# 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.

• I had fascinating conversations with Neeman about this. See arxiv.org/pdf/1406.7599.pdf. I am given to understand that the adjoint (which Neeman denotes $f^{\times}$) exists in great generality (e.g. separated, Noetherian, finite type is enough), and is technically very useful. The main issue is that it is not "local", but otherwise is quite well behaved. But it is much better to read the paper! – Geordie Williamson Oct 27 '17 at 5:02

Provided that $X$ is quasi-compact and separated and $f$ is separated then what is true is that $Rf_\ast \colon \operatorname{D}(X) \to \operatorname{D}(Y)$ 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 $Rf_\ast$ 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 $f \colon X \to Y$ is an isomorphism then the adjoint pair you know gives an equivalence so that $f^\ast$ is also right adjoint to $f_\ast$.

• 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

If $f_\ast$ 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 $X=\operatorname{Spec} B$ and $Y=\operatorname{Spec} A$, then I believe the adjoint exists and is given by $M \mapsto \operatorname{Hom}_A(B,M)$.

• 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

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: $\mathcal{Hom}(f_*F,G)=\mathcal{Hom}(F,f^!G)$.

Set $F=O_X$. Then $(f^!G)(U)=\mathcal{Hom}(O_X(U)$, $f^!G(U))=\mathcal{Hom}((f_* O_X)(U), G(U))$. 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.

• 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