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Let $A$ be a finitely generated $\mathbb{Z}$-algebra and let $f: \operatorname{Spec} A \rightarrow \operatorname{Spec} \mathbb{Z}$ be the canonical map. On pg. 53, Thm. 8.2 of https://www.math.uni-bonn.de/people/scholze/Condensed.pdf
one defines a functor $$f_!: D(A_\blacksquare) \rightarrow D(\mathbb{Z}_\blacksquare)$$ where $A_\blacksquare$ and $\mathbb{Z}_\blacksquare$ denotes certain categories of solid modules.

On the same page, Scholze remarks that $f_!$ does not preserve discrete objects in general?

Is there an easy way to see that $f_!$ can not possibly preserve discrete objects for general morphisms $f?$

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    $\begingroup$ If it did it would be left adjoint to $f^*$ on discrete objects, but the latter does not preserve infinite products. $\endgroup$
    – Marc Hoyois
    Commented Apr 11, 2022 at 5:05
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    $\begingroup$ @MarcHoyois Sorry for my ignorance, but in the global case, if $f$ is proper, then if I am not mistaken, $f_!$ seems to preserve discrete objects, but not a left adjoint to $f^*$ (rather, $f^!$)? $\endgroup$
    – Z. M
    Commented Apr 11, 2022 at 5:38
  • $\begingroup$ @MarcHoyois Is it clear that $f_!$ is a left adjoint to $f^*,$ in the condensed framework? More precisely, in Thm 8.2 of the notes, it is rather the case that $f^!$ is the right adjoint. $\endgroup$
    – Crystallineperiodic
    Commented Apr 11, 2022 at 7:33
  • $\begingroup$ Maybe one could argue using j from Thm 8.1? Suppose that $f!$ preserves discrete objects, then the same is true for $j!.$ Thus the induced functor $j^* :D((A,\mathbb{Z})_{\blacksquare,disc}) \rightarrow D((A_\blacksquare)_{disc})$ would have a right adjoint, implying that it preserves products. But this is not true, I think. It seems to me as if $ j^*(\Pi_{I} A^{disc})$ for I infinite is not discrete. $\endgroup$
    – Crystallineperiodic
    Commented Apr 11, 2022 at 9:02
  • $\begingroup$ @Z.M In my comment $f$ should be an open immersion, sorry (this is the case where $f_!$ does not exist classically). $\endgroup$
    – Marc Hoyois
    Commented Apr 11, 2022 at 18:26

1 Answer 1

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You can see this by a direct calculation, for example in the most basic case $A=\mathbb{Z}[T]$. When you apply $f_!$ to $A=\mathbb{Z}[T]$ itself, you get the object represented by the two-term complex $$\mathbb{Z}[T]\to \mathbb{Z}((T^{-1})).$$ with $\mathbb{Z}[T]$ in degree $0$. (This is the "compactly supported cohomology of the structure sheaf" on $\mathbb{A}^1$; the Laurent series in $T^{-1}$ represents "functions defined in a neighborhood of $\infty$").

The map in this complex is injective, and the cokernel is $T^{-1}\mathbb{Z}[[T]]$, which is not discrete. In fact, $f_!A$ is predual to an infinite discrete object, namely the usual algebraic Grothendieck dualizing complex of f. This is a general feature of the situation, and follows from the adjunction between $f_!$ and $f^!$. Actually, $f_!$ has a complementary property to preservation of discreteness: it preserves pseudocoherent objects.

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  • $\begingroup$ Can you describe the differential? Is it $T \mapsto T^{-1}$? $\endgroup$
    – Leo Alonso
    Commented Apr 11, 2022 at 15:37
  • $\begingroup$ No, the differential is just the natural inclusion, $T\mapsto T$. By the way, there's a typo above: the cokernel is actually $T^{-}\mathbb{Z}[[T^{-1}]]$. $\endgroup$
    – Dustin
    Commented Apr 11, 2022 at 19:09

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