Morphism between Fourier-Mukai functors implies the morphism between kernels?

Suppose $X,Y$ are smooth varieties over $\mathbb{C}$, and let $K_i \in D^b(X \times Y), i=1,2$ be objects in the derived category of bounded complex of coherent sheaves on $X \times Y$. Then there are Fourier-Mukai functors (FM functors) $\Phi_{K_i}$ from $D^{b}(X)$ to $D^b(Y)$ associated to kernel $K_i$:

$$F \mapsto q_{*}(K_i\otimes p^*(F)).$$ (here pullback, tensor and pusshfoward are all in the derived sense, and $p: X\times Y \to X$, $q: X \times Y \to Y$ are projections.)

My question is: if there is a natural transformation $G: \Phi_{K_1} \to \Phi_{K_2}$ respect derived categories, then does $G$ necessary come from a morphism of kernel $K_1 \to K_2$?

Somehow, I feel if one can write $\Phi_{K_1} \circ \Phi^{-1}_{K_2}$ (I use $\Phi^{-1}_{K_2}$ to denote some adjunct functor of $\Phi_{K_2}$), then it is also a FM transform, and by the uniqueness of FM kernel, one should get some morphism between the kernels. But I am not quite sure...

• The FM kernel is not always unique for functors which are not fully faithful. See, for example, the article arxiv.org/abs/1009.5577.
– naf
May 4 '14 at 7:54
• The functor $\Phi:D^b(X\times Y) \to Fun(D^b(X),D^b(Y))$ is known to be neither faithful nor full. May 4 '14 at 8:36
• However, the corresponding statement at the level of dg-categories is true: $\mathbf{D}(X \times Y) = \underline{\mathbf{Hom}}(\mathbf{D}(X), \mathbf{D}(Y))$, so that morphisms of integral kernels correspond to morphisms of the induced dg-functors (note though that the Hom means the internal Hom in the localization of the category of dg-categories with respect to the quasi-equivalences).
– AAK
May 4 '14 at 8:45

• @Adeel: well, looking at your posts it's obvious that you are well versed in the dark $\infty$-arts. People like me tend to use a differet point of view. I am aware that $D(X)$ is actually the shadow of a stable $\infty$-category and that problems like descent and uniqueness of cones are solved if you take into account this higher structure. However, when I work on a problem I try to be as 1-categorical as I can. Since Toen has proved that $dg-Fun(D(X),D(Y))=D(X \times Y)$, I might as well think of everything as a kernel. That way I can bypass dg-land entirely. (but you are quite right) May 25 '14 at 12:30