# Are there (enough) injectives in condensed abelian groups?

The question is very simple : does $$Cond(\mathbf{Ab})$$, the category of condensed abelian groups (as defined in Scholze's Lectures in Condensed Mathematics), have enough injectives ? Does it, in fact, have any nontrivial injective ?

Recall that $$Cond(\mathbf{Ab})$$ is defined to be the colimit over strong limit cardinals $$\kappa$$ of $$Sh(*_{\kappa-proet}, \mathbf{Ab})$$, where $$*_{\kappa-proet}$$ is the site of $$\kappa$$-small profinite sets (it's easier to just use extremally disconnected spaces, and it is in fact equivalent)

Each of these sheaf-categories has enough injectives, but it's not clear that the colimit does, because a priori, the left Kan extension functor (along the inclusion of $$\kappa$$-small extremally disconnected spaces to $$\kappa'$$-small ones) $$Sh(*_{\kappa-proet}, \mathbf{Ab}) \to Sh(*_{\kappa'-proet}, \mathbf{Ab})$$ ($$\kappa<\kappa'$$) has no reason to preserve injectives, and any injective comes from one of these categories.

A lot of the time one can make do without actual injectives (for instance to define $$R\hom$$, use projectives; or you can also a lot of time use the injectives of one of the sheaf-categories to get what you need), but I suspect that they might be useful at some point; and the question seems relevant regardless

One could argue that we don't care about set-theoretic complications but it seems to me that this is one situation where they're actually non-stupid complications (that can't be solved by just saying "fix a universe"), but maybe someone can explain why they don't matter here ?

EDIT : in this comment, Scholze seems to claim that there are not enough injectives : he says "A few things that exist in pyknotic condensed sets but not in condensed abelian groups (e.g., injective pyknotic abelian groups)". So the question could become : what would be a proof of that ?

• Made the tag. Suggest that if you know other questions that qualify you edit the tag in. – David Roberts Mar 31 at 7:37
• @DavidRoberts : thanks ! – Maxime Ramzi Mar 31 at 8:04
• But just to emphasize, pyknotic abelian groups do have enough injectives (because they are the category of sheaves of abelian groups on a site). – Tim Campion Apr 1 at 14:05
• @TimCampion : I don't know a lot about those, but it seems from what I've heard that they just form the abelian group objects of a certain Grothendieck topos, in which case it's obvious, right ? (my intuition for the lack of injectives in condensed abelian groups is essentially that if you're coming from $\kappa$-condensed people, there's no way you're going to stay injective in the colimit, since the left Kan extension functors have no reason to preserve those ) – Maxime Ramzi Apr 1 at 14:07
• @Tim Campion: Pyknotic abelian groups are just $\kappa$-condensed abelian groups ($Sh(\ast_{\kappa-proet},\mathrm{Ab})$) for some choice of (strongly inaccessible) cardinal $\kappa$, so arguably Maxime Ranzi stated this right away in this question. – Peter Scholze Apr 1 at 20:38

Indeed, there are no nonzero injective condensed abelian groups.

Let $$I$$ be an injective condensed abelian group. We can find some surjection $$\bigoplus_{j\in J} \mathbb Z[S_j]\to I$$ for some index set $$J$$ and some profinite sets $$S_j$$, where $$\mathbb Z[S_j]$$ is the free condensed abelian group on $$S_j$$ -- this is true for any condensed abelian group. But now we can find an injection $$\bigoplus_{j\in J} \mathbb Z[S_j]\hookrightarrow K$$ into some compact abelian group $$K$$, for example a product of copies of $$\mathbb Z_p$$ for any chosen prime $$p$$. Indeed, it suffices to do this for any summand individually (embedding into a product in the end), and each factor embeds into a product of copies of $$\mathbb Z$$ (by choosing many maps $$S\to \mathbb Z$$), thus into a product of copies of $$\mathbb Z_p$$. We remark that it in this step that we need to work in the condensed setting: In the pyknotic setting, $$J$$ can be larger than the relevant cutoff cardinal for the profinite sets, so $$K$$ would not be in the site of $$\kappa$$-small compact Hausdorff spaces. By injectivity of $$I$$, we get a surjection $$K\to I$$. In particular, the underlying condensed set of $$I$$ is quasicompact. Now assume that $$I$$ is $$\kappa$$-condensed for some $$\kappa$$, and pick a set $$A$$ of cardinality bigger than $$\kappa$$, and consider the injection $$\bigoplus_A I\hookrightarrow \prod_A I.$$ The sum map $$\bigoplus_A I\to I$$ extends to $$\prod_A I\to I$$ by injectivity of $$I$$. I claim that the map $$\prod_A I\to I$$ necessarily factors over a map $$\prod_{A'} I\to I$$ for some subset $$A'\subset A$$ where the cardinality of $$A'$$ is less than $$\kappa$$. To check this, we use the surjection $$K\to I$$; then it is enough to prove that the map $$\prod_A K\to I$$ factors over $$\prod_{A'} K\to I$$ for some such $$A'$$. But this follows from $$I$$ being $$\kappa$$-condensed and $$\prod_A K$$ being profinite. Thus, the sum map $$\bigoplus_A I\to I$$ factors over $$\prod_{A'} I\to I$$ for some $$A'\subset A$$. But then restricting the sum map along the inclusion $$I\to \bigoplus_A I$$ given by some $$a\in A\setminus A'$$ gives both the identity and the zero map, finally showing that $$I=0$$.

I hope I didn't screw something up.

• Thanks for answering ! The result (and the proof) seem to indicate that my intuition that set-theoretic issues aren't silly here isn't completely off. There's just a point in your argument that I'm not entirely sure about : I'm not entirely sure how you embed $\mathbb Z[S]$ into a product of $\mathbb Z$'s - I don't understand how you guarantee that you get an injection from sufficiently many maps $S\to \mathbb Z$ (surely you can get an injection $S\to \prod_A\mathbb Z$, but I'm not sure how you get to $\mathbb Z[S]$). One possible way would be – Maxime Ramzi Apr 1 at 14:01
• to use the map to the solidification (which is itself a product of $\mathbb Z$'s), but I'm not sure that map is injective (my scribblings don't get me anywhere, I seem to need to rely on $\mathbb Z^{presheaf}[S]$ being a seprated presheaf, which isn't clear to me). Could you explain that ? – Maxime Ramzi Apr 1 at 14:01
• See Proposition 2.1 in math.uni-bonn.de/people/scholze/Analytic.pdf – Peter Scholze Apr 1 at 15:37
• Ah ok, I had missed that one, great ! Thanks a lot ! – Maxime Ramzi Apr 1 at 15:45