A simplicial structure on symmetric groups Do the symmetric groups admit a simplicial structure?
By this, I mean a functor $X: \Delta^{op} \to \text{Sets} $ such that $X(n) = S_n$. More explicitly, one has to find functions (not necessarily group morphisms) $s_j, d_i$ that respect simplicial identities.
I'd be particularly interested in a simplicial structure such that $d_i$'s are given in the following way: if we identify permutations of $n$ elements with linear orders on a labeled set $1.. n$, the faces $d_i: S_{n+1} \to S_n$ , where $ i\in \{0, \ldots,n\}$ remove the element with label $i+1$,  while $d_{n+1} = d_n$.
 A: Well, in the end I figured this out. There is the following lemma:
Consider a semisimplicial set $X:\Delta_0^{op} \to \text{Sets} $, where $\Delta_0$ is the subcategory of $\Delta$ spanned by the face maps $d_i$. Suppose $X$ is \textit{degenerately acyclic}, i.e. wherever you assign $v_0, \ldots, v_n \in X_{n-1}$ such that:

*

*For any $i< j$, it holds $d_i v_j = d_{j-1} v_i $;

*There exist an index $k<n$ such that $v_k = v_{k+1}$
There exist a unique $w \in X_n$ such that $d_i w = v_i$.
Then there exist a unique extension $\hat{X} : \Delta^{op} \to \text{Sets}$ such that $\hat{X} | \Delta_0^{op} = X$.
Proof. Let's define by induction on $n$ the degeneracy maps $s_j : X_n \to X_{n+1}$. For $n=0$ and $x \in X_0$, note that $s_0(x) $ would satisfy
$$d_0 s_0(x) d_1 s_0(x) = x$$
By degenerate aciclicity, there exist a unique element with such property. There are no degeneracy relations to show because there is just one degeneracy map, and all the degeneracy-face relations are satisfied by hp.
Inductive step. Suppose we know all degeneracy maps up to degree n, and we want to find $s_j(x) $ for all $x \X_{n+1}, j=0, \ldots, n+1$. The degeneracy-face relations assign all the faces of an hypothetical $s_j(x) $ with the j-th and (j+1)-th faces equal to $x$, because in all other cases $d_i s_j$ will be of the form $ s_* d_*$: but $d_*(x) \in X_n$ so it's degeneracies are known by inductive hypothesis.
We are left with showing that such $s_j(x) $ satisfy the degeneracy relations: for $i\le j \le n$, we should have $s_i s_j(x) = s_{j+1}s_i(x)$. By looking at faces of the two terms, with a few case-by-case verification, we reduce to show the same relation applied to some $d_*(x) $, which is verified by inductive hypothesis.  Now if two things have same faces two of which are equal, they must coincide by degenerate acyclicity, and we conclude.

A step back in our original problem. It is enough to show that such $d_i : S_{n+1} \to S_n$ definite a degenerate acyclic semisimplicial set. Let us take $v_0, .., v_n \in S_{n-1} $ linear orders, such that $v_k = v_{k+1}$ and $d_i v_j = d_{j-1} v_i$ for $i<j$. We want to define a new linear order $w$ on $1, \ldots, n$ such that $d_k w= v_k$.
Note that formally  $d_p x = y $ when $ a\le b $ in $y$ iff $s_{p+1} a \le s_{p+1} b$. Indeed, $s_{p+1}$ is the function that "jumps" the $p+1$ element and we obtain the right order.
Define $a \le b$ in $w$ if there exist $p+1 \neq a, b $ such that $s_{p+1}(a') = a, s_{p+1}(b') =b$ and $a' \le b'$ in $v_p$. Note that the compatibility condition implies that for any other $q+1 \neq a, b$, the resulting inequality will be the same, by reducing to the order on $d_p v_q = d_{q-1} v_p $ (supposing WLOG $p< q$). This is a bit weird formally but clear if one has in mind a linear order with labels.
By definition, this $w$ will satisfy the equations, and it is also unique, because the equations we used to define the linear order were also necessary.
