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The $\kappa$-condensed sets are defined as the sheaves on the site of profinite spaces of cardinality less than $\kappa$ (with $\kappa$ an uncountable strong limit cardinal) with morphisms the continuous maps, and whose covers are finite collections of jointly surjective maps.

I understand that you get the same category of condensed sets if you instead take the larger category of compact hausdorff spaces of cardinality less than $\kappa$, or the smaller category of Stone-Cech compactifications of sets of cardinality less than $\kappa$.

What happens if you use the "same site" on the category of all topological spaces of cardinality less than $\kappa$? Does this also yield the condensed sets? If not, what goes wrong?

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The question is not precise enough: it depends which topology you chose on the category of topological spaces. You will get the same category of sheaves if you are in a situation where Grothendieck's comparison lemma applies.

That is, you need to chose a topology on the category of all topological spaces, that induces the correct topology when restricted to compact spaces, and such that for each topological space X, the set of all maps $K \rightarrow X$ with $K$ compact are a covering.

There definitely is a (actually unique) topology that satisfies this, and others that don't, and this is a necessary and sufficient conditions to get the equivalence of the categories of sheaves.

The unique topology for which this is going to work can be described as follow:

A collection of maps $X_i \to X$ is a covering (generates a covering sieve) if and only for each profinite space K and each continuous map $K \to X$, there is a finite covering $K_j \to K$ by compact spaces $(K_j)_{j \in J}$ such that each map $K_j \to X$ factors through one of the $X_i$.

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  • $\begingroup$ Thanks for this answer. I am definitely a noob at all of this stuff (I have not thought about a topos since I was an undergraduate). Would you be willing to spell out the details a little more? $\endgroup$
    – Steven Gubkin
    Commented Dec 24, 2020 at 2:13
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    $\begingroup$ I've added a spelled out description of the topology one need to use one the category of spaces for the equivalence to work, but I'm not sure which kind of details you want. $\endgroup$
    – Simon Henry
    Commented Dec 24, 2020 at 2:55
  • $\begingroup$ Thanks, this is very useful (especially the link to the comparison lemma). So we would have (for instance), the injections of intervals $(n,n+2) \subset \mathbb{R} $ as a valid covering of the reals. $\endgroup$
    – Steven Gubkin
    Commented Dec 24, 2020 at 16:12
  • $\begingroup$ It seems so yes. More generally, every open covering. $\endgroup$
    – Simon Henry
    Commented Dec 24, 2020 at 16:37
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    $\begingroup$ My previous description of the topology was wrong: I had the wrong definition of condensed sets in mind. Now that I checked the definition, it is fixed. $\endgroup$
    – Simon Henry
    Commented Dec 24, 2020 at 17:07

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