Computations in condensed mathematics II, page 43 This is a follow up on my previous question for Lectures on Condensed Mathematics. I am reading ahead at page 43. But it is not directly clear to me from the results that:

*

*How do we know $\Bbb Z[[T]], \Bbb R, \Bbb Z_p$ are solid modules?


*How did we obtain that $\Bbb R^{L \blacksquare}=0?$


*For the last equation 6.4 to holds, does  $-\otimes^{L\blacksquare}-$ commute with filtered limits in each variable?
Some elaborations would help.
I'd also like to know what are the formal aspects of the deductions (which I suppose is the great part this new category).

Basically all I know which are solid: are

*

*those of the form $(-)^{L\blacksquare}$,

*Those which are local objects. Characterization 5.8.

*$\prod_I \Bbb Z$, the cpt. projective generators.


 A: Let me basically repeat what R. says in the comments.
For 1, $\mathbb{Z}[[T]]$ and $\mathbb{Z}_p$ are solid becuase $\mathbb{Z}$ is solid and solid abelian groups are closed under all limits and colimits.  But $\mathbb{R}$ is not solid, see 2.
For 2, the first point is that $\mathbb{R}$ is pseudocoherent as a condensed abelian group, i.e. the following equivalent conditions are satisfied:

*

*$Ext^i(\mathbb{R},-)$ commutes with filtered colimits for all i;

*$\mathbb{R}$ admits a projective resolution where the terms are compact projective condensed abelian groups.

(The equivalence of 1 and 2 is valid in any abelian category generated by compact projectives.  It's a good exercise in homological algebra if you'd like to try.)
The pseudocoherence of $\mathbb{R}$ follows from the short exact sequence $\mathbb{Z}\rightarrow\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}$.  Indeed, $\mathbb{Z}$ is clearly pseudo-coherent (it is projective and compact) and the Breen-Deligne resolution implies that any compact abelian group is psuedo-coherent.  Equivalent condition 1) shows that pseudocoherent modules have the 2 out of 3 property in short exact sequences, so this implies that $\mathbb{R}$ is pseudo-coherent.
The second point is that for a pseudocoherent condensed abelian group $M$ the derived solidification of $M$ identifies with $\underline{RHom}(\underline{RHom}(M,\mathbb{Z}),\mathbb{Z})$.  Indeed, using condition 2 one reduces to checking that when $M=\mathbb{Z}[S]$ for $S$ extr. disconnected we have that $\underline{RHom}(\mathbb{Z}[S],\mathbb{Z})$ lives in degree $0$ and $\underline{RHom}(-,\mathbb{Z})$ on it is the derived solidification of $\mathbb{Z}[S]$.  But these we verified in producing the solid theory (they follow from Specker's theorem and the definition of solidification).
Thus it suffices to see that $\underline{RHom}(\mathbb{R},\mathbb{Z})=0$.  But this was verified in the proof of the existence of the solid theory.  It follows from the Breen-Deligne resolution and the fact that $\mathbb{R}$ is contractible (so its cohomology with $\mathbb{Z}$-coefficients vanishes).
For 3, the answer is yes (Edit: no, I misread and thought the question was about filtered colimits.  See comments) .  For ordinary tensor product this is clear, and it follows for derived tensor product because filtered colimits are exact so they have vanishing higher derived functors.
