Lemma 57.12.4. Let k be a field. Let X, Y be proper schemes over k. Assume X is regular. Let F, G : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ Y) be k-linear exact functors such that
F(\mathcal{F}) \cong G(\mathcal{F}) for any coherent \mathcal{O}_ X-module \mathcal{F} with \dim (\text{Supp}(\mathcal{F})) = 0,
F is fully faithful.
Then the essential image of G is contained in the essential image of F.
Proof.
Recall that F and G have both adjoints, see Lemma 57.7.1. In particular the essential image \mathcal{A} \subset D_{perf}(\mathcal{O}_ Y) of F satisfies the equivalent conditions of Derived Categories, Lemma 13.40.7. We claim that G factors through \mathcal{A}. Since \mathcal{A} = {}^\perp (\mathcal{A}^\perp ) by Derived Categories, Lemma 13.40.7 it suffices to show that \mathop{\mathrm{Hom}}\nolimits _ Y(G(M), N) = 0 for all M in D_{perf}(\mathcal{O}_ X) and N \in \mathcal{A}^\perp . We have
\mathop{\mathrm{Hom}}\nolimits _ Y(G(M), N) = \mathop{\mathrm{Hom}}\nolimits _ X(M, G_ r(N))
where G_ r is the right adjoint to G. Thus it suffices to prove that G_ r(N) = 0. Since G(\mathcal{F}) \cong F(\mathcal{F}) for \mathcal{F} as in (1) we see that
\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, G_ r(N)) = \mathop{\mathrm{Hom}}\nolimits _ Y(G(\mathcal{F}), N) = \mathop{\mathrm{Hom}}\nolimits _ Y(F(\mathcal{F}), N) = 0
as N is in the right orthogonal to the essential image \mathcal{A} of F. Of course, the same vanishing holds for \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, G_ r(N)[i]) for any i \in \mathbf{Z}. Thus G_ r(N) = 0 by Lemma 57.11.3 and we win.
\square
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