More on profinite sets

Stream: condensed mathematics

[Adam Topaz]

Thinking about how to finish off the last few sorries for 8.19, it seems we will need the following: If $X$ is a profinite set which happens to be a cofiletered limit $X = \varprojlim_i X_i$ with $X_i$ profinite, and $S$ is a discrete quotient of $X$, then there exists some $i$ such that $X \to S$ factors through $X_i \to S$. I have a sketch of an argument in mind which is probably much more complicated than it should be. Does anyone see a slick way to prove this using what we already have?

[Johan Commelin]

Didn't you already exhibit every profinite set as cofiltered limit of all its discrete quotients?

[Johan Commelin]

Is that somehow useful here?

[Adam Topaz]

Yeah that's right, but the actual limit involved is not the limit over all discrete quotients.

[Peter Scholze]

I think you even know $X_i$ finite, right? (Not that it matters mathematically, but maybe it makes the argument easier.)

[Adam Topaz]

Yes, in my case the $X_i$ are indeed finite.

[Peter Scholze]

Maybe one could again try to prove that the functor from profinite sets to sets, sending $X$ to $\mathrm{Hom}(X,S)$, is Kan extended from finite sets?

[Peter Scholze]

(Hmm maybe it's not clear that this is enough.)

[Adam Topaz]

Maybe worth mentioning that if the $X_i$ are acctually all the discrete quotients of $X$, then the proof is easy with what's already done. I think, at least a priori, saying that $\mathrm{Hom}(X,S)$ is a Kan extension would boil down to that case where the $X_i$ are all the discrete quotients

[Peter Scholze]

Yeah I guess that's right...