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 $\underline{S}$.

Adjointness gives us that

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

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.


1 Answer 1


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.


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