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Thm B.3 of Nikolaus and Scholze shows that the natural inclusion $\Delta \to \Lambda_\infty$, from the simplex category to the paracycle category, is an initial functor, i.e. satisfies the hypotheses of Quillen's Theorem A. That is, they show that the category $\mathcal C = \Delta \times_{\Lambda_\infty} (\Lambda_\infty)_{/(1/n)\mathbb Z}$ is contractible for each $n \geq 1$. They do this by covering $\mathcal C$ with intervals $\mathcal C[t,t+1)$ comprising those maps $f : (1/k)\mathbb Z \to (1/n)\mathbb Z$ such that $f(\{0/k,1/k,\dots,k-1/k\}) \subseteq [t,t+1)$ and performing an induction over the lattice of subcategories generated by these ones. In order for their induction to work, they need to know that $\mathcal C[t,t+1) \cap \mathcal C[s,s+1)$ is contractible for all $t,s \in (1/n) \mathbb Z$. I agree that this intersection is contractible when nonempty. However it seems to me that this intersection is often empty. For example, it's empty when $s$ and $t$ are distinct integers.

Question: Have I misunderstood something? Is $\mathcal C[t,t+1) \cap \mathcal C[s,s+1)$ always nonempty? If I've understood correctly, is there some tweak which will allow this argument to go through?

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    $\begingroup$ I think you are right, and I think you can tweak it. The point is that empty intersections is not a big deal because your "opens" are ordered, and so you are like in a "$\mathbb R$ covered by $[k,k+1)$'s" type of situation. I'll try to write this out when I get the time if no one does $\endgroup$ Commented Dec 19, 2023 at 8:26

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As predicted by Maxime, the argument in Nikolaus and Scholze is indeed incorrect, but it can be fixed as follows. Note that since each map $(1/k)\mathbb Z \to (1/n)\mathbb Z$ factors through some $\mathcal C_{[s,s+1)}$, we have that for $b>a+1$ a pushout of simplicial sets $C_{[a,b)} = C_{[a,b-1/n)} \cup_{C_{[b-1,b-1/n)}} C_{[b-1,b)}$, which allows to show by induction on $b-a$ that $C_{[a,b)}$ is contractible. Then we can exhaust $C$ by intervals of increasing size to see that $C$ is contractible too.

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  • $\begingroup$ I might be confused, but I don't believe any map (1/k)Z -> (1/n)Z has to live in some C[s,s+1). For a simple example, let's do k=2. I can send 0 to 0, 1/2 to 1, and more generally k to k, k +1/2 to k+1 for k an integer. The point is that sure, all of the interval [0,1) in (1/k)Z has to be sent to something leq the image of 0 + 1, but it's only \leq, not <. $\endgroup$ Commented May 30 at 9:48
  • $\begingroup$ I think it is not a problem though: the pushout argument works with $[c,d]$ everywhere in place of $[c,d)$, so then you have to prove that for $b\leq a+1$, $C[a,b]$ is contractible. But for this the Nikolaus-Scholze argument essentially works $\endgroup$ Commented May 30 at 10:17

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