I have a projective variety $X$ and an open immersion $j : U \to X$.


Say I have a sheaf, locally free in my case of interest, $\mathcal{S}$ on $U$. Is there any reasonable relationship between $H^i(X,j\_! \mathcal{S})$ and $H^i(U,\mathcal{S})$? What if I add that I know that $H^i(U,\mathcal{S}) = 0$ for $i>0$. I'm hopeful the latter can imply that $H^i(X,j\_! \mathcal{S}) = 0 $ for $i>0$.



cohomology and $j_!$
Consider $X = \mathbb{P}^{1}$, $U =\mathbb{A}^{1}$ and j the inclusion of a affine chart in the projective line. The complement of the affine chart is a point P, let i be the inclusion of this point in the projective line.
Consider the sheaf $\mathcal{F} = \mathcal{O}(-2)$ on the projective line. One has $dim H^{1}( \mathbb{P}^{1}, \mathcal{F} ) =1$.
One has an exact sequence :


$0 \to j\_{!} ( \mathcal{F}\_{U} ) \to \mathcal{F} \to i\_{\*} ( \mathcal{F}\_{P} ) \to 0$


$i\_{\*} ( \mathcal{F}|\_{P} ) $ is a skyscraper sheaf over P so its $H^{1}$ is 0.


$H^{1}$ of $\mathcal{F}$ is not 0. By long exact sequence in cohomology, we have 
$H^{1}$ of $j\_{!}( \mathcal{F}|\_{U} )$ is not 0.


But on the other hand $\mathcal{F}|\_{U}$ is a trivial sheaf over a affine scheme so its 
$H^{1}$ is 0.


This example seems to show that the expected relation is not true.


