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Suppose $S$ and $T$ are simplicial sets such that

  • $S_0 = T_0$, $S_1 = T_1$, and $S_2 = T_2$,
  • the maps $d_0, d_1 : S_1 \to S_0$ agree with the maps $d_0, d_1 : T_1 \to T_0$ respectively,
  • the map $s_0 : S_0 \to S_1$ agrees with $s_0 : T_0 \to T_1$,
  • the maps $d_0, d_1, d_2 : S_2 \to S_1$ agree with the maps $d_0, d_1, d_2 : T_2 \to T_1$ respectively, and, finally,
  • the maps $s_0, s_1: S_1 \to S_2$ agree with the maps $s_0, s_1: T_1 \to T_2$ respectively.

That is, $S$ and $T$ are the same simplicial set as far as the "levels" $0$, $1$, and $2$ are concerned. Furthermore, suppose that $S$ and $T$ are simplicial sets of dimension $\leq 2$.

Question 1: Does it follow that $S$ and $T$ are isomorphic?

The motivation of that question is the following: When the $2$-skeleton of $S$ is isomorphic to the $2$-skeleton of $T$, then $S$ and $T$ are isomorphic, clearly. So in this sense, no information is lost when truncating all above level $2$. But the $2$-skeleton of $S$ is the same as $S$ (same with $T$)! So the $2$-skeleton doesn't truncate all above level $2$, but only deletes all nondegenerate simplices. My question is exactly this: do we loose information when we really delete all simplices (and face and degeneracy maps) above level $2$, instead of just considering the $2$-skeleton?

Question 2: Suppose we are given the following data:

  • sets $S_0$, $S_1$, and $S_2$,
  • maps $d_0, d_1 : S_1 \to S_0$,
  • a map $s_0 : S_0 \to S_1$,
  • maps $d_0, d_1, d_2 : S_2 \to S_1$, and
  • maps $s_0, s_1: S_1 \to S_2$.

If the answer to question 1 is "yes", then we know that there is at most one simplicial set $S$ whose restriction to the levels $0$, $1$, $2$ is given by these data.

Under which assumption does there exist a simplicial set with these data? In particular, does it suffice that these data satisfy all the simplicial identities that can be formulated with these data?

(One can ask the same question with any $n$ instead of $2$.)

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  • $\begingroup$ What do you mean by dimension? I can think of a definition that makes the answers to your questions yes. $\endgroup$
    – Zhen Lin
    Commented Dec 2, 2021 at 22:27
  • $\begingroup$ A simplicial set has dimension $\leq 2$ if for every $n>2$ each $n$-simplex in it is degenerate. Can you give me a hint for the proofs? $\endgroup$
    – user997814
    Commented Dec 3, 2021 at 0:11
  • $\begingroup$ I guess the first question can be proved using the fact that to show that a simplicial map is an isomorphism it suffices that it's bijective on the nondegenerate simplices. Is that true, and what about the second question? $\endgroup$
    – user997814
    Commented Dec 3, 2021 at 14:07

1 Answer 1

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Let $\mathbf{\Delta}_{\le n}$ be the full subcategory of the usual simplex category $\mathbf{\Delta}$ with objects $[0], \ldots, [n]$. Restricting along the inclusion $\mathbf{\Delta}_{\le n} \hookrightarrow \mathbf{\Delta}$ and then taking the left Kan extension along the same inclusion gives the $n$-skeleton functor $\operatorname{sk}_n : \textbf{sSet} \to \textbf{sSet}$. (The Kerodon uses a different construction, but the result is the same.) This is precisely the sense in which the $n$-skeleton of a simplicial set depends only on the simplices of dimension $\le n$. So the answer to your first question is yes (and in fact no assumption about dimension is necessary if you ask the question in the form "does the $n$-skeleton depend only on the simplices in dimension $\le n$"), and the answer to your second question is that it suffices that the simplicial identities are satisfied.

It is true that a morphism of simplicial sets is determined by its action on non-degenerate simplices. This is because every simplex, degenerate or otherwise, can be obtained by applying some simplicial operator to some non-degenerate simplex. Since simplicial maps are compatible with simplicial operators, it follows that the action of a map on degenerate simplices can be recovered from the action on non-degenerate simplices (together with the simplicial operators in the codomain).

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  • $\begingroup$ I don't know about Kan extensions, but the construction of the $n$-skeleton given in the Kerodon depends not only on the simplices of dimension $\leq n$. (See "denote the subset of $S_n$ consisting of those $n$-simplices [of $S_\bullet$] which satisfy conditions (a) and (b)" in the above link. You have to know the full $S_\bullet$ to build its skeleton using this definition.) $\endgroup$
    – user997814
    Commented Dec 3, 2021 at 14:26
  • $\begingroup$ "So the answer to your first question is yes" - can this also be proved along the lines sketched in my comment: "I guess the first question can be proved using the fact that to show that a simplicial map is an isomorphism it suffices that it's bijective on the nondegenerate simplices." (In your answer you remark that that statement is true, but I want to know whether it can be used to prove Question 1.) $\endgroup$
    – user997814
    Commented Dec 3, 2021 at 14:27
  • $\begingroup$ "and in fact no assumption about dimension is necessary" -- that's wrong: if $S$ and $T$ aren't of dimension $\leq 2$, then they aren't determined by the $n$-simplices for $n\in\{0, 1, 2\}$. $\endgroup$
    – user997814
    Commented Dec 3, 2021 at 14:29
  • $\begingroup$ " the answer to your second question is that it suffices that the simplicial identities are satisfied" -- you mean only the simplicial identities that can be formulated using the data on the levels 0, 1, 2, right? But how do you prove that? $\endgroup$
    – user997814
    Commented Dec 3, 2021 at 14:31
  • $\begingroup$ "Since simplicial maps are compatible with simplicial operators, it follows that the action of a map on degenerate simplices can be recovered from the action on non-degenerate simplices" -- is that a new statement or part of your intuitive proof sketch of the fact mentioned in the previous sentence? $\endgroup$
    – user997814
    Commented Dec 3, 2021 at 14:33

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