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.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.