**Proof.**
Let $E$ be an object of $D(\mathcal{O}_ X)$ whose cohomology sheaves are supported on $T$. Then we see that $E|_ U = 0$ and $E|_{U \times _ X V} = 0$ as $T$ doesn't meet $U$ and $j^{-1}T$ doesn't meet $U \times _ X V$. Thus (1)(a) follows from Lemma 75.10.2. In exactly the same way (1)(b) follows from Lemma 75.10.1.

Let $F$ be an object of $D(\mathcal{O}_ V)$ whose cohomology sheaves are supported on $j^{-1}T$. By Lemma 75.3.1 we have $(Rj_*F)|_ U = Rj_{W, *}(F|_ W) = 0$ because $F|_ W = 0$ by our assumption. Similarly $(j_!F)|_ U = j_{W!}(F|_ W) = 0$ by Lemma 75.10.6. Thus $j_!F$ and $Rj_*F$ are supported on $T$ and $(j_!F)|_ V$ and $(Rj_*F)|_ V$ are supported on $j^{-1}(T)$. To check that the maps (2)(a), (b), (c) are isomorphisms in the derived category, it suffices to check that these map induce isomorphisms on stalks of cohomology sheaves at geometric points of $T$ and $j^{-1}(T)$ by Properties of Spaces, Theorem 66.19.12. This we may do after replacing $X$ by $V$, $U$ by $U \times _ X V$, $V$ by $V \times _ X V$ and $F$ by $F|_{V \times _ X V}$ (restriction via first projection), see Lemmas 75.3.1, 75.10.6, and 75.9.2. Since $V \times _ X V \to V$ has a section this reduces (2) to the case that $j : V \to X$ has a section.

Assume $j$ has a section $\sigma : X \to V$. Set $V' = \sigma (X)$. This is an open subspace of $V$. Set $U' = j^{-1}(U)$. This is another open subspace of $V$. Then $(U' \subset V, V' \to V)$ is an elementary distinguished square. Observe that $F|_{U'} = 0$ and $F|_{V' \cap U'} = 0$ because $F$ is supported on $j^{-1}(T)$. Denote $j' : V' \to V$ the open immersion and $j_{V'} : V' \to X$ the composition $V' \to V \to X$ which is the inverse of $\sigma $. Set $F' = \sigma ^*F$. The distinguished triangles of Lemmas 75.10.1 and 75.10.2 show that $F = j'_!(F|_{V'})$ and $F = Rj'_*(F|_{V'})$. It follows that $j_!F = j_!j'_!(F|_{V'}) = j_{V'!}F = F'$ because $j_{V'} : V' \to X$ is an isomorphism and the inverse of $\sigma $. Similarly, $Rj_*F = Rj_*Rj'_*F = Rj_{V', *}F = F'$. This proves (2)(c). To prove (2)(a) and (2)(b) it suffices to show that $F = F'|_ V$. This is clear because both $F$ and $F'|_ V$ restrict to zero on $U'$ and $U' \cap V'$ and the same object on $V'$.
$\square$

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