Lemma 57.13.4. 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

1. $F(\mathcal{F}) \cong G(\mathcal{F})$ for any coherent $\mathcal{O}_ X$-module $\mathcal{F}$ with $\dim (\text{Supp}(\mathcal{F})) = 0$,

2. $F$ 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 57.8.1. In particular the essential image $\mathcal{A} \subset D_{perf}(\mathcal{O}_ Y)$ of $F$ satisfies the equivalent conditions of Derived Categories, Lemma 13.40.5. We claim that $G$ factors through $\mathcal{A}$. Since $\mathcal{A} = {}^\perp (\mathcal{A}^\perp )$ by Derived Categories, Lemma 13.40.5 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 57.12.3 and we win. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).