# Question about adjoint of forgetful functor from condensed abelian groups to condensed sets

There is a forgetful functor from condensed abelian groups to condensed sets. According to Scholze's notes, this has an adjoint $$T \mapsto \mathbb{Z}[T]$$ (which is the sheafification of the functor sending any extremally disconnected set $$S$$ to the free abelian group $$\mathbb{Z}[T(S)]$$).

Now in Scholze's notes (proof of Theorem 2.2) it states that using this adjunction, for any extremally disconnected set $$S$$ we have a condensed abelian group $$\mathbb{Z}[S]$$ satisfying that for any condensed abelian group $$M$$, $$\text{Hom}(\mathbb{Z}[S], M) = M(S)$$ and I do not understand why this is true.

First of all, I assume that they are taking $$S$$ as a condensed set (this would be the sheaf taking any profinite set $$X$$ to the set of continuous maps from $$X$$ to $$S$$). From now on I will be referring to this condensed set as $$\underline{S}$$.

$$\text{Hom} (\mathbb{Z}[\underline{S}], M) = \text{Hom} (\underline{S}, M)$$

However, I don't see how it follows that this is equal to $$M(S)$$. If $$M$$ was a group and we were talking about $$\underline{M}$$ (note that the continuous maps $$X \to M$$ form a group so this is a condensed group) we would have that

$$\text{Hom} (\underline{S}, \underline{M}) = \text{Hom}(S, M) = \underline{M}(S)$$

since the functor from sets to condensed sets is fully faithful and then we just use the definition of $$\underline{M}$$.

However, if $$M$$ is just any condensed abelian group, it might not be representable in this way and I don't see how to get that equality.

Noting that $$\underline{S} = \operatorname{Cont}(\cdot, S) = \operatorname{Hom}_{\operatorname{ProFin}}(\cdot, S)$$, this reduces to the Yoneda lemma:
$$\operatorname{Nat}(h_S, M) \xrightarrow{\sim} M(S), \eta \mapsto \eta_{S}(\operatorname{id}_S)$$
where $$h_S = \operatorname{Hom}(\cdot, S)$$ denotes the contravariant $$\operatorname{Hom}$$-functor.