Derived pushforward of a projection

Question

Given two smooth projective varieties, $X,Y$, consider their derived categories $D^b(X), D^b(Y)$. Let $\mathcal{F}$ a complex of coherent sheaves in $D^b(X \times Y)$, why the derived pushforward of the projection $R\pi_{Y*}$ should be regarded as "integration along the fibers"?

As a first approximation I know there is the following isomorphism: let $U \subseteq Y$ open

$$\Gamma(U,\pi_{Y*}(\mathcal{G} )) \simeq \bigoplus_{x \in X}\Gamma(U \times \{ x \}, \mathcal{G}_{x})$$

Where $ \mathcal{G}_{x}$ denotes

$$ \mathcal{G}|_{ x \times Y}$$

seen as a sheaf on $Y$.

This actually holds for any sheaf of modules $\mathcal{G}$ over $ X\times Y$, where $X,Y$ are topological spaces and $X$ is equipped with the discrete topology.

Is there a better justification for this interpretation?

Answer 1

I think the idea is that you have to keep in mind what happens with manifolds, if you have something like an oriented vector bundle of dimension n $\pi:E\to M$ then there is a pushforward of differential forms $\pi_{\ast}:\Omega^{\bullet}_{cv}(E)\to \Omega^{\bullet-n}(M)$ where $\Omega_{cv}(E)$ is the space of forms of vertical-compact support. This morphism is given precisely the integration along the fibers. This can then pass to a morphism between the de Rham cohomologies and gives you things like the Thom isomorphism.

I guess the context for smooth projective varieties can be formally extended for some sheaves other than differential forms, but I think the point is that working even purely by analogy, thinking of the derived pushforward as integrating along the fibers fits rather well with the idea that the situation you describe is a categorification of general integral transforms in analysis.