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For a flat finite surjective morphism of smooth varieties $f : X \rightarrow Y$ we have the pushforward functor $f_* : \mathcal{S}h (X) \rightarrow \mathcal{S}h (Y)$ and its left adjoint $f^* : \mathcal{S}h (Y) \rightarrow \mathcal{S}h (X)$ between coherent sheaves of $\mathcal{O}_X$-modules and $\mathcal{O}_Y$-modules. Is it true that $f^*$ is a faithful functor? It doesn't seem obvious to me...I don't mind if it is full or not. Thanks!

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It's true at the level of toposes and at the level of sheaves of abelian groups as soon as the map is surjective, but I think it is also true at level of quasicoherent sheaves if the morphism is faithfully flat. So what kind of sheaves are you asking about? – Zhen Lin May 22 '13 at 14:34
I was considering coherent sheaves of $\mathcal{O}_X$-modules, sorry I should have put that in. I've just edited the question to include it. – nan May 22 '13 at 14:57
up vote 3 down vote accepted

If $f : X \to Y$ is a flat and surjective, i.e. faithfully flat morphism, then $f^* : \mathsf{Qcoh}(Y) \to \mathsf{Qcoh}(X)$ is faithful. In fact, it is exact since $f$ is flat, so that it remains to prove $f^* M = 0 \Rightarrow M = 0$. But this can be checked locally, and is one of the well-known characterizations of faithfully flat ring homomorphisms: $A \to B$ is faithfully flat iff it is flat and $M \otimes_A B = 0 \Rightarrow M=0$.

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Thank you! Very clear explanation. – nan May 23 '13 at 15:09

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