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Let $X,Y$ be finite etale $T$ schemes for some scheme $T$ (assume the maps $X\rightarrow T,Y\rightarrow T$ are surjective). Then the sheaf $\mathcal{Hom}_T(X,Y)$ on $\text{Sch}/T$ (with the etale topology) sending $$(U\rightarrow T)\mapsto\text{Hom}_U(X_U,Y_U)$$ is representable, since on each component of $T$ it is finite locally constant. Let $H\stackrel{f}{\rightarrow}T$ be the representing $T$-scheme.

This means that we have a natural isomorphism of functors $\mathcal{Hom}_T(X,Y)\cong\text{Hom}_T(\underline{\;\;},H)$, so that for any $T$-scheme $U$, $U$-valued points of $H$ are given the interpretation of morphisms $X_U\rightarrow Y_U$.

On the other hand, suppose $T\rightarrow S$ is a covering in the etale topology (not sure if this important), then you can consider $H$ as an $S$-scheme. If $V\rightarrow S$ is also an $S$-scheme, then is there a natural interpretation for the $V$-valued points of $H$ (ie morphisms $V\rightarrow H$ over $S$)?

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    $\begingroup$ Isn't this what Grothendieck denotes by the symbol $\Pi_{T/S}$ in "Fondements de la Geometrie Algebrique"? $\endgroup$ Aug 7, 2014 at 21:02
  • $\begingroup$ @JasonStarr No, SGA3 Exp 1 defines $\Pi_{T/S}$ as (Weil) restriction of scalars, so $V$-points of $\Pi_{T/S} \mathcal{H}om_T(X,Y)$ are $V \times_S T$-points of $\mathcal{H}om_T(X,Y)$. $\endgroup$
    – S. Carnahan
    Aug 9, 2014 at 9:42
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    $\begingroup$ If you want a reference for writing a $T$-functor as an $S$-functor (using disjoint unions), see SGA 3 Exp. 1, the paragraph after the statement of Proposition 1.4.1. $\endgroup$
    – S. Carnahan
    Aug 9, 2014 at 10:13
  • $\begingroup$ @S.Carnahan: I could not find the reference that you mention in SGA 3 Exp 1 for $\Pi_{T/S}$. My understanding of Weil restriction is that it is defined for finite extensions $T/S$, whereas $\Pi_{T/S}$ in FGA is for more general morphisms, mainly projective flat morphisms. However, I completely agree that $\Pi_{T/S}\textit{Hom}_T(X,Y)$ contains many more points than the $S$-valued points of $\textit{Hom}_T(X,Y)$. The $S$-valued points are only those points that are constant on fibers of $T$ over $S$. $\endgroup$ Aug 9, 2014 at 10:25
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    $\begingroup$ @JasonStarr Sorry, I should have said Exp. 2 section 1 shortly before Remark 1.2. $\endgroup$
    – S. Carnahan
    Aug 9, 2014 at 12:44

1 Answer 1

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This might be a disappointing kind of answer and maybe there are better ways, but what is straightforward is this: morphisms $V\to H$ over $S$ are in one-to-one correspondence with pairs $(p,\varphi)$ where $p:V\to T$ is a $V$-valued point of $T$ over the $V$-valued point $V\to S$ of $S$ and $\varphi:p^*X\to p^*Y$ is a morphism over $V$ between $p$-fibres of $X$ and $Y$.

In other words, given $f:T\to S$ and $s:V\to S$, $$ \mathrm{Hom}_S(V,H)\approx\coprod_{\substack{t:V\to T\\ft=s}}\mathrm{Hom}_V(t^*X,t^*Y) $$

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  • $\begingroup$ Its a little disappointing but I suppose it doesn't make sense to hope for any more. Nonetheless, it does clear a few things up :-D $\endgroup$
    – Will Chen
    Aug 8, 2014 at 9:02

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