The Stacks project

Proof. Equivalently, we are saying that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, \mathcal{O}_ X)$ has a unique nonzero cohomology sheaf in degree $1$ and that this sheaf is isomorphic to $\mathcal{N}$. Since $i_*$ is exact and fully faithful, it suffices to prove that $i_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, \mathcal{O}_ X)$ is isomorphic to $i_*\mathcal{N}[-1]$. We have $i_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, \mathcal{O}_ X) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ D, \mathcal{O}_ X)$ by Lemma 48.9.3. We have a resolution

where $\mathcal{I}$ is the ideal sheaf of $D$ which we can use to compute. Since $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ X, \mathcal{O}_ X) = \mathcal{O}_ X$ and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{I}, \mathcal{O}_ X) = \mathcal{O}_ X(D)$ by a local computation, we see that

where on the right hand side we have $\mathcal{O}_ X$ in degree $0$ and $\mathcal{O}_ X(D)$ in degree $1$. The result follows from the short exact sequence

coming from the fact that $D$ is the zero scheme of the canonical section of $\mathcal{O}_ X(D)$ and from the fact that $\mathcal{N} = i^*\mathcal{O}_ X(D)$. $\square$