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Let $X,Y $ be smooth projective varieties over a field $k$. The following example is taken from Huybrecht's "Fourier Mukai Transforms in Algebraic Geometry" [example 5.4.vi]:

"Suppose $\mathcal{P}$ is a coherent sheaf on $X\times Y$ flat over $X$ and consider the Fourier Mukai Transform $\Phi_\mathcal{P}$. If $x\in X$ is a closed point with $k(x)\cong k$, then $\Phi_\mathcal{P}(k(x))\cong P_{|x\times Y}$ where $\mathcal{P}_{|x\times Y}$ is considered as a sheaf on $Y$ via the second projection."

Let $q: X\times Y\to X$ and $p: X\times Y\to Y$ denote the projections. Let $i: x\times Y\to X\times Y$ denote the closed immersion obtained from $x\to X$ by base change.

I calculated $q^*k(x)\cong i_*i^*\mathcal{O}_{X\times Y}$ and then I got

$q^*k(x)\otimes\mathcal{P}\cong i_*i^*\mathcal{O}_{X\times Y}\otimes \mathcal{P}\cong i_*(i^*\mathcal{O}_{X\times Y}\otimes i^*\mathcal{P})\cong i_*i^*\mathcal{P}$ via the projection formula.

Applying the right derived functor $Rp_*$ yields $Rp_*i_*i^*\mathcal{P}\cong R(p_*i_*)(i^*\mathcal{P})\cong q'_*i^*\mathcal{P}$ where $q':x\times Y\to Y$ is the second projection. This is the result from the example. I do not see where I have used flatness of $\mathcal{P}$ over $X$ though. Where do I need it ? Thanks a lot .

EDIT: Flatness over $X$ should play a role in calculating the derived tensor. I actually assumed the derived tensor product equals the usual tensor product when calculating the FMT. Is this implied by $\mathcal{P}$ being flat over $X$?

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    Did you figure this out? $i^* O_{X \times Y} = O_{x \times Y}$, and tensoring against this (derived or not) is the identity, I think...2017-01-29

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I know it’s a very old question, but I think I figured out what’s the solution, so I post the answer.

The final answer you get is $q’_* i^* \mathcal{P} $, however, you don’t know whether this is a sheaf or a true complex. First we notice that $q’_*$ need not be derived being an isomorphism, as the closed point was chosen such that $k(x) \simeq k$ (I’m quite sure about this, as it works affine locally, but correct me if I’m wrong). As $\mathcal{P}$ is flat over $X$ we know that the functor $\mathcal{G} \mapsto q^*\mathcal{G} \otimes \mathcal{P}$ from modules over $X$ to modules over $X \times Y$ is exact, where $q$ is the projection. Therefore, $q^* k(x) \otimes \mathcal{P}$ is a sheaf and the tensor product is not derived. Hence, we can use the stand projection formula (not the derived one) as $i_*$ is also exact, and therefore not derived, to get $$ q^* k(x) \otimes \mathcal{P} \simeq i_* i^* \mathcal{P} $$ as sheaves. We now conclude applying the functor $Rp_*$, which yields $$ \Phi_{\mathcal{P}} (k(x)) \simeq q’_* (i^* \mathcal{P}) $$ as a sheaf.