My question is about Corollary 6.1(ii) in [Lectures on Condensed Mathematics](https://www.math.uni-bonn.de/people/scholze/Condensed.pdf) by Scholze (page 41). Here is the claim:



> 
> The derived category $D(\mathrm{Solid})$ is compactly generated, and the full subcategory $D(\mathrm{Solid})^{\omega}$ of compact objects consists of the bounded complexes all of whose terms are of the form $\Pi\_{I} \mathbb{Z}$. The category $D(\mathrm{Solid})^{\omega}$ is contravariantly equivalent to the category $D^{b}(\mathbb{Z})$ via the functor
> $$
> C \mapsto R \underline{\operatorname{Hom}}(C, \mathbb{Z}) : D^{b}(\mathbb{Z})^{\mathrm{op}} \rightarrow D(\mathrm{Solid})
> $$
> 
> 
> 


I'm not quite understanding how to interpret this internal hom. I know how to view $R\underline{\operatorname{Hom}}(C, \mathbb{Z})$ as an object in $D^b(\mathbb{Z})$, but I'm expecting to get an object in $D(\mathrm{Solid})$. One way to get from $D^b(\mathbb{Z})$ to $D(\mathrm{Solid})$ is to first move to $D(\mathrm{Cond}(Ab))$, then apply the derived solidification functor $(-)^{L\blacksquare}$. So should I think of this as
$$C \mapsto \left(\underline{R \underline{\vphantom{\\_}\operatorname{Hom}}(C, \mathbb{Z})}\right)^{L\blacksquare}?$$
Alternatively, is there some reason $\underline{R \underline{\vphantom{\\_}\operatorname{Hom}}(C, \mathbb{Z})}$ is solid? Or am I completely off base?


Edit: As mentioned in the lectures and by Dan below, the fact that every bounded complex of abelian groups can be resolved by a complex of free abelian groups means that the objects in the internal hom are direct products of $\mathbb{Z}$. This roughly answered my question about $\underline{R \underline{\vphantom{\\_}\operatorname{Hom}}(C, \mathbb{Z})}$ being solid, but made me realize I don't know exactly what I mean by $\underline{R \underline{\vphantom{\\_}\operatorname{Hom}}(C, \mathbb{Z})}$.



Duality between $D^b(\mathbb{Z})$ and $D(\mathrm{Solid})^\omega$
The notation is actually amibiguous which I guess causes the confusion. In fact this $\underline{RHom}(C,\mathbb{Z})$ stands for the internal hom from $C$ to $\mathbb{Z}$ in derived $\textit{condensed}$ abelian groups. Here $C$ and $\mathbb{Z}$ start in $D(\mathbb{Z})$ but we can view that as the ``discrete" full subcategory of $D(Cond(\mathbb{Z}))$. As Dan indicates in the comments, this internal RHom is calculated by representing $C$ by a bounded complex of free $\mathbb{Z}$-modules, and then noting that $\underline{RHom}(\oplus\_I \mathbb{Z},\mathbb{Z})=\prod\_I\mathbb{Z}$.


