I started reading the notes on Condesned Mathematics. I am looking at the material at page 32-34. I have three fundamental computation questions:

1. At the last line of pg 32 - it seems to imply that for finite sets $S$, $\Bbb Z[S] \simeq \underline{Hom}(C(S, \Bbb Z ), \Bbb Z) $?

2. In pg 33 line 4 how is $Hom(C(S,\Bbb Z), \Bbb Z)$ identified with measure on $S$?

3. In proof of Proposition 5.7, there was the following equivalence,

$$ RHom(\prod_J \Bbb R, \Bbb Z) \simeq RHom_{\Bbb R} (\prod_J \Bbb R, R\underline{Hom}(\Bbb R, \Bbb Z) )=0$$

why is this true and what exactly does $RHom_{\Bbb R}$ mean? Is there some adjunction happening here from $\Bbb Z$ modules of $Cond(Set)$ to $\Bbb R$-modules of $Cond(Set)$?

I wuold appreciate if there are some related references for the general set up in 2 and 3.

For 1, I would like to compute the $T$ points for $T$ extremely. disconnected. What i don't see is an easy expression for lhs.
$$\Bbb Z[S] (T)= \bigoplus_{C(T,S)}\Bbb Z$$ Conversely for rhs we have

$$\underline{Hom}(C(S,\Bbb Z), \Bbb Z)(T)=Hom(\Bbb Z[T] \otimes C(S,\Bbb Z), \Bbb Z) $$ This doesnt seem an easy expression to handle too.

I started reading the Lectures on Condensed Mathematics. I am looking at the material at page 32-34. I have three fundamental computation questions:

1. At the last line of pg 32 - it seems to imply that for finite sets $S$, $\Bbb Z[S] \simeq \underline{Hom}(C(S, \Bbb Z ), \Bbb Z) $?

2. In pg 33 line 4 how is $Hom(C(S,\Bbb Z), \Bbb Z)$ identified with measure on $S$?

3. In proof of Proposition 5.7, there was the following equivalence,

$$ RHom(\prod_J \Bbb R, \Bbb Z) \simeq RHom_{\Bbb R} (\prod_J \Bbb R, R\underline{Hom}(\Bbb R, \Bbb Z) )=0$$

why is this true and what exactly does $RHom_{\Bbb R}$ mean? Is there some adjunction happening here from $\Bbb Z$ modules of $Cond(Set)$ to $\Bbb R$-modules of $Cond(Set)$?

I would appreciate if there are some related references for the general set up in 2 and 3.

For 1, I would like to compute the $T$ points for $T$ extremely. disconnected. What i don't see is an easy expression for lhs.
$$\Bbb Z[S] (T)= \bigoplus_{C(T,S)}\Bbb Z$$ Conversely for rhs we have

$$\underline{Hom}(C(S,\Bbb Z), \Bbb Z)(T)=Hom(\Bbb Z[T] \otimes C(S,\Bbb Z), \Bbb Z) $$ This doesn't seem an easy expression to handle too.