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Suppose we have two schemes $X, Y$ and a map $f\colon X\to Y$. Then we know that $\operatorname{Hom}_X(f^*\mathcal{G}, \mathcal{F})\simeq \operatorname{Hom}_Y(\mathcal{G}, f_*\mathcal{F})$, where $\mathcal{F}$ is an $\mathcal{O}_X$-module and $\mathcal{G}$ an $\mathcal{O}_Y$-module (and the Homs are in the category of $\mathcal{O}_X$-modules etc). This gives a natural map $f^* f_* \mathcal{F}\to \mathcal{F}$, just by setting $\mathcal{G}=f_* \mathcal{F}$ and looking at where the identity map goes.

Are there any well-known conditions on the map or sheaves that give this is an isomorphism? For instance, I was looking through a book and saw that the map is surjective if $\mathcal{F}$ is a very ample invertible sheaf (and maybe some more hypothesis on the map and $X$ and $Y$ were assumed as well).

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    Here is a condition when that is true, but the reason is stupid. Let's assume your sheaf upstairs $\mathcal{F}$ is a pullback of a sheaf $\mathcal{G}$ downstairs, and assume moreover that the map $f: X \to Y$ has the property that $f_{*}\mathcal{O}_X=\mathcal{O}_Y$ which will be implied, for example, by requiring the map to be proper with geometrically connected fibres (as Charles Staats) said in his comment. Then your unlikely-happening-cancelation actually hold. And it's just projection formula.2017-02-04

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