Lemma 52.15.10. Let $I \subset \mathfrak a \subset A$ be ideals of a Noetherian ring $A$. Let $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Let $\mathcal{V}$ be the set of open subschemes of $U$ containing $U \cap V(I)$ ordered by reverse inclusion. Let $\mathcal{F}$ and $\mathcal{G}$ be coherent $\mathcal{O}_ V$-modules for some $V \in \mathcal{V}$. The map

$\mathop{\mathrm{colim}}\nolimits _{V' \geq V} \mathop{\mathrm{Hom}}\nolimits _ V(\mathcal{G}|_{V'}, \mathcal{F}|_{V'}) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{\textit{Coh}(U, I\mathcal{O}_ U)}(\mathcal{G}^\wedge , \mathcal{F}^\wedge )$

is bijective if the following assumptions hold:

1. $A$ is $I$-adically complete and has a dualizing complex,

2. if $x \in \text{Ass}(\mathcal{F})$, $x \not\in V(I)$, $\overline{\{ x\} } \cap V(I) \not\subset V(\mathfrak a)$ and $z \in \overline{\{ x\} } \cap V(\mathfrak a)$, then $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) > \text{cd}(A, I) + 1$.

Proof. We may choose coherent $\mathcal{O}_ U$-modules $\mathcal{F}'$ and $\mathcal{G}'$ whose restriction to $V$ is $\mathcal{F}$ and $\mathcal{G}$, see Properties, Lemma 28.22.5. We may modify our choice of $\mathcal{F}'$ to ensure that $\text{Ass}(\mathcal{F}') \subset V$, see for example Local Cohomology, Lemma 51.15.1. Thus we may and do replace $V$ by $U$ and $\mathcal{F}$ and $\mathcal{G}$ by $\mathcal{F}'$ and $\mathcal{G}'$. Set $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{G}, \mathcal{F})$. This is a coherent $\mathcal{O}_ U$-module. We have

$\mathop{\mathrm{Hom}}\nolimits _ V(\mathcal{G}|_ V, \mathcal{F}|_ V) = H^0(V, \mathcal{H}) \quad \text{and}\quad \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{H}/\mathcal{I}^ n\mathcal{H}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Coh}(U, I\mathcal{O}_ U)} (\mathcal{G}^\wedge , \mathcal{F}^\wedge )$

See Cohomology of Schemes, Lemma 30.23.5. Thus if we can show that the assumptions of Proposition 52.12.2 hold for $\mathcal{H}$, then the proof is complete. This holds because $\text{Ass}(\mathcal{H}) \subset \text{Ass}(\mathcal{F})$. See Cohomology of Schemes, Lemma 30.11.2. $\square$

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