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Let $f:X\rightarrow Y$ be a finite morphism of smooth projective varieties over the field of complex numbers. Let $E$ be a locally free sheaf over $X$. We have the natural morphism $\phi:f^*f_*E\rightarrow E$. Does the finiteness of $f$ imply that $\phi$ is surjective? If $f$ is a closed immersion, then the above morphism is an isomorphism. If it is an arbitrary finite map, then is $\phi$ onto?

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Since $f$ is finite, the functor $f_*$ is exact and conservative, so surjectivity of $f^*f_*E \to E$ is equivalent to surjectivity of $$ f_*f^*f_*E \to f_*E. $$ On the other hand, by adjunction there is also a natural morphism $$ f_*E \to f_*f^*f_*E $$ and the composition $f_*E \to f_*f^*f_*E \to f_*E$ is the identity. This proves required surjectivity.

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  • $\begingroup$ thank you for the answer. Why is the direct image functor $f_*$ conservative when $f$ is finite. $\endgroup$
    – user52991
    Commented Feb 4, 2019 at 9:24
  • $\begingroup$ Since $f_*$ is exact, for conservativity it is enough to note that $f_*F = 0$ implies $F = 0$. $\endgroup$
    – Sasha
    Commented Feb 4, 2019 at 9:34

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