Email from Noah Olander of Jun 9, 2020

Lemma 57.12.6. 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 57.12.5 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

$F(\mathcal{O}_ x) \to F(\mathcal{F}) \to F(\mathcal{O}_ x) \xrightarrow {F(\xi )} F(\mathcal{O}_ x)$

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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