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Let $\pi : X' \rightarrow X$ be a morphism of schemes (or sites).

  1. For $X$ scheme $T$, what is the pull back $\pi^*h_T$ of the representable functor $h_T$? Is it the fiber product $T\times_{X} X'$?

  2. For $X'$ scheme $T'$, what is the push foward $\pi_*h_{T'}$? When is it representable?

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    $\begingroup$ 1. Yes. 2. It's called the Weil restriction. Certainly it exists if $\pi$ is finite flat. Perhaps it always exists (at least as an algebraic space) if $\pi$ is proper flat. $\endgroup$
    – Moosbrugger
    Commented Dec 1, 2011 at 3:38
  • $\begingroup$ 2. It is also called "restriction of scalars". Wikipedia has an article about it, with some references. If the morphism $h$ isn't finite, you can expect the pushforward to be quite large. $\endgroup$
    – S. Carnahan ♦
    Commented Dec 1, 2011 at 4:18

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This question has been answered in the comments. The former is the fibered product and the latter is the Weil restriction. The former always exists, but the latter does not. I am reposting this as a CW answer; if it gets upvoted, this question will not reappear on the front page.

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  • $\begingroup$ Dear @Dan, Are you saying that the pullback $\pi^{-1}h_T$ is always $h_{T_{X^\prime}}$? This is true, for example, if we're talking about the small etale site and $T$ is etale over $X$, or if $X^\prime\rightarrow X$ is etale, but is it really true in general? $\endgroup$
    – Keenan Kidwell
    Commented Oct 16, 2012 at 22:36
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In any category $\mathcal{C}$ the Yoneda immersion $Y: \mathcal{C}\to \mathcal{C}^>: x\mapsto h_x$ (some notation of SGA:Seminar of Algebraic Geometry ..by Grothendieck ecc) preserve (and lift) limits (then Pullback too), then the firs assertion if true. ABout pushout isnt true in general categories.

In the following I'm no entirely sure (I have no literature, only an attempt to demonstrate based on the thoughts of the moment)

ABout pushout, I think that is true for affine schemas and sum (i.e. pushout on $Sch(\mathbb{Z})$) this means that the contravariant Yoneda immersion for commutative rings $Y: Ring^{op}\to Ring^<: R\mapsto h^R$ preserve sums i.e

$Ring(R\times S, T)\cong Ring(R, T)\coprod Ring(S, T)$ infact:

if $Sch(T), Sch(R), Sch(S)$ are connected, then $Sch(R\times S)=Sch(R)\coprod Sch(S)$) has (only) two connected components, then (let $S$ the schemas category):

$Ring(R\times S, T)\cong S(Sch(T), Sch(R)\coprod Sch(S))\cong$

$S(Sch(T), Sch(R))\coprod S(Sch(T),Sch(S))\cong Ring(R, T)\coprod Ring (S, T)$

by a decomposition of schemas on connected components (that are associated to othogonals idempotents then these make a decompositions on prodocts of the ring) follow the assertion.

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    $\begingroup$ To Sergio: you should avoid posting answers when you barely understand the question. $\endgroup$
    – Angelo
    Commented Dec 1, 2011 at 19:39
  • $\begingroup$ To Angelo Vistoli, il mio inglese è carente (spesso anche nella comprensione), ma stavolta la domanda mi sembrava piuttosto chiara. Del resto accetto e mi aspetto di essere biasimato per errori vari, ma non scriverei mai niente altrimenti, ne leggerei le risposte (data che come maestro ho me sesso, non migliore dell'alunno). [email protected] $\endgroup$
    – Buschi Sergio
    Commented Dec 1, 2011 at 21:18
  • $\begingroup$ Caro Sergio, la domanda era chiara, la risposta molto meno. C'erano già due commenti che rispondevano in modo soddisfacente. $\endgroup$
    – Angelo
    Commented Dec 1, 2011 at 22:16
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    $\begingroup$ I agree with Angelo's first comment. $\endgroup$
    – Martin Brandenburg
    Commented Dec 2, 2011 at 8:36
  • $\begingroup$ Non capisco la struttura della frase data che come maestro ho me sesso, non migliore dell'alunno. Può spiegarla, Sergio o @Angelo, per favore ? $\endgroup$
    – Georges Elencwajg
    Commented Dec 2, 2011 at 9:42

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