# Limits and colimits in the category of condensed abelian groups

Sheafification is needed in limits and colimits of condensed abelian groups? If I have a functor $$T: i \mapsto T_i$$ from an index category to condensed abelian groups the limit and colimit of this functor are just $$S \mapsto \lim_i T_i (S)$$ and $$S \mapsto \text{colim}_i T_i (S)$$ or sheafification is needed for it to be a condensed abelian group?

In particular if I have a map $$\phi: T \rightarrow Q$$ of condensed sets the kernel and the cokernel are just $$S \mapsto \ker \phi_S$$ and $$S \mapsto \text{coker} \phi_S$$?

Thank you!

I'll ignore the set-theoretic issues since I don't understand them well enough to say anything about them.

As with any site, the limits (and in particular, the kernel) of sheaves may be computed pointwise. On the other hand, the colomits are usually not computed pointwise (cokernels included). For example, an exact sequence of condensed Abelian groups $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$ gives a long exact sequence of cohomology $$0 \rightarrow A(S) \rightarrow B(S) \rightarrow C(S) \rightarrow H^1(S,A) \rightarrow \dotsc$$

This means that the cokernels should be given by the pointwise quotient if the map $$H^1(S,A) \rightarrow H^1(S,B)$$ is injective (this happens, for example, when $$H^1(S,A)=0$$).

• In the case of condensed abelian groups, it turns out that evaluation at extremally disconnected spaces preserves all colimits : this is because condensed abelian groups are of the form "finite product preserving functors from something (that has finite products) to abelian groups", and finite products of abelian groups are also coproducts, so the inclusion preserves all colimits Jan 17, 2022 at 18:00
• Not if you restrict to extremally disconnected spaces. It is needed if you want to take $S$ a general compact Hausdorff space. I forget what happens for profinite sets, someone will have to fill that in, but it is definitely somewhere in Scholze's condensed notes, in terms of $H^1$ (as in Thiago's answer) Jan 17, 2022 at 18:20
• Sheafification is not needed to evaluate a colimt on an extremally disconnected space. More precisely, for a diagram $F_i$ of condensed sets and an extremally disconnected space $X$, we have $(\operatorname{colim}_i F_i)(X) = \operatorname{colim}_i (F_i(X))$. If $X$ is not assumed extremally disconnected, then the formula can fail. (edit: too slow :( ) Jan 17, 2022 at 18:25
• Indeed, in general $H^1(S,M)\neq 0$ for profinite $S$ and condensed abelian groups $M$ (otherwise $\mathbb Z[S]$ would always be projective, which it isn't -- this isn't obvious, but can be deduced from Proposition 4.8 of [Analytic.pdf]). Jan 17, 2022 at 22:10
• I should say that I still don't have a good feeling for what this "cohomology of profinite sets" "measures", in particular I think I don't know any example of a nonzero such higher cohomology group that I can actually compute (as opposed to merely showing it's nonzero)... Jan 17, 2022 at 22:12

Let me complement Thiago's answer with what we said in the comments below. This is just a very long comment.

Note that if we restrict our attention to extremally disconnected sets, the situation is much nicer. Indeed, $$Cond(Ab) \simeq Fun^\times(ExDisc^{op}, Ab)$$, where $$Fun^\times$$ is the category of finite product preserving functors.

In particular, because in $$Ab$$, finite products are coproducts, the inclusion $$Fun^\times(ExDisc^{op},Ab) \subset Fun(ExDisc^{op},Ab)$$ preserves all colimits, and because in the latter, colimits are pointwise, the same holds for the later. In particular, if $$S$$ is extremally disconnected, the formula $$(colim_i F_i)(S) = colim_i F_i(S)$$ always holds for diagrams of condensed abelian groups (in particular, for cokernels).

In general, you have to sheafify, including for cokernels.

However, there are special cases where you can forget about it : for instance, as proved in theorem 3.2 of Scholze's notes, if $$A$$ is discrete, then $$H^1(S,A) = 0$$ for $$S$$ profinite too, so the formula $$coker(A\to B)(S) = coker(A(S)\to B(S))$$ holds also for profinite $$S$$.

This formula also holds if $$S$$ is profinite and $$A,B$$ are solid (which is a kind of "completion" requirement). Indeed, solid modules are closed under colimits, so the cokernel will be solid too, and $$\mathbb Z[S]^\blacksquare$$ is projective in solid modules. This implies the claim as by definition, solidity implies that $$\hom(\mathbb Z[S]^\blacksquare, C) \cong C(S)$$ .

In fact, as Zhouhang points out in the comments below this answer, lemma 5.9 in the condensed notes shows that solid implies "derived solid", and so in particular $$H^1(S,A) = 0$$ if $$A$$ is solid and $$S$$ profinite, so really it suffices to assume $$A$$ is solid. I prefer to leave the previous argument as is, because it's easier to prove.

Now let $$X$$ be a profinite set. Projectivity of $$\mathbb Z[X]$$ amounts exactly to $$H^1(X,-)$$ vanishing, and so to the coker formula holding in full generality when evaluating at $$X$$. Peter Scholze mentioned in the comments that there are nontrivial examples of such a situation, i.e. where $$\mathbb Z[X]$$ is projective but $$X$$ is not extremally disconnected. These apparently always have a dense open subset which is extremally disconnected, and so that rules out many (most ?) examples.

The situation is even worse if you try to evaluate at a general compact Hausdorff space, and you should definitely not expect it to commute with colimits, or even specifically cokernels.

• Solidity of $A$ implies that $H^1(S;A)\cong\operatorname{Ext}^1(\mathbb Z[S]^\blacksquare,A)=0$, I think.
– Z. M
Jan 17, 2022 at 20:41
• Doesn't that require $A$ to be solid as an object of $D(Cond)$ ? (which is the case for $A = \mathbb Z[S]^\blacksquare$, but I don't think it's the case for all solid $A$'s) Jan 17, 2022 at 20:45
• It seems better to isolate the last into a new question. I tend to believe that $\mathbb Z[X]$ is compact projective if and only if $X$ is extremally disconnected.
– Z. M
Jan 17, 2022 at 21:10
• There are non-extremally disconnected profinite sets for which $\mathbb Z[X]$ is projective. See the examples of injective Banach spaces of the form $C(X)$ in the book "Separably Injective Banach Spaces". (Fact: For $X$ profinite, $\mathbb Z[X]$ projective implies $C(X)$ injective as Banach space, and in practice the converse seems to hold (in the sense that all arguments proving injectivity of $C(X)$ even seem to prove projectivity of $\mathbb Z[X]$).) Jan 17, 2022 at 22:16
• Here's an explicit example: Take any extremally disconnected space and identify two nonisolated points. All known examples are variations on this theme, I think. In particular, I think all known examples of projective condensed abelian groups are isomorphic to $\mathbb Z[X]$ with $X$ extremally disconnected. Jan 17, 2022 at 22:57