# When does the sheaf direct image functor f_* have a right adjoint?

Say f: X → Y is a morphism of schemes. The sheaf direct image functor f always has a left adjoint, namely the sheaf inverse image functor f (with tensoring).

Under what (sufficient) conditions do we know that f has a right adjoint? What is it?

Answer to a related question (edit): If f preserves quasicoherence, then its restriction to quasicoherents f: QCoh(X) → QCoh(Y) has a right adjoint when f is affine (in particular, any closed immersion or finite morphism will do). The basic idea is to globalize the affine case (see Eric Wofsey's answer below); thanks to Pablo Solis for pointing me to page 6 of Ravi Vakil's notes explaining this.

In this question, I'm not restricting to the quasi-coherent categories. One reason for working with non-quasicoherents is that j! , the "extension by zero" right adjoint to j for an open immersion j, doesn't take qcoh to qcoh.

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If f_* has a right adjoint, it must preserve colimits and hence be right-exact. Thus a necessary condition is that the higher derived functors vanish. In particular, when everything is affine and we have X=Spec B, Y=Spec A, then I believe the adjoint exists and is given by M \mapsto Hom_A(B,M).

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Are you sure about the affine case even when the modules aren't quasi-coherent? I know the corresponding result for A-modules is true, but a non-quasicoherent O_A-Module (sheaf on Spec A) won't be the "tilde" of any A-module... – Andrew Critch Oct 28 '09 at 2:07
Oh, I was assuming we were in the category of quasicoherent sheaves. – Eric Wofsey Oct 28 '09 at 2:56

Provided that X is quasi-compact and separated and f is separated then what is true is that Rf_* : D(X) -> D(Y) has a right adjoint f^! where these are the unbounded derived categories of sheaves of modules with quasicoherent cohomology. This is the Grothendieck duality functor. Its existence can be viewed as a consequence of the fact that Rf_* in such a situation preserves coproducts. It is worth mentioning I guess that sometimes one does not need such big derived categories to produce an adjoint (for instance if X and Y are smooth and projective over some field).

One gets a right adjoint on the level of abelian categories of all sheaves of modules corresponding to the inclusion of a closed subscheme as well namely the inverse image of the subsheaf with supports.

There is the obvious cheat that if f:X -> Y is an isomorphism then the adjoint pair you know gives an equivalence so that f^* is also right adjoint to f_*.

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Are you saying "f is a closed immersion" is a sufficient condition? – Andrew Critch Oct 28 '09 at 2:11
Yes, I believe that is all one needs and then one can take the open complement and get a six functor diagram (I hope there is not some hypothesis I am forgetting). This is discussed in Artin's book on Grothendieck Topologies I believe. – Greg Stevenson Oct 28 '09 at 2:25
Really? I believe it if you restrict attention to the qcoh categories (in fact all you need then is for f to be affine), but for the larger categories I'm unconvinced... I also couldn't find Artin's book :( – Andrew Critch Oct 29 '09 at 5:12
Really... it works in the generality of sheaves of modules on ringed spaces and sheaves of abelian groups on topological spaces. See for instance Dan Murfet's notes therisingsea.org/notes/RingedSpaceModules.pdf the relevant bit is Proposition 97 on page 38 for the sheaves of modules. – Greg Stevenson Oct 29 '09 at 6:17

We sometimes (when !!??) have a second adjoint pair (f_!,f^!) between the sheaf categories where f_! is direct image with proper support and f^! is a right adjoint. Now when f is proper on has f_!=f_* , so f^! is right adjoint to f_* .

You can find out what it does by adjoint yoga with the sheaf-Homs: Hom(f_*F,G)=Hom(F,f^!G).

Set F=O_X. Then (f^!G)(U)=Hom(O_X(U), f^!G(U))=Hom((f_* O_X)(U), G(U)). If you can determine the latter you know more. This is a very general answer, but it can help in concrete situations, boiling down the question to the knowledge of f_*O_X...

If you don't know whether the right adjoint exists, you can also try to define one via this equation.

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Enclosing them in a pair of  works – Greg Stevenson Oct 27 '09 at 20:59
The underscores only make italics in the preview; in the actual post it works fine. – Eric Wofsey Oct 27 '09 at 21:00
You were right, thanks – Peter Arndt Oct 27 '09 at 21:06