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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.