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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 ?

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    $\begingroup$ Made the tag. Suggest that if you know other questions that qualify you edit the tag in. $\endgroup$ – David Roberts Mar 31 at 7:37
  • $\begingroup$ @DavidRoberts : thanks ! $\endgroup$ – Maxime Ramzi Mar 31 at 8:04
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    $\begingroup$ But just to emphasize, pyknotic abelian groups do have enough injectives (because they are the category of sheaves of abelian groups on a site). $\endgroup$ – Tim Campion Apr 1 at 14:05
  • $\begingroup$ @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 ) $\endgroup$ – Maxime Ramzi Apr 1 at 14:07
  • $\begingroup$ @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. $\endgroup$ – Peter Scholze Apr 1 at 20:38
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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.

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  • $\begingroup$ 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 $\endgroup$ – Maxime Ramzi Apr 1 at 14:01
  • $\begingroup$ 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 ? $\endgroup$ – Maxime Ramzi Apr 1 at 14:01
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    $\begingroup$ See Proposition 2.1 in math.uni-bonn.de/people/scholze/Analytic.pdf $\endgroup$ – Peter Scholze Apr 1 at 15:37
  • $\begingroup$ Ah ok, I had missed that one, great ! Thanks a lot ! $\endgroup$ – Maxime Ramzi Apr 1 at 15:45

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