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The question is related to Morphism between Fourier-Mukai functors implies the morphism between kernels?

Consider a complex smooth projective variety $X$ and the bounded derived category $D^b(X)$, it is known that non-trivial morphism $P_1\rightarrow P_2$ in $D^b(X\times X)$ will give a trivial natural transformation $\Phi_{P_1}\Rightarrow \Phi_{P_2}$ of Fourier--Mukai functors.

For example, let $X$ be an elliptic curve and $P_1=\mathcal{O}_{\Delta},P_2=\mathcal{O}_{\Delta}[2]$, one has a zero transformation $$\Phi_{P_1}=Id\Rightarrow \Phi_{P_2}=[2]$$

One the other hand, we know that the morphisms of integral kernels correspond to morphisms of the induced dg-functors by a famous result of Toën. So in particular, a non-trivial map $P_1\rightarrow P_2$ should generate a non-trivial dg-natural transformation $$\Phi_{P_1}\Rightarrow \Phi_{P_2}$$

What different will this statement make for the above exampe?

Moreover, assume that our $P_i$ are (quasi-isomorphic to) coherent sheaves, does the non-trivial dg-natural transformation $\Phi_{P_1}\Rightarrow \Phi_{P_2}$ descends to a non-trivial natural transformation between the Fourier--Mukai functors $\Phi_{P_1}\Rightarrow \Phi_{P_2}$?

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  • $\begingroup$ What do you mean exactly by the corresponding dg-natural transformation? What you get by Toën is a iso in the homotopy category of dg-cats inverting quasi-equivalences, so you only get a quasi-iso between the hom chain complexes between the (dg-)category of kernels and the ones in the internal hom in this homotopy category. This is different than a straight iso between dg-categories, so how do you get this dg-nat transformation. But maybe I might be misunderstanding something. $\endgroup$
    – AT0
    Commented Jan 15, 2023 at 21:02

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