This is about a rather concrete problem that occurs in the middle of a lecture by Scholze. First I'll refer to the lecture, but then I'll state the problem.


In <https://www.youtube.com/watch?v=q6Tv2vJJShg> , at around the 22 minute mark, Peter Scholze claims that if you have a simplicial hypercover of a profinite set by profinite sets, then you can write it as a cofiltered limit of hypercovers of finite sets by finite sets. After a lot time of grappling with this technicality, I have finally given up and am hoping someone can help me.


**Question:** $S$ is a profinite set. $T\_{\bullet}\to S$ is a simplicial hypercover of $S$ by profinite sets $T\_i.$ Can we find some cofiltered poset $J,$ and some simplicial hypercovers $T\_{\bullet, j} \to S\_j,$ where $S\_j$ is finite and each $T\_{n,j}$ is finite, such that the cofiltered limit of the hypercovers $T\_{\bullet, j} \to S\_j$ is precisely our original hypercover $T\_{\bullet}\to S$?




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Currently, I can prove that $T\_{\bullet}\to S$ is a cofiltered limit of certain simplicial objects; however, I do not know that these objects are hypercovers. My main difficulty is that the hypercover condition, namely that the map $T\_{n+1, j} \to (\operatorname{cosk}\_n\operatorname{sk}\_n T\_{\bullet, j})\_{n+1}$ is surjective, seems to be very hard to keep true when constructing these sets $T\_{n, j}.$


To elaborate a little more, let's think just about the case of $T\_{1, j} \to (\operatorname{cosk}\_0 \operatorname{sk}\_0T\_{\bullet, j})\_1 = T\_{0,j}\times\_{S\_j} T\_{0, j}$ is surjective.


Assume that there is some index set $I$ so that $S = \lim\_i S\_i$ and $T\_0 = \lim\_i T\_{0, i}.$ For an index $i,$ one might get the idea to try producing a simplicial hypercover of $S\_i.$ A natural choice would be to first assume each projection map $S \to S\_i$ and $T\_0 \to T\_{0, i}$ is surjective (assuming this loses no generality, up to changing how we present $S$ and $T\_0$ as limits). Then surjectivity of $T\_0\to S$ implies that $T\_0\to S\_i$ is surjective, which implies that there's some index $i'$ so that $T\_0\to S\_i$ factors through $T\_{0, i'}.$


So, we try building a hypercover of $S\_i$ by starting with $T\_{0, i'} \to S\_i.$ To go to the next stage, there's some list of conditions we want to satisfy; the first is that whatever object $X$ we use as our next term, there should be a surjection $X \to T\_{0,i'} \times\_{S\_i} T\_{0, i'}.$ I cannot seem to easily construct such an $X,$ mainly because the map $T\_0 \times\_S T\_0 \to T\_{0,i'} \times\_{S\_i} T\_{0, i'}$ is not always surjective! (Imagine for example that $S\_i = \{\*\}$ is a singleton and $T\_0 = T\_{0, i'}$; then the latter fiber product is just $T\_0 \times T\_0,$ but the fiber product $T\_0 \times\_S T\_0$ might end up being some proper subset of $T\_0\times T\_0$ is $S$ is bigger than $S\_i$).



A hypercover of profinite sets as a limit of hypercovers of finite sets
I'm sorry for being cryptic.


The subtle point in the construction is that the maps $T\_n\to T\_{n,j}$ are not all surjective, i.e. one cannot construct this pro-system as a system of quotients.


By induction on $n$, one constructs a cofiltered system of $n$-skeletal simplicial hypercovers $T\_{\bullet,j}^{(n)}\to S\_j$ (of finite sets by finite sts) whose limit is the $n$-skeleton of $T\_\bullet\to S$, where each inductive step is refining the previous one; in the end, one takes the limit over $n$ of the induced coskeletal simplicial hypercovers.


In the first step, one needs to write the surjection $T\_0\to S$ as a limit of surjections of finite sets; this can be done by simply taking the cofiltered system of compatible finite quotients. Next, one needs to write $T\_1$ as a cofiltered limit of surjections $T\_{1,j}\to T\_{0,j}\times\_{S\_j} T\_{0,j}$. For this, we prove



> 
> Lemma. Let $X=\mathrm{lim}\_i X\_i$ be a profinite set written as a limit of finite sets $X\_i$ along a cofiltered category $I$. Let $Y\to X$ be a surjection of profinite sets. Then there is a cofiltered category $J$ with a functor $g: J\to I$ and (functorial) surjections of finite sets $Y\_j\to X\_{g(j)}$ whose limit is $Y\to X$.
> 
> 
> 


Admitting the lemma for the moment, one can find the desired $T\_{1,j}\to T\_{0,j}\times\_{S\_0} T\_{0,j}$. But for higher $n$, one is in the similar situation, for the surjection $T\_n\to (\mathrm{cosk}\_{n-1} \mathrm{sk}\_{n-1} T\_\bullet)\_n$ where the target already has a given presentation as a cofiltered limit of finite sets.


Thus, it remains to prove the lemma. For this, one considers the category $J$ whose objects $j\in J$ consist of an object $i\in I$ and a factorization of $Y\to X\to X\_i$ into $Y\to Y\_j\to X\_i$ where $Y\_j$ is finite and $Y\_j\to X\_i$ is surjective. A morphism $j\to j'$ is given by a map $i\to i'$ and a map $Y\_j\to Y\_{j'}$ making the obvious diagrams (from $Y$, and to $X\_i$) commute.


One needs to see that $J$ is cofiltered, and that $Y$ is the limit of the $Y\_j$'s. Given two $j,j'\in J$, one can first arrange via cofilteredness of $I$ that they map to the same $i\in I$. Then we have two maps $Y\to Y\_j\to X\_i$ and $Y\to Y\_{j'}\to X\_i$ where $Y\_j\to X\_i$ and $Y\_{j'}\to X\_i$ are surjective. We get a similar factorization over $Y\_j\times\_{X\_i} Y\_{j'}$, which is still finite and surjective over $X\_i$, so we find some $j''$ dominating $j$ and $j'$. Now, given two morphisms $f,g: j\to j'$, we need to find some $j''$ mapping to $j$ equalizing $f$ and $g$. Again, we can assume that the diagram $f,g: j\to j'$ gets contracted to $i\in I$ (using that $I$ is cofiltered). Then we get two maps $Y\_j\to Y\_{j'}$ over $X\_i$, and $Y$ factors over the equalizer $Z\subset Y\_j$. This is finite, but not necessarily surjective over $X\_i$! Still, as $Y\to X$ is surjective, there is some $i'\to i$ such that $Z\times\_{X\_i} X\_{i'}\to X\_{i'}$ is surjective. This defines the desired object $j''\in J$.


It remains to see that $Y\to \mathrm{lim}\_j Y\_j$ is bijective. Injectivity is easy to see (for any finite quotient $Y'$ of $Y$, $Y\to Y'\times X\_i\to X\_i$ defines an object of $J$). For surjectivity, the argument at the end of the last paragraph applies (one needs to see that for any $j$ there is some $j'$ such that $Y\_{j'}\to Y\_j$ factors over the image of $Y\to Y\_j$, which was proved there).


