Some lemma which were originally part of the chapter on derived categories of varieties but are no longer needed.
Proof.
By Derived Categories of Varieties, Lemma 57.11.2 we conclude that for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ whose support is a closed point there are isomorphisms
\[ H^0(X, M \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{F}) = H^0(X, F(M) \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{F}) \]
functorial in $M$.
Let $x \in X$ be a closed point and apply the above with $\mathcal{F} = \mathcal{O}_ x$ the skyscraper sheaf with value $\kappa (x)$ at $x$. We find
\[ \dim _{\kappa (x)} \text{Tor}^{\mathcal{O}_{X, x}}_ p(M_ x, \kappa (x)) = \dim _{\kappa (x)} \text{Tor}^{\mathcal{O}_{X, x}}_ p(F(M)_ x, \kappa (x)) \]
for all $p \in \mathbf{Z}$. In particular, if $H^ i(M) = 0$ for $i > 0$, then $H^ i(F(M)) = 0$ for $i > 0$ by Derived Categories of Varieties, Lemma 57.11.3.
If $\mathcal{E}$ is locally free of rank $r$, then $F(\mathcal{E})$ is locally free of rank $r$. This is true because a perfect complex $K$ over $\mathcal{O}_{X, x}$ with
\[ \dim _{\kappa (x)} \text{Tor}^{\mathcal{O}_{X, x}}_ i(K, \kappa (x)) = \left\{ \begin{matrix} r
& \text{if}
& i = 0
\\ 0
& \text{if}
& i \not= 0
\end{matrix} \right. \]
is equal to a free module of rank $r$ placed in degree $0$. See for example More on Algebra, Lemma 15.75.6.
If $M$ is supported on a closed subscheme $Z \subset X$, then $F(M)$ is also supported on $Z$. This is clear because we will have $M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ x = 0$ for $x \not\in Z$ and hence the same will be true for $F(M)$ and hence we get the conclusion from Derived Categories of Varieties, Lemma 57.11.3.
In particular $F(\mathcal{O}_ x)$ is supported at $\{ x\} $. Let $i \in \mathbf{Z}$ be the minimal integer such that $H^ i(\mathcal{O}_ x) \not= 0$. We know that $i \leq 0$. If $i < 0$, then there is a morphism $\mathcal{O}_ x[-i] \to F(\mathcal{O}_ x)$ which contradicts the fact that all morphisms $\mathcal{O}_ x[-i] \to \mathcal{O}_ x$ are zero. Thus $F(\mathcal{O}_ x) = \mathcal{H}[0]$ where $\mathcal{H}$ is a skyscraper sheaf at $x$.
Let $\mathcal{G}$ be a coherent $\mathcal{O}_ X$-module with $\dim (\text{Supp}(\mathcal{G})) = 0$. Then there exists a filtration
\[ 0 = \mathcal{G}_0 \subset \mathcal{G}_1 \subset \ldots \subset \mathcal{G}_ n = \mathcal{G} \]
such that for $n \geq i \geq 1$ the quotient $\mathcal{G}_ i/\mathcal{G}_{i - 1}$ is isomorphic to $\mathcal{O}_{x_ i}$ for some closed point $x_ i \in X$. Then we get distinguished triangles
\[ F(\mathcal{G}_{i - 1}) \to F(\mathcal{G}_ i) \to F(\mathcal{O}_{x_ i}) \]
and using induction we find that $F(\mathcal{G}_ i)$ is a coherent sheaf placed in degree $0$.
Let $\mathcal{G}$ be a coherent $\mathcal{O}_ X$-module. We know that $H^ i(F(\mathcal{G})) = 0$ for $i > 0$. To get a contradiction assume that $H^ i(F(\mathcal{G}))$ is nonzero for some $i < 0$. We choose $i$ minimal with this property so that we have a morphism $H^ i(F(\mathcal{G}))[-i] \to F(\mathcal{G})$ in $D_{perf}(\mathcal{O}_ X)$. Choose a closed point $x \in X$ in the support of $H^ i(F(\mathcal{G}))$. By More on Algebra, Lemma 15.100.2 there exists an $n > 0$ such that
\[ H^ i(F(\mathcal{G}))_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X, x}/\mathfrak m_ x^ n \longrightarrow \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(F(\mathcal{G})_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \]
is nonzero. Next, we take $m \geq 1$ and we consider the short exact sequence
\[ 0 \to \mathfrak m_ x^ m \mathcal{G} \to \mathcal{G} \to \mathcal{G}/\mathfrak m_ x^ m\mathcal{G} \to 0 \]
By the above we know that $F(\mathcal{G}/\mathfrak m_ x^ m\mathcal{G})$ is a sheaf placed in degree $0$. Hence $H^ i(F(\mathfrak m_ x^ m \mathcal{G})) \to H^ i(F(\mathcal{G}))$ is an isomorphism. Consider the commutative diagram
\[ \xymatrix{ H^ i(F(\mathfrak m_ x^ m\mathcal{G}))_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X, x}/\mathfrak m_ x^ n \ar[r] \ar[d] & \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(F(\mathfrak m_ x^ m\mathcal{G})_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \ar[d] \\ H^ i(F(\mathcal{G}))_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X, x}/\mathfrak m_ x^ n \ar[r] & \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(F(\mathcal{G})_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) } \]
Since the left vertical arrow is an isomorphism and the bottom arrow is nonzero, we conclude that the right vertical arrow is nonzero for all $m \geq 1$. On the other hand, by the first paragraph of the proof, we know this arrow is isomorphic to the arrow
\[ \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(\mathfrak m_ x^ m\mathcal{G}_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \longrightarrow \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(\mathcal{G}_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \]
However, this arrow is zero for $m \gg n$ by More on Algebra, Lemma 15.102.2 which is the contradiction we're looking for.
Thus we know that $F$ preserves coherent modules. By Derived Categories of Varieties, Lemma 57.12.2 we find $F$ is a sibling to the Fourier-Mukai functor $F'$ given by a coherent $\mathcal{O}_{X \times X}$-module $\mathcal{K}$ flat over $X$ via $\text{pr}_1$ and finite over $X$ via $\text{pr}_2$. Since $F(\mathcal{O}_ X)$ is an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ placed in degree $0$ we see that
\[ \mathcal{L} \cong F(\mathcal{O}_ X) \cong F'(\mathcal{O}_ X) \cong \text{pr}_{2, *}\mathcal{K} \]
Thus by Functors and Morphisms, Lemma 56.7.6 there is a morphism $s : X \to X \times X$ with $\text{pr}_2 \circ s = \text{id}_ X$ such that $\mathcal{K} = s_*\mathcal{L}$. Set $f = \text{pr}_1 \circ s$. Then we have
\begin{align*} F'(M) & = R\text{pr}_{2, *}(L\text{pr}_1^*K \otimes \mathcal{K}) \\ & = R\text{pr}_{2, *}(L\text{pr}_1^*M \otimes s_*\mathcal{L}) \\ & = R\text{pr}_{2, *}(Rs_*(Lf^*M \otimes \mathcal{L})) \\ & = Lf^*M \otimes \mathcal{L} \end{align*}
where we have used Derived Categories of Schemes, Lemma 36.22.1 in the third step. Since for all closed points $x \in X$ the module $F(\mathcal{O}_ x)$ is supported at $x$, we see that $f$ induces the identity on the underlying topological space of $X$. We still have to show that $f$ is an isomorphism which we will do in the next paragraph.
Let $x \in X$ be a closed point. For $n \geq 1$ denote $\mathcal{O}_{x, n}$ the skyscaper sheaf at $x$ with value $\mathcal{O}_{X, x}/\mathfrak m_ x^ n$. We have
\[ \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, \mathcal{O}_{x, n}) \cong \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, F(\mathcal{O}_{x, n})) \cong \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, f^*\mathcal{O}_{x, n} \otimes \mathcal{L}) \]
functorially with respect to $\mathcal{O}_ X$-module homomorphisms between the $\mathcal{O}_{x, n}$. (The first isomorphism exists by assumption and the second isomorphism because $F$ and $F'$ are siblings.) For $m \geq n$ we have $\mathcal{O}_{X, x}/\mathfrak m^ n = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, \mathcal{O}_{x, n})$ via the action on $\mathcal{O}_{x, n}$ we conclude that $f^\sharp : \mathcal{O}_{X, x}/\mathfrak m_ x^ n \to \mathcal{O}_{X, x}/\mathfrak m_ x^ n$ is bijective for all $n$. Thus $f$ induces isomorphisms on complete local rings at closed points and hence is étale (Étale Morphisms, Lemma 41.11.3). Looking at closed points we see that $\Delta _ f : X \to X \times _{f, X, f} X$ (which is an open immersion as $f$ is étale) is bijective hence an isomorphism. Hence $f$ is a monomorphism. Finally, we conclude $f$ is an isomorphism as Descent, Lemma 35.25.1 tells us it is an open immersion.
$\square$
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