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?



Are pullbacks from a factor of a product scheme flat over the other factor?
Certainly it is flat, and you don't even have to assume that $M$ is torsion free. Indeed, the question is local, so you can assume that $X$ and $Y$ are affine, so the question is: given $k$-algebras $A$ and $B$ and an $A$-module $M$ check that $M\otimes\_A(A\otimes\_k B)$ is flat over $B$. But this is easy --- $M\otimes\_A(A\otimes\_k B) \cong M\otimes\_k B$, so it is a free $B$-module. 


