Let $X=Spec(R)$ be an affine scheme. Let $Y$ be a closed subset of $X$ and denote by $U$ its complement. Assume $U$ is quasicompact. Then $U= \cup\_{i=1}^{n} D(f\_i)$, where $f\_i \in R$.
Denote the inclusions by $j: U \to X$ and $i:Y \to X$. Let $F \in D\_{qcoh}(X)$, where $D\_{qcoh}(X)$ stands for the derived category of complex with quasicoherent cohomology (If you don't like $D\_{qcoh}(X)$ replace it with $D^{b}\_{qcoh}(X)$). Assume that $j^{\ast} F=0$.


Is the cohomology of $F$ $(f\_1,\ldots,f\_n)$-torsion? And how to prove it?


A qick way to get convince (not a proof). Assume that $U=D(f)$. Since $j^{\ast} G=0$ where $G$ is a quasicoherent sheaf, $j^{\ast} G=0$ is equivalent to $G(X)\_f=0$. which implies that $G(X)$ is $(f)$-torsion.


Thanks



Sheaf cohomology and torsion
The answer is yes, assuming by $(f\_1,\ldots, f\_n)$-torsion you mean that each element of each cohomology group of $F$ is killed by some power of this ideal.


There are several ways to see this. The most straightforward is probably to note that the inclusion $j\_i\colon D(f\_i) \to X$ factors via $j\colon U \to X$ so $j\_i^\*F \cong F\_{f\_i} \cong 0$ for each $i=1,\ldots,n$. Your example, the fact that localization commutes with taking cohomology, and that there are finitely many $f\_i$ shows that this is enough.


