TITLE: Kodaira-Spencer map as a "differential"
QUESTION [5 upvotes]: Using the laguage of derived category, the Kodaira-Spencer map
$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$
can be described as a Fourier-Mukai transformation $\Phi_{\mathcal F}$.
$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $
Here, $x \in X$ is a point of a smooth projective variety over an algebraically closed field. 
$\mathcal F$ is a coherent sheaf on $X \times X$ which is flat over the first factor, and $\mathcal F_x$ means a $\mathcal {i}^*_{x\times X} \mathcal F$. 
(which is equal to $\Phi_{\mathcal F} (k(x))$)
Of course, KS map is not defined as a differential of some morphism, but please consider the following argument ;

Suppose $\mathcal F_x$  is concentrated at $x$ for all $x \in X$. It means the map $f:x \mapsto \mathcal F_x$ is injective. Hence $\kappa(x) := df(x)$ is injective at generic point. 

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)
Here, the definition of $f$ is a nonsense. It is not even clear where the image of $f$ lives. But the argument is quite persuasive. And it is even geometrically intuitive, because it describes a KS map as a differential of some "function". 
I think this must be a shadow of rigorous mathematical contents(probably a deformation theory), but I failed to make it so. Could someone explain to me what's going on? Thanks in advance.

REPLY [4 votes]: You can view the sheaf $\mathcal{F}$ on $X \times X$ instead as a map from $X$ (the first factor) to the moduli stack $M$ of sheaves on $X$ (the second factor).  This induces a map on the tangent spaces.  The tangent space to $M$ at a sheaf $F$ is the space of first order deformations of $F$, which is $\mathrm{Ext}^1(F, F)$.
Using the exact sequence
$0 \rightarrow I \rightarrow \mathcal{O}_X \rightarrow k(x) \rightarrow 0$
you get an identification between $T_x X = \mathrm{Hom}_{\mathcal{O}_X}(I, k(x)) = \mathrm{Ext}^1(k(x),k(x))$.  Therefore the differential gives a map
$T_x X = \mathrm{Ext}^1(k(x),k(x)) \rightarrow T_{\mathcal{F}_x} M = \mathrm{Ext}^1(\mathcal{F}_x, \mathcal{F}_x)$.