Lemma 20.33.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $j : U \to X$ be an open subspace. Let $T \subset X$ be a closed subset contained in $U$.
If $E$ is an object of $D(\mathcal{O}_ X)$ whose cohomology sheaves are supported on $T$, then $E \to Rj_*(E|_ U)$ is an isomorphism.
If $F$ is an object of $D(\mathcal{O}_ U)$ whose cohomology sheaves are supported on $T$, then $j_!F \to Rj_*F$ is an isomorphism.
Proof. Let $V = X \setminus T$ and $W = U \cap V$. Note that $X = U \cup V$ is an open covering of $X$. Denote $j_ W : W \to V$ the open immersion. Let $E$ be an object of $D(\mathcal{O}_ X)$ whose cohomology sheaves are supported on $T$. By Lemma 20.32.4 we have $(Rj_*E|_ U)|_ V = Rj_{W, *}(E|_ W) = 0$ because $E|_ W = 0$ by our assumption. On the other hand, $Rj_*(E|_ U)|_ U = E|_ U$. Thus (1) is clear. Let $F$ be an object of $D(\mathcal{O}_ U)$ whose cohomology sheaves are supported on $T$. By Lemma 20.32.4 we have $(Rj_*F)|_ V = Rj_{W, *}(F|_ W) = 0$ because $F|_ W = 0$ by our assumption. We also have $(j_!F)|_ V = j_{W!}(F|_ W) = 0$ (the first equality is immediate from the definition of extension by zero). Since both $(Rj_*F)|_ U = F$ and $(j_!F)|_ U = F$ we see that (2) holds. $\square$
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