcal F$ does interpolate the cohomology of the fibres: if the map $f$ is proper, than the proper base-change theorem says that the stalk of $R^if_*\mathcal F$ at $s$ is the cohomology of $\mathcal F$ along the fibre of $s$. (One good place for these kinds of facts is the beginning of Borel's book on Intersection Cohomology.)

Also, in the context of maps of varieties, if $f$ is proper and $\mathcal F$ is a coherent sheaf, then the completed stalk of $R^if_* \mathcal F$ at $s$ coicides with the cohomology of the pull-back of $\mathcal F$ to the formal completion of $f^{-1}(s)$ in $X$. (This is Grothendieck's proper base-change theorem, proved in some form in Hartshorne, Ch. III.)

Answer 3

I would think -- but this is for the moment a thought, not meant as an authorative answer -- that every notion of cohomology whatsoever has a relative version in this sense.

I am thinking here of the general abstract definition of cohomology as exposed at nLab:cohomology. This includes in particular the special case of sheaf cohomology as described there in some detail.

What I expect the fully general notion of relative cohomology from this point of view to be I have now briefly indicated at relative cohomology. I'd believe this does in particular reproduce the definition given in Emerton's answer.