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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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    $\begingroup$ Have you thought about the case when $Y$ is just a point, say Spec $k$, so that $f_*$ is the same as computing global sections. You are then asking when the natural map $\mathcal O_X\otimes_k H^0(X,\mathcal F)\to \mathcal F$ is an isomorphism. This then becomes a useful exercise; once you solve it, you will see that the answer to your question as to when this map is an isomorphism is "not often". $\endgroup$
    – Matt E
    Oct 10, 2010 at 3:39
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    $\begingroup$ I'm the last person in this Grothendieck-universe which can talk about schemes, but if you write $Hom(f^* \mathcal F,\mathcal G)\cong Hom(\mathcal G,f_*\mathcal F)$ I think about some kind of adjunction. Then you're looking for its counity to be an isomorphism, am I right? (sorry for the stupid contro-question) $\endgroup$
    – fosco
    Oct 23, 2010 at 17:28
  • $\begingroup$ If $\mathcal F = \mathcal O_X$, then the stereotypical example of when this is true is when $X = \mathbb P^n_Y$. More generally (but still with the $\mathcal F = \mathcal O_X$), the condition that "$f$ is proper and all fibers are geometrically connected" may suffice, although I am not at all confident of that. $\endgroup$
    – Charles Staats
    Aug 14, 2012 at 1:25
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    $\begingroup$ 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. $\endgroup$
    – lee
    Feb 4, 2017 at 3:24
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    $\begingroup$ In general the counit is an isomorphism if and only if the right adjoint is fully faithful. So you have to ask yourself when the pushforward is fully faithful. I cannot come up and prove any good conditions at the moment for when this is true, but on first glance it apperas to be very rare $\endgroup$
    – Rene Recktenwald
    Aug 17, 2019 at 13:09

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