Category whose classifying space is $S^n$

Question

Suppose one wants a finite category $C_n$ (finite sets of objects and morphisms), for which $\left\vert NC_n \right\vert\simeq S^n$. I think one such $C_n$ can be built as follows : Take $\partial \Delta^n$, we get its poset of faces, i.e. we order its set of simplices $\coprod_{i=0}^n (\partial\Delta^n)_i $, taking $x\leq y$ iff $x$ is a face of $y$ (i.e. $d^i(y)= x$ for some $i$ ). Then this poset in particular can be seen as a category, and I believe we do get the desired property, i.e. the classifying space of this category is $S^n$. But the poset of faces is particularly large, and I suspect there should be a category with a much nicer presentation that has $S^n$ as classifying space, for one, $S^n$ has a very simple $CW$-decomposition, which leads me to think that surely there is a category whose classifying space will be precisely be that CW-decomposition. But not sure how to go about describing it.

Answer 1

This is related to the subdivision functor. I don't know if there's an easier way but given any simplicial complex $Y$ we can, by "subdividing", find a poset $C$ whose classifying space is $Y$.

Let $Y\cong|K|$ for some simplicial "polyhedral" set $K$. By this I mean a simplex of $K$ is uniquely determined by its vertices i.e. $K_n\to(K_0)^{n+1},\,x\mapsto(x[0],x[1],\cdots)$ is injective. It can be shown that the canonical map $\mathsf{sd}(K)\to SK$ is an isomorphism in this case and $SK$ is the nerve of the poset of nondegenerate simplices of $K$. It can also be shown that (unnaturally) $|\mathsf{sd}(K)|\cong|K|\cong Y$ so it follows $Y\cong|\mathcal{N}C|$ where $C:=\mathscr{P}^{\mathsf{nd}}(K)$.

In particular you can take $Y=S^n$ and $K=\partial\Delta^n$. A (sometimes rather sketchy) reference for this can be found here and the notation I use is all explained in my question here.

Answer 2

Edit: Now that I had the time to look at the other answer properly, it seems to me that this is a special case of a poset of non-degen. faces as in FShrike's answer.

Take the poset $\lbrace a_1,b_1 \leq a_2,b_2 \leq \dotsc \leq a_n,b_n \rbrace$ by which I mean the following. Consider $2n$ elements $a_1,\dotsc,a_n,b_1,\dotsc,b_n$ and define a partial order on the set $\lbrace a_1,\dotsc,a_n,b_1,\dotsc,b_n \rbrace$ by demanding $a_i \leq a_j,b_j$ and $b_i \leq a_j,b_j$ for all $j > i$ and then take the reflexive hull. In other words, the above poset is supposed to be read as:

the elements $a_1$ and $b_1$ are smaller or equal than all the $a_i$'s and $b_i$'s with $i \geq 2$ (and than themselves). Then $a_2$ and $b_2$ are smaller or equal than all the $a_i$'s and $b_i$'s with $i \geq 3$ (and than themselves) and so on. I do not want to compare $a_i$ with $b_j$ though for $i = j$.

Let's have a look at the ones for $n \leq 3$ as concrete examples:

In general we have that $B\lbrace a_1,b_1 \leq a_2,b_2 \leq \dotsc \leq a_n,b_n \rbrace \cong S^{n-1}$ and I think this category is not too bad size-wise.