Is the Fourier-Mukai functor fully faithful?

Question

Let $X$ and $Y$ be smooth projective varieties over a field $k$. Then for each $E \in D^b(X \times Y)$, we get a functor $\Phi_E : D^b(X) \to D^b(Y)$. I believe the assignment $E \to \Phi_E$ is functorial -- essentially this is the statement that the natural square built from $f : M \to M' $ and $g : N \to N'$ with objects $M \otimes N$, $M' \otimes N$, etc. commutes.

Question: Is this faithful? Fully faithful? In other words, do we get an embedding of $D^b(X \times Y)$ into the category of exact functors from $D^b(X)$ to $D^b(Y)$. Is Orlov's theorem the statement that this is an equivalence of categories onto the subcategory consisting of fully faithful exact functors from $D^b(X)$ to $D^b(Y)$?

Answer 1

No, the functor $E \to \Phi_E$ is not faithful. The following counter-example is usually attributed to A. Caldararu. Let $X=Y=E$ where $E$ is a smooth elliptic curve. $\operatorname{Ext}^2(\mathcal{O}_\Delta, \mathcal{O}_\Delta)$ is one dimensional by Serre duality, thus we get a non-zero map $$ f: \mathcal{O}_\Delta \to \mathcal{O}_\Delta[2]. $$ But $\Phi_{f}=0$, because $\operatorname{Ext}_E^2(F,F)=0$ for any sheaf $F$ on $E$. Any complex $C$ of sheaves on any smooth curve is formal therefore $\Phi_f(C):C \to C[2]$ is trivial for any $C \in D^b(E)$.

Orlov's theorem states that any fully faithful functor has a unique kernel, it does not tell anything about presenting morphisms of functors as morphisms of kernels.