In Proposition 5.7 on page 34 in [lectures on condensed mathematics](https://www.math.uni-bonn.de/people/scholze/Condensed.pdf) Peter Scholze shows that $\mathbb{Z}[S]^\blacksquare$ is solid. He shows that the two relevant expressions are isomorphic, however, in the definition of solidity it says that the isomorphism has to come from the natural map $\mathbb{Z}[S]\to\mathbb{Z}[S]^\blacksquare$. Can somebody tell me where this is proved?



Isomorphism of RHoms in condensed mathematics
I guess that this basically follows from keeping track of isomorphisms. Seemingly it is a bit harder to trace the following isomorphisms, due to the occurrence of the shift $\require{AMScd}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\iHom{\underline{Hom}}\DeclareMathOperator\fin{fin}\DeclareMathOperator\colim{colim}\mathbb Z[-1]$:
\begin{align\*}
R\Hom(\prod\_J\mathbb R/\mathbb Z,\mathbb Z)&=\bigoplus\_J\mathbb Z[-1]\\
R\Hom(\prod\_J\mathbb Z,\mathbb Z)&=\bigoplus\_J\mathbb Z
\end{align\*}
I am not sure what the author of these notes had in mind then, but one can cook up an argument by adapting the proof of Thm 4.3 as follows.


**Claim.** The canonical map $\bigoplus\_J\mathbb Z\to R\iHom(\prod\_J\mathbb Z,\mathbb Z)$ is an isomorphism. Namely, $\bigoplus\_J\mathbb Z$ is (derived condensed) reflexive.


**Proof.** We examine the morphism
\begin{CD}
\colim\_{I\subseteq\_{\fin}J}R\Hom(\prod\_I\mathbb R/\mathbb Z,\mathbb Z)@>>>\colim\_{I\subseteq\_{\fin}J}R\Hom(\prod\_I\mathbb R,\mathbb Z)@>>>\colim\_{I\subseteq\_{\fin}J}R\Hom(\prod\_I\mathbb Z,\mathbb Z)\\
@VVV@VVV@VVV\\
R\Hom(\prod\_J\mathbb R/\mathbb Z,\mathbb Z)@>>>R\Hom(\prod\_J\mathbb R,\mathbb Z)@>>>R\Hom(\prod\_J\mathbb Z,\mathbb Z)
\end{CD}
of fiber sequences. By the proof Thm 4.3, the middle terms are all zeros, and the left-most vertical map is an isomorphism, thus so is the right-most map.


In Prop 5.7, note that, for profinite sets $T$, the map $\mathbb Z[T]\to\mathbb Z[T]^\blacksquare$ can be rewritten as the canonoical double dual map $\mathbb Z[T]\to \mathbb Z[T]^{\vee\vee}$, where $(-)^\vee:=R\iHom(-,\mathbb Z)$, and it suffices to check that this induces an equivalence $\mathbb Z[T]^{\vee\vee\vee}\to\mathbb Z[T]^\vee$ after taking a further dual $(-)^\vee$, and this should follow formally from the reflexivity of $\mathbb Z[T]^\vee$.


