Lemma 51.5.8. Let $A$ be a ring and let $T, T' \subset \mathop{\mathrm{Spec}}(A)$ subsets stable under specialization. For $K \in D^+(A)$ there is a spectral sequence

$E_2^{p, q} = H^ p_ T(H^ p_{T'}(K)) \Rightarrow H^{p + q}_{T \cap T'}(K)$

as in Derived Categories, Lemma 13.22.2.

Proof. Let $E$ be an object of $D_{T \cap T'}(A)$. Then we have

$\mathop{\mathrm{Hom}}\nolimits (E, R\Gamma _ T(R\Gamma _{T'}(K))) = \mathop{\mathrm{Hom}}\nolimits (E, R\Gamma _{T'}(K)) = \mathop{\mathrm{Hom}}\nolimits (E, K)$

The first equality by the adjointness property of $R\Gamma _ T$ and the second by the adjointness property of $R\Gamma _{T'}$. On the other hand, if $J^\bullet$ is a bounded below complex of injectives representing $K$, then $H^0_{T'}(J^\bullet )$ is a complex of injective $A$-modules representing $R\Gamma _{T'}(K)$ and hence $H^0_ T(H^0_{T'}(J^\bullet ))$ is a complex representing $R\Gamma _ T(R\Gamma _{T'}(K))$. Thus $R\Gamma _ T(R\Gamma _{T'}(K))$ is an object of $D^+_{T \cap T'}(A)$. Combining these two facts we find that $R\Gamma _{T \cap T'} = R\Gamma _ T \circ R\Gamma _{T'}$. This produces the spectral sequence by the lemma referenced in the statement. $\square$

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