Is there a simplicial set $S_\bullet$ which has a nondegenerate $k+1$-simplex but all whose $k$-simplices are degenerate?
I tried proving that whenever all faces $d_i\sigma\in S_k$ of a $k+1$-simplex $\sigma\in S_{k+1}$ are degenerate, then so is $\sigma$. This would answer the above question with "no". But I wasn't able to to show that using the simplicial identities.
On the other hand, at least for singular simplicial sets and nerves of categories I can prove that whenever all faces $d_i\sigma\in S_k$ of a $k+1$-simplex $\sigma\in S_{k+1}$ are degenerate, then so is $\sigma$. That led me to think that this statement is true in general.
Any hints?
Best Answer
Indeed quotients exist in simplicial sets and are computed levelwise. If you take $$S^k = \Delta^k/\partial \Delta^k$$ to be the simplicial set obtained by collapsing the boundary of the standard simplicial set $\Delta^k$ (with $\Delta^k_n$ the set of order-preserving maps $\{1, \cdots, n\} \to \{1, \cdots, k\}$).
Then $S^k$ has exactly two nondegenerate simplices, one in degree zero and one in degree k.