Lemma 48.14.2. As above, let $X$ be a scheme and let $D \subset X$ be an effective Cartier divisor. Then (48.14.1.1) combined with Lemma 48.14.1 defines an isomorphism

functorial in $K$ in $D(\mathcal{O}_ X)$.

Lemma 48.14.2. As above, let $X$ be a scheme and let $D \subset X$ be an effective Cartier divisor. Then (48.14.1.1) combined with Lemma 48.14.1 defines an isomorphism

\[ Li^*K \otimes _{\mathcal{O}_ D}^\mathbf {L} \mathcal{N}[-1] \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, K) \]

functorial in $K$ in $D(\mathcal{O}_ X)$.

**Proof.**
Since $i_*$ is exact and fully faithful on modules, to prove the map is an isomorphism, it suffices to show that it is an isomorphism after applying $i_*$. We will use the short exact sequences $0 \to \mathcal{I} \to \mathcal{O}_ X \to i_*\mathcal{O}_ D \to 0$ and $0 \to \mathcal{O}_ X \to \mathcal{O}_ X(D) \to i_*\mathcal{N} \to 0$ used in the proof of Lemma 48.14.1 without further mention. By Cohomology, Lemma 20.52.4 which was used to define the map (48.14.1.1) the left hand side becomes

\[ K \otimes _{\mathcal{O}_ X}^\mathbf {L} i_*\mathcal{N}[-1] = K \otimes _{\mathcal{O}_ X}^\mathbf {L} (\mathcal{O}_ X \to \mathcal{O}_ X(D)) \]

The right hand side becomes

\begin{align*} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ D, K) & = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}((\mathcal{I} \to \mathcal{O}_ X), K) \\ & = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}((\mathcal{I} \to \mathcal{O}_ X), \mathcal{O}_ X) \otimes _{\mathcal{O}_ X}^\mathbf {L} K \end{align*}

the final equality by Cohomology, Lemma 20.48.5. Since the map comes from the isomorphism

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}((\mathcal{I} \to \mathcal{O}_ X), \mathcal{O}_ X) = (\mathcal{O}_ X \to \mathcal{O}_ X(D)) \]

the lemma is clear. $\square$

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