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 $f?$**



What is the reason for $f_!$ not preserving discrete objects?
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.


