Lemma 101.4.1. If $f : \mathcal{X} \to \mathcal{Y}$ is a flat morphism of algebraic stacks then $f^* : \mathit{QCoh}(\mathcal{O}_\mathcal {Y}) \to \mathit{QCoh}(\mathcal{O}_\mathcal {X})$ is an exact functor.

Proof. Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose a scheme $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. Then $U \to \mathcal{X}$ is still smooth and surjective as a composition of two such morphisms. From the commutative diagram

$\xymatrix{ U \ar[d] \ar[r]_{f'} & V \ar[d] \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} }$

we obtain a commutative diagram

$\xymatrix{ \mathit{QCoh}(\mathcal{O}_ U) & \mathit{QCoh}(\mathcal{O}_ V) \ar[l] \\ \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \ar[u] & \mathit{QCoh}(\mathcal{O}_\mathcal {Y}) \ar[l] \ar[u] }$

of abelian categories. Our proof that the bottom two categories in this diagram are abelian showed that the vertical functors are faithful exact functors (see proof of Sheaves on Stacks, Lemma 94.14.1). Since $f'$ is a flat morphism of schemes (by our definition of flat morphisms of algebraic stacks) we see that $(f')^*$ is an exact functor on quasi-coherent sheaves on $V$. Thus we win. $\square$

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