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Let $K$ be simplicial set and $d_i:K_n\rightarrow K_{n-1}$, $s_i: K_n\rightarrow K_{n+1}$ ($i = 0,...,n$) face and degeneracy maps respectively.

Suppose we have some $x\in K_n$ with $d_0x = ... = d_n x = y$. How to prove that $s_0y=...=s_{n-1}y$? I think that $y$ should be an iterated degeneracy of some vertex $v\in K_0$ (probably $v = d_i^{n-1}y$), but I cannot prove it.

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Could you expand on your motivation for this question? I think I have found a counterexample, essentially because if this identity was true, it would have to be an easy consequence of the simplicial identities.


The functor $F$ that sends a simplicial set $K_*$ to the $n$-simplices satisfying this condition is corepresented by some simplicial set $X^n$, i.e. there is a natural isomorphism $F(K) \cong \operatorname{sSet}(X^n,K)$, and by the Yoneda lemma it suffices to check this identity for the universal example $X^n$. Explicitly, it is given by the pushout $$ \begin{array}{ccc} \bigsqcup_{0}^n \Delta^{n-1} &\xrightarrow{\bigsqcup i\text{-th face}} & \Delta^n\\ \downarrow & & \downarrow\\ \Delta^{n-1} & \xrightarrow{\qquad} & X^n \end{array} $$ i.e. we identify all the faces of the "free" $n$-simplex.

By the simplicial identities, one can show that all iterated faces of the non-degenerate $n$-simplex of $X_n$ are equal. This shows that in the above pushout description, we can glue the faces of the $\Delta^{n-1}$'s together to obtain a pushout diagram $$ \begin{array}{ccc} \partial \Delta^{n} &\hookrightarrow & \Delta^n\\ \downarrow & & \downarrow\\ X^{n-1} & \to & X^n \end{array} $$ where the left-hand downward map is constructed inductively from the right-hand one. Since injections of simplicial sets are stable under pushouts, $X^{n-1}\to X^n$ is injective, and we see that $X^n$ has exactly one non-degenerate $k$-simplex for $0\le k\le n$, all of whose faces are equal to the non-degenerate $k-1$-simplex. This shows already that your simplex $y$ need not be degenerate.

There is an obvious map $X_n\to S^n = \Delta^n/\partial \Delta^n$ by collapsing $X^{n-1}$ inside $X^n$ which sends the non-degenerate $n$-simplex of $X^n$ to that of $S^n$ - on corepresented functors, this just says that an $n$-simplex all of whose faces are degeneracies of a $0$-vertex must have all faces equal. For $S^n$, it is not that difficult to see that all degeneracies of the nondegenerate $n$-simplex are pairwise different (each one has exactly two non-degenerate faces which are at different positions), so this must hold for $X^n$ as well.

Now $X^{n-1}\to X^n$ is injective, so the degeneracies of the nondegenerate $(n-1)$-simplex of $X^n$ are pairwise different. But this is "the" face of the nondegenerate $n$-simplex. So this is a counterexample once $n\ge 2$.


Here is an explicit description of the simplicial set $X^n$: its $k$-simplices are given by partitions of $[k]$ into at most $n+1$ nonempty intervals. The preimage of a nonempty interval under a monotonous function is a (possibly empty) interval; throwing away the empty preimages, this defines a functor $X^n:\Delta^{op}\to \mathrm{Set}$. Now consider the partition of $[n]$ into singletons. Each of its faces is the partition of $[n-1]$ into singletons, so these are all equal. Its degeneracies are the various partitions of $[n]$ into $n-1$ singletons and one pair, and these are all different for $n\ge 2$.


In the comments, the question was raised whether an $n$-simplex is a degeneracy of (one of) its $0$-faces if all of its degeneracies $s_k x$ are equal. More formally, this asks whether $$ \begin{array}{ccc} \bigsqcup_{0}^n \Delta^{n+1} &\xrightarrow{\bigsqcup i\text{-th degeneracy}} & \Delta^n\\ \downarrow & & \downarrow\\ \Delta^{n+1} & \xrightarrow{\qquad} & \Delta^0 \end{array} $$ is a pushout diagram. This is indeed true: The simplicial identities imply that $d_0s_0 = \operatorname{id}$ and $d_0^ks_k = s_0d_0^k$, thus $d_0^k x = d_0^{k+1} s_0 x = d_0^{k+1} s_{k+1} x = s_0 d_0^{k+1} x$, and an easy induction shows that $x = d_0^0 x = s_0^n d_0^n x$, so $x$ is an $n$-fold degenerate simplex.

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  • $\begingroup$ Ok, let $\delta[n]$ be a simplicial set with vertices $0,1,2,...,n$ and $\delta[n]_q = \{(v_0,...,v_q)| 0\leq v_0 \leq ... \leq v_q \leq n\}$. Then simplicial sphere $S^n = \delta[n]/\delta[n]^{n-1}$, where $\delta[n]^{n-1}$ is $(n-1)$-skeleton of $\delta[n]$. Let $i_n = (0, ..., n) \in \delta[n]_n$. $\endgroup$ – Samarkand Feb 2 '16 at 18:55
  • $\begingroup$ Curtis in his book Simplicial homotopy theory makes the following assertion (Proposition 1.5). Let $K$ be a simplicial set, $x\in K_n$. Then there is a unique simplicial map $f_x:\delta[n]\rightarrow K$ which sends $i_n$ to $x$. Then he says that if all $d_i x = *$ then $f_x$ passes to the quotient simplicial map $\bar{f}_x: S^n \rightarrow K$, which sends the image of $i_n$ in $S^n$ to $x$. But this map $\bar{f}_x$ is correctly defined if all degeneracies of $d_i x$ are equal. Is it the wrong statement made by Curtis at the very beginning of his book? $\endgroup$ – Samarkand Feb 2 '16 at 18:58
  • $\begingroup$ @Samarkand The condition $d_i x = *$ is much stronger than all faces of $x$ being equal - they have to be degenerate $0$-simplices, too! It turns out that a simplex is a degeneracy of a $0$-simplex iff all of its degeneracies are equal (I updated my answer with a proof), but that is not the question you asked, i.e. this is not a consequence of the fact that all faces $d_i x$ are equal. In other words, the map $X^n\to S^n$ I have constructed above is not an isomorphism. So Curtis's statement should be true, but not be applicable under the hypothesis of the question. $\endgroup$ – Bertram Feb 3 '16 at 12:09
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    $\begingroup$ I hate situations like that. Whoa, he meant that $*$ is degenerate vertex. With no apparent convention. $\endgroup$ – Samarkand Feb 3 '16 at 12:13
  • $\begingroup$ It's definitely abuse of notation, but using $*$ for $\Delta^0$ (or more generally, any terminal object) is standard. This simplicial set has one (degenerate) $k$-simplex for all $k>0$, so it's reasonable to use the same notation for these degenerate simplices. But then, he also defines $*$ immediately before this as the $0$-simplex of $S^n$... $\endgroup$ – Bertram Feb 3 '16 at 12:28

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