57.11 Deducing fully faithfulness
It will be useful for us to know when a functor is fully faithful we offer the following variant of [Lemma 2.15, Orlov-K3].
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Lemma 57.11.1. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of triangulated categories. Let $S \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ be a set of objects. Assume
$F$ has both right and left adjoints,
for $K \in \mathcal{D}$ if $\mathop{\mathrm{Hom}}\nolimits (E, K[i]) = 0$ for all $E \in S$ and $i \in \mathbf{Z}$ then $K = 0$,
for $K \in \mathcal{D}$ if $\mathop{\mathrm{Hom}}\nolimits (K, E[i]) = 0$ for all $E \in S$ and $i \in \mathbf{Z}$ then $K = 0$,
the map $\mathop{\mathrm{Hom}}\nolimits (E, E'[i]) \to \mathop{\mathrm{Hom}}\nolimits (F(E), F(E')[i])$ induced by $F$ is bijective for all $E, E' \in S$ and $i \in \mathbf{Z}$.
Then $F$ is fully faithful.
Proof.
Denote $F_ r$ and $F_ l$ the right and left adjoints of $F$. For $E \in S$ choose a distinguished triangle
\[ E \to F_ r(F(E)) \to C \to E[1] \]
where the first arrow is the unit of the adjunction. For $E' \in S$ we have
\[ \mathop{\mathrm{Hom}}\nolimits (E', F_ r(F(E))[i]) = \mathop{\mathrm{Hom}}\nolimits (F(E'), F(E)[i]) = \mathop{\mathrm{Hom}}\nolimits (E', E[i]) \]
The last equality holds by assumption (4). Hence applying the homological functor $\mathop{\mathrm{Hom}}\nolimits (E', -)$ (Derived Categories, Lemma 13.4.2) to the distinguished triangle above we conclude that $\mathop{\mathrm{Hom}}\nolimits (E', C[i]) = 0$ for all $i \in \mathbf{Z}$ and $E' \in S$. By assumption (2) we conclude that $C = 0$ and $E = F_ r(F(E))$.
For $K \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ choose a distinguished triangle
\[ F_ l(F(K)) \to K \to C \to F_ l(F(K))[1] \]
where the first arrow is the counit of the adjunction. For $E \in S$ we have
\[ \mathop{\mathrm{Hom}}\nolimits (F_ l(F(K)), E[i]) = \mathop{\mathrm{Hom}}\nolimits (F(K), F(E)[i]) = \mathop{\mathrm{Hom}}\nolimits (K, F_ r(F(E))[i]) = \mathop{\mathrm{Hom}}\nolimits (K, E[i]) \]
where the last equality holds by the result of the first paragraph. Thus we conclude as before that $\mathop{\mathrm{Hom}}\nolimits (C, E[i]) = 0$ for all $E \in S$ and $i \in \mathbf{Z}$. Hence $C = 0$ by assumption (3). Thus $F$ is fully faithful by Categories, Lemma 4.24.4.
$\square$
Lemma 57.11.2. Let $k$ be a field. Let $X$ be a scheme of finite type over $k$ which is regular. Let $x \in X$ be a closed point. For a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ supported at $x$ choose a coherent $\mathcal{O}_ X$-module $\mathcal{F}'$ supported at $x$ such that $\mathcal{F}_ x$ and $\mathcal{F}'_ x$ are Matlis dual. Then there is an isomorphism
\[ \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, M) = H^0(X, M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{F}'[-d_ x]) \]
where $d_ x = \dim (\mathcal{O}_{X, x})$ functorial in $M$ in $D_{perf}(\mathcal{O}_ X)$.
Proof.
Since $\mathcal{F}$ is supported at $x$ we have
\[ \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, M) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\mathcal{F}_ x, M_ x) \]
and similarly we have
\[ H^0(X, M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{F}'[-d_ x]) = \text{Tor}^{\mathcal{O}_{X, x}}_{d_ x}(M_ x, \mathcal{F}'_ x) \]
Thus it suffices to show that given a Noetherian regular local ring $A$ of dimension $d$ and a finite length $A$-module $N$, if $N'$ is the Matlis dual to $N$, then there exists a functorial isomorphism
\[ \mathop{\mathrm{Hom}}\nolimits _ A(N, K) = \text{Tor}^ A_ d(K, N') \]
for $K$ in $D_{perf}(A)$. We can write the left hand side as $H^0(R\mathop{\mathrm{Hom}}\nolimits _ A(N, A) \otimes _ A^\mathbf {L} K)$ by More on Algebra, Lemma 15.74.15 and the fact that $N$ determines a perfect object of $D(A)$. Hence the formula holds because
\[ R\mathop{\mathrm{Hom}}\nolimits _ A(N, A) = R\mathop{\mathrm{Hom}}\nolimits _ A(N, A[d])[-d] = N'[-d] \]
by Dualizing Complexes, Lemma 47.16.4 and the fact that $A[d]$ is a normalized dualizing complex over $A$ ($A$ is Gorenstein by Dualizing Complexes, Lemma 47.21.3).
$\square$
Lemma 57.11.3. Let $k$ be a field. Let $X$ be a scheme of finite type over $k$ which is regular. Let $x \in X$ be a closed point and denote $\mathcal{O}_ x$ the skyscraper sheaf at $x$ with value $\kappa (x)$. Let $K$ in $D_{perf}(\mathcal{O}_ X)$.
If $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathcal{O}_ x, K) = 0$ then there exists an open neighbourhood $U$ of $x$ such that $H^{i - d_ x}(K)|_ U = 0$ where $d_ x = \dim (\mathcal{O}_{X, x})$.
If $\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_ x, K[i]) = 0$ for all $i \in \mathbf{Z}$, then $K$ is zero in an open neighbourhood of $x$.
If $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(K, \mathcal{O}_ x) = 0$ then there exists an open neighbourhood $U$ of $x$ such that $H^ i(K^\vee )|_ U = 0$.
If $\mathop{\mathrm{Hom}}\nolimits _ X(K, \mathcal{O}_ x[i]) = 0$ for all $i \in \mathbf{Z}$, then $K$ is zero in an open neighbourhood of $x$.
If $H^ i(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ x) = 0$ then there exists an open neighbourhood $U$ of $x$ such that $H^ i(K)|_ U = 0$.
If $H^ i(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ x) = 0$ for $i \in \mathbf{Z}$ then $K$ is zero in an open neighbourhood of $x$.
Proof.
Observe that $H^ i(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ x)$ is equal to $K_ x \otimes _{\mathcal{O}_{X, x}}^\mathbf {L} \kappa (x)$. Hence part (5) follows from More on Algebra, Lemma 15.76.4. Part (6) follows from part (5). Part (1) follows from part (5), Lemma 57.11.2, and the fact that the Matlis dual of $\kappa (x)$ is $\kappa (x)$. Part (2) follows from part (1). Part (3) follows from part (5) and the fact that $\mathop{\mathrm{Ext}}\nolimits ^ i(K, \mathcal{O}_ x) = H^ i(X, K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ x)$ by Cohomology, Lemma 20.50.5. Part (4) follows from part (3) and the fact that $K \cong (K^\vee )^\vee $ by the lemma just cited.
$\square$
Lemma 57.11.4. Let $X$ be a Noetherian scheme. Let $x \in X$ be a closed point and denote $\mathcal{O}_ x$ the skyscraper sheaf at $x$ with value $\kappa (x)$. Let $K$ in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$. Let $b \in \mathbf{Z}$. The following are equivalent
$H^ i(K)_ x = 0$ for all $i > b$ and
$\mathop{\mathrm{Hom}}\nolimits _ X(K, \mathcal{O}_ x[-i]) = 0$ for all $i > b$.
Proof.
Consider the complex $K_ x$ in $D^ b_{\textit{Coh}}(\mathcal{O}_{X, x})$. There exist an integer $b_ x \in \mathbf{Z}$ such that $K_ x$ can be represented by a bounded above complex
\[ \ldots \to \mathcal{O}_{X, x}^{\oplus n_{b_ x - 2}} \to \mathcal{O}_{X, x}^{\oplus n_{b_ x - 1}} \to \mathcal{O}_{X, x}^{\oplus n_{b_ x}} \to 0 \to \ldots \]
with $\mathcal{O}_{X, x}^{\oplus n_ i}$ sitting in degree $i$ where all the transition maps are given by matrices whose coefficients are in $\mathfrak m_ x$. See More on Algebra, Lemma 15.75.5. The result follows easily from this (and the equivalent conditions hold if and only if $b \geq b_ x$).
$\square$
Lemma 57.11.5. Let $k$ be a field. Let $X$ and $Y$ be proper schemes over $k$. Assume $X$ is regular. Then a $k$-linear exact functor $F : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ Y)$ is fully faithful if and only if for any closed points $x, x' \in X$ the maps
\[ F : \mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathcal{O}_ x, \mathcal{O}_{x'}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^ i_ Y(F(\mathcal{O}_ x), F(\mathcal{O}_{x'})) \]
are isomorphisms for all $i \in \mathbf{Z}$. Here $\mathcal{O}_ x$ is the skyscraper sheaf at $x$ with value $\kappa (x)$.
Proof.
By Lemma 57.7.1 the functor $F$ has both a left and a right adjoint. Thus we may apply the criterion of Lemma 57.11.1 because assumptions (2) and (3) of that lemma follow from Lemma 57.11.3.
$\square$
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Lemma 57.11.6. Let $k$ be a field. Let $X$ be a proper scheme over $k$ which is regular. Let $F : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ X)$ be a $k$-linear exact functor. Assume for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ with $\dim (\text{Supp}(\mathcal{F})) = 0$ there is an isomorphism $\mathcal{F} \cong F(\mathcal{F})$. Then $F$ is fully faithful.
Proof.
By Lemma 57.11.5 it suffices to show that the maps
\[ F : \mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathcal{O}_ x, \mathcal{O}_{x'}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^ i_ X(F(\mathcal{O}_ x), F(\mathcal{O}_{x'})) \]
are isomorphisms for all $i \in \mathbf{Z}$ and all closed points $x, x' \in X$. By assumption, the source and the target are isomorphic. If $x \not= x'$, then both sides are zero and the result is true. If $x = x'$, then it suffices to prove that the map is either injective or surjective. For $i < 0$ both sides are zero and the result is true. For $i = 0$ any nonzero map $\alpha : \mathcal{O}_ x \to \mathcal{O}_ x$ of $\mathcal{O}_ X$-modules is an isomorphism. Hence $F(\alpha )$ is an isomorphism too and so $F(\alpha )$ is nonzero. Thus the result for $i = 0$. For $i = 1$ a nonzero element $\xi $ in $\mathop{\mathrm{Ext}}\nolimits ^1(\mathcal{O}_ x, \mathcal{O}_ x)$ corresponds to a nonsplit short exact sequence
\[ 0 \to \mathcal{O}_ x \to \mathcal{F} \to \mathcal{O}_ x \to 0 \]
Since $F(\mathcal{F}) \cong \mathcal{F}$ we see that $F(\mathcal{F})$ is a nonsplit extension of $\mathcal{O}_ x$ by $\mathcal{O}_ x$ as well. Since $\mathcal{O}_ x \cong F(\mathcal{O}_ x)$ is a simple $\mathcal{O}_ X$-module and $\mathcal{F} \cong F(\mathcal{F})$ has length $2$, we see that in the distinguished triangle
\[ F(\mathcal{O}_ x) \to F(\mathcal{F}) \to F(\mathcal{O}_ x) \xrightarrow {F(\xi )} F(\mathcal{O}_ x)[1] \]
the first two arrows must form a short exact sequence which must be isomorphic to the above short exact sequence and hence is nonsplit. It follows that $F(\xi )$ is nonzero and we conclude for $i = 1$. For $i > 1$ composition of ext classes defines a surjection
\[ \mathop{\mathrm{Ext}}\nolimits ^1(F(\mathcal{O}_ x), F(\mathcal{O}_ x)) \otimes \ldots \otimes \mathop{\mathrm{Ext}}\nolimits ^1(F(\mathcal{O}_ x), F(\mathcal{O}_ x)) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^ i(F(\mathcal{O}_ x), F(\mathcal{O}_ x)) \]
See Duality for Schemes, Lemma 48.15.4. Hence surjectivity in degree $1$ implies surjectivity for $i > 0$. This finishes the proof.
$\square$
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