Let $X,Y$ be smooth varieties defined over $k$. Suppose $P$ is a coherent sheaf on $X \times Y$ flat over $X$, considering the Fourier-Mukai transform 
$$\Phi\_P : D^{b}(X) \to D^{b}(Y)$$
which is defined as 
$$F \mapsto q\_\* (P \otimes p^{\*}F).$$


Suppose $k(x)$ is the skyscraper sheaf on the closed point $x \in X$, and $k(x) \cong k$. Then how to justify the following claim:
$$\Phi(k(x)) \cong P\_x$$ where $P\_x := P \mid\_{ \{ x \} \times Y}$ is considered as a sheaf on $Y$ via the second projection: $\{x\} \times Y \to Y$ ?


my difficulties comes at two places: (1)All the definition here are in the derived sense, so I have no idea of compute them efficiently. (2) It seems $P \mid\_{ \{ x \} \times Y}$ is a pullback rather than pushforward which is what the claim looks like.


Moreover, where is the condition $P$ is flat used in the calculation?



Skyscraper sheaf under Fourier-Mukai transform
$p^\*$ for smooth $p$ is exact, so $p^\*$ of a skyscraper sheaf in the derived and regular senses are identical. So now we just have a pullback of a skyscraper sheaf - the structure sheaf of a single fiber and zero elsewhere.


We want to check that the higher tensor products vanish, so that the derived tensor product is just the regular tensor product, which is of course the sheaf $P\_x$ on $x \times Y$ and $0$ elsewhere.


This is where flatness comes in. It ensures that the tensor product is exact, so its higher derived functors vanish. 


So the tensor product is just the pushforward of $P\_x$ along the inclusion $Y \to X \times Y$ defined by $x$. Since this is a closed immersion the higher pushforwards vanish. If we pushforward again to $Y$, this is the same as the pushforward along the composition morphism, which is the identity, so we just end up with $P\_x$.


