Given two smooth projective surfaces $X$ and $Y$ over some algebraically closed field. Given a torsion free coherent sheaf $M$ on $X$. One has the projections $\pi_X$ and $\pi_Y$ from the product $X\times Y$. Then we have $\pi_X^{\*}M$ on $X\times Y$.

Question: Is $\pi_X^{\*}M$ flat over $Y$?

One has to show that for each $z\in X\times Y$ the $O_{X\times Y,z}$-module $\pi_X^{\*}M_{z}$ is a flat $O_{Y,\pi_Y(z)}$-module. Now $\pi_X^{\*}M_z=M_{\pi_X(z)} \otimes_{O_{X,\pi_X(z)}} O_{X\times Y,z}$. I don't see why this should be flat over $O_{Y,\pi_Y(z)}$. I just know that $O_{X\times Y}$ is flat over $O_X$ and $O_Y$, but that doesn't seem to help me.

Background: I'm reading www.imsc.res.in/library/pdf/shaves.pdf, page 144, Theorem 6.1.8. When they want to compute the relative $Ext$-sheaves explicitely, they choose a certain complex of sheaves for whose existence they cite an article by Banica, Putinar and Schumacher. Now i looked that article up, and the complex is constructed using that both sheaves are flat over the base scheme. In this proof the sheaf $\mathcal{E}$ is a quasi universal family, so it is flat by definition, but what about the pullback sheaf?