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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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    $\begingroup$ 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! $\endgroup$
    – Geordie Williamson
    Commented Oct 27, 2017 at 5:02

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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 $M \mapsto \operatorname{Hom}_A(B,M)$.

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  • $\begingroup$ 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... $\endgroup$
    – Andrew Critch
    Commented Oct 28, 2009 at 2:07
  • $\begingroup$ Oh, I was assuming we were in the category of quasicoherent sheaves. $\endgroup$
    – Eric Wofsey
    Commented Oct 28, 2009 at 2:56
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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 $f_\ast$.

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  • $\begingroup$ Are you saying "f is a closed immersion" is a sufficient condition? $\endgroup$
    – Andrew Critch
    Commented Oct 28, 2009 at 2:11
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    $\begingroup$ 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. $\endgroup$
    – Greg Stevenson
    Commented Oct 28, 2009 at 2:25
  • $\begingroup$ 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 :( $\endgroup$
    – Andrew Critch
    Commented Oct 29, 2009 at 5:12
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    $\begingroup$ 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. $\endgroup$
    – Greg Stevenson
    Commented Oct 29, 2009 at 6:17
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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.

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  • $\begingroup$ Enclosing them in a pair of `` works $\endgroup$
    – Greg Stevenson
    Commented Oct 27, 2009 at 20:59
  • $\begingroup$ The underscores only make italics in the preview; in the actual post it works fine. $\endgroup$
    – Eric Wofsey
    Commented Oct 27, 2009 at 21:00
  • $\begingroup$ Thanks for the answer! Where can I read about this? $\endgroup$
    – Andrew Critch
    Commented Oct 27, 2009 at 21:19
  • $\begingroup$ The general categorial picture is analyzed in here: tac.mta.ca/tac/volumes/11/4/11-04abs.html For the scheme part I am not sure, I have been skimming through EGA, Hartshorne's Residues and Duality and Konrad's Grothendieck Duality and Base Change but without success... - now I should go back to work! $\endgroup$
    – Peter Arndt
    Commented Oct 27, 2009 at 22:17
  • $\begingroup$ Minor issue: your adjoint pair is backwards. $\endgroup$
    – S. Carnahan ♦
    Commented Oct 28, 2009 at 2:04
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For your question.

i agree with the answer from Greg, but in a different formalism. He used derived category. I did not. If f:X--->Y is a morphism of scheme. Then if f is affine morphism. Then direct image functor f_* :Cx---->Cy, has right adjoint functor f^!. Where Cx and Cy are category of sheaves or in particular, category of quasi coherent sheaves. So, in general, what we need is only the scheme is quasi compact and quasi separated.(I believe the quasi compact can be dropped, but I need some time to check globalization, I believe the flag variety of affine Kac-Moody algebra which is not quasi compact lies in this case). The reference is M.Kontsevich and A.Rosenberg Noncommutative spaces and flat descent. MPIM preprint

There is another related question. In category of quasi coherent sheaves. Say, if we have scheme morphism X---->Y, we always can get inverse image functor f^*: QcohY--->QcohX. But the direct image functor does not always exist. But if the scheme we are talking about is quasi compact and quasi separated. It exists. There is of course weaker condition. For this case, one can see the following papers: 1 D.Orlov Quasi coherent sheaves in commutative and noncommutative geometry 2 M.Kontsevich, A.Rosenberg. Noncommutative stack MPIM preprint 3 SGA 6

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  • $\begingroup$ I find your answer difficult to understand... twice you say "the scheme is quasi compact and quasi separated", but which scheme are you talking about? X? Y? Or are you talking about the morphism X --> Y? $\endgroup$
    – Andrew Critch
    Commented Nov 17, 2009 at 3:29
  • $\begingroup$ Oh, I am sorry for not pointing out. I mean the quasi compactness and quasi separtedness of scheme Y implies the existence of direct image functor which is right adjoint to f^*. For your original question, I mean the scheme X should be quasi compact and quasi separated. $\endgroup$
    – Shizhuo Zhang
    Commented Nov 17, 2009 at 6:41
  • $\begingroup$ Shizhuo, the question in the title is about the sheaf direct image, not about the direct image for quasicoherent modules (the latter direct image may even not exist for a general morphism of schemes). $\endgroup$
    – Zoran Skoda
    Commented May 15, 2011 at 19:49

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