Lemma 29.25.13. Let h : X \to Y be a morphism of schemes over S. Let \mathcal{G} be a quasi-coherent sheaf on Y. Let x \in X with y = h(x) \in Y. If h is flat at x, then
\mathcal{G}\text{ flat over }S\text{ at }y \Leftrightarrow h^*\mathcal{G}\text{ flat over }S\text{ at }x.
In particular: If h is surjective and flat, then \mathcal{G} is flat over S, if and only if h^*\mathcal{G} is flat over S. If h is surjective and flat, and X is flat over S, then Y is flat over S.
Proof.
You can prove this by applying Algebra, Lemma 10.39.9. Here is a direct proof. Let s \in S be the image of y. Consider the local ring maps \mathcal{O}_{S, s} \to \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}. By assumption the ring map \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x} is faithfully flat, see Algebra, Lemma 10.39.17. Let N = \mathcal{G}_ y. Note that h^*\mathcal{G}_ x = N \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x}, see Sheaves, Lemma 6.26.4. Let M' \to M be an injection of \mathcal{O}_{S, s}-modules. By the faithful flatness mentioned above we have
\begin{align*} \mathop{\mathrm{Ker}}( M' \otimes _{\mathcal{O}_{S, s}} N \to M \otimes _{\mathcal{O}_{S, s}} N) \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x} \\ = \mathop{\mathrm{Ker}}( M' \otimes _{\mathcal{O}_{S, s}} N \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x} \to M \otimes _{\mathcal{O}_{S, s}} N \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x}) \end{align*}
Hence the equivalence of the lemma follows from the second characterization of flatness in Algebra, Lemma 10.39.5.
\square
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