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 20.54.4 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 20.50.5. 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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)