The Stacks project

Proof. In (the proof of) Cohomology of Schemes, Lemma 30.8.4 we constructed a canonical isomorphism

Moreover, $Rf_*(f^*\mathcal{L}(-n - 1))[n] = R^ nf_*(f^*\mathcal{L}(-n - 1))$, i.e., the other higher direct images are zero. Thus we find an isomorphism

This isomorphism determines $c$ as in the statement of the lemma because $a$ is the right adjoint of $Rf_*$. By Lemma 48.4.4 construction of the $a$ is local on the base. In particular, to check that $c$ is an isomorphism, we may work locally on $Y$. In other words, we may assume $Y$ is affine and $\mathcal{E} = \mathcal{O}_ Y^{\oplus n + 1}$. In this case the sheaves $\mathcal{O}_ X, \mathcal{O}_ X(-1), \ldots , \mathcal{O}_ X(-n)$ generate $D_\mathit{QCoh}(X)$, see Derived Categories of Schemes, Lemma 36.16.3. Hence it suffices to show that $c : \mathcal{O}_ X(-n - 1)[n] \to a(\mathcal{O}_ Y)$ is transformed into an isomorphism under the functors

for $i \in \{ -n, \ldots , 0\} $ and $p \in \mathbf{Z}$. For $F_{0, p}$ this holds by construction of the arrow $c$! For $i \in \{ -n, \ldots , -1\} $ we have

by the computation of cohomology of projective space (Cohomology of Schemes, Lemma 30.8.1) and we have

because $Rf_*\mathcal{O}_ X(i) = 0$ by the same lemma. Hence the source and the target of $F_{i, p}(c)$ vanish and $F_{i, p}(c)$ is necessarily an isomorphism. This finishes the proof. $\square$